PULLOUT WORKSHEETS Material...2015/12/29 · Mathematics New Saraswati House (India) Pvt. Ltd. New...

Mathematics

New Saraswati House (India) Pvt. Ltd.New Delhi-110002 (INDIA)

Solutions to

PULLOUT WORKSHEETSFOR CLASS IXSecond Term

Kusum Wadhwa(PGT Mathematics)

Delhi Public School, Mathura Road, New Delhi

Anju Loomba(HOD Mathematics)

Apeejay School, Noida

By

Second Floor, MGM Tower, 19 Ansari Road, Daryaganj, New Delhi-110002 (India) Phone : +91-11-43556600Fax : +91-11-43556688E-mail : [email protected] : www.saraswatihouse.comCIN : U22110DL2013PTC262320Import-Export Licence No. 0513086293

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First published 2016

ISBN: ?

Published by: New Saraswati House (India) Pvt. Ltd.19 Ansari Road, Daryaganj, New Delhi-110002 (India)

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CONTENTCONTENTCONTENTCONTENTCONTENTSSSSS

1. Linear Equations in two Variables

Worksheets (1 to 6) ..................................................................................................... 6

• Assessment Sheets (1 and 2) ...................................................................................15

• Chapter Test ............................................................................................................. 17

2. Quadrilaterals

Worksheets (12 to 16) ...............................................................................................18


• Chapter Test ............................................................................................................. 27

3. Areas of Parallelograms and Triangles

Worksheets (21 to 25) ...............................................................................................29


• Chapter Test ............................................................................................................. 39

4. Circles

Worksheets (28 to 33) ...............................................................................................41


• Chapter Test ............................................................................................................. 54

5. Constructions

Worksheet 36 ........................................................................................................... 56

• Assessment Sheet 9 ..................................................................................................58

• Chapter Test ............................................................................................................. 60

6. Surface Areas and Volumes

Worksheets (40 to 47) ...............................................................................................63

– 3 –

• Assessment Sheets (10 and 11) ................................................................................75

• Chapter Test ............................................................................................................. 79

7. Probability

Worksheets (51 and 52) ............................................................................................81

• Assessment Sheets (12 and 13) ............................................................................... 83

• Chapter Test ............................................................................................................. 85

8. Statistics

Worksheets (56 to 63) ...............................................................................................87

• Assessment Sheets (14 and 15) ............................................................................... 98

• Chapter Test ...........................................................................................................100

PRACTICE PAPERS (1 to 5) ................................................................................... 102

– 4 –

��

��

��

��

��

��

� ��

��

LINEAR EQUATIONS IN TWO VARIABLES

��

�� (D) 8x = 6y can be expressed as8x – 6y + 0 = 0

Comparing with ax + by + c = 0, we get a = 8, b = – 6 and c = 0.

�� 5x + 7 = 0� 5x + 0 + 7 = 0�� 5x + 0.y + 7 = 0

�� Let us take a, b, c as three constants.Then required equation is ax + by + c = 0.

�� Let the cost of a notebook be � x and thecost of a pencil be � y.We are given that the cost of a notebookis four times that of a pencil. So, we havethe equation x = 4y. This can be expressedas x – 4y = 0 or x – 4.y + 0 = 0 which isthe required equation in two variables.

�� Let the cost of a pen be � x and the costof a book be � y.We are given that, the cost of a book isthree times that of the pen. So, we havethe equation y = 3x. This can be expressedas 3x – y = 0 or 3x – 1.y + 0 = 0, which isthe required equation in two variables.

�� Let the cost of a pen be � x and that of abook be � y.We are given that, the cost of a book istwo times that of a pen. Thus, y = 2x. Thisequation can be written as 2x – y = 0 or2 .x + (– 1) .y + 0 = 0 which is linearequation in two variables.Answers will vary. But sample solutionis x = 10; y = 20. No neither x nor y cantake negative values; as the cost can'tbe negative.

� (i) Putting a = 0, b = – 3 and c = 0 in thestandard equation ax + by + c = 0, we get

0.x + (– 3). y + 0 = 0� 0.x – 3y + 0 = 0.

(ii) Putting a = 12

, b = 1

3−

and c =16

in the standard linear equationax + by + c = 0, we get

12

.x + –13

� ��

.y +16

= 0

Multiplying both sides by 6, we get3x – 2y + 1 = 0.

� (i) As given the perimeter p of a squareis four times its side s.p = 4s or p – 4s + 0 = 0 which is a linearequation in two variables.(ii) Let the two numbers be x and y.As the given ratio of two numbers is 2 : 3.

� x : y = 2 : 3 or xy = 2

3 ��3x = 2y

� 3x – 2y + 0 = 0 which is a linearequation in two variables.

�� (i) Given equation of the line isx = y …(i)

This line passes through a point whose x-coordinate is 20.5

� x = 20.5Substituting this value of x in (i), we get

y = 20.5So the required coordinates are (20.5, 20.5)(ii) The given of the line is

x + y = 0 …(ii)This line passes through a point whose y-coordinate is – 19.5

� y = – 19.5Substituting this value of y in equation(ii), we get

x – 19.5 = 0� x = 19.5So the required coordinates are (19.5,– 19.5)(iii) The coordinates of all the points lyingon a line satisfy the equation of the line.(iv) Punctuality; respect for elders orparents.

��

��

�� (C)�� Every solution has the values of two

variables that follow an ordered pair.�� An order pair that satisfies an equation

in two variables is called solution ofthat equation.

�� p + 9 = 9� p = 9 – 9 = 0.

Thus, p = 0 is a root of the given equation.�� x + y = 9 can be written as y = 9 – �x

For x = 1, y = 9 – �.1 = 9 – �For x = 0, y = 9 – �.0 = 9 – 0 = 9For x = 2, y = 9 – �.2 = 9 – 2�For x = 3, y = 9 – �.3 = 9 – 3�

x 1 0 2 3

y 9 – � 9 9 – 2� 9 – 3�

Thus, (0, 9); (1, 9 – �); (2, 9 – 2�);(3, 9 – 3�) are four different solutionsfor �x + y = 9.Answers may vary.

�� x = 2 213

and no y, means y = 0

So, only solution is 2 2

, 013

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

.

� Putting x = 2 and y = 1 in the givenequation, we get

2.2 + 3.1 = kThus, k = 4 + 3 = 7.

� (i) Putting (0, 2), i.e., x = 0 and y = 2,we getx – 2y = 4 �� 0 – 2 × 2 = 4� – 4 = 4 which is false.Thus, (0, 2) is not a solution of thegiven equation.(ii) Putting (4, 0), i.e., x = 4 and y = 0,we get4 – 2 × 0 = 4 � 4 = 4 which is true.

Thus, (4, 0) is a solution of the givenequation.

(iii) Putting ( )2, 4 2 , i.e., x = 2 and

y = 4 2 , we get

2 – 2 × 4 2 = 4

�� 2 – 8 2 = 4 � – 7 2 = 4which is false

Thus, � �2,4 2 is not the solution of

the given equation.

(iv) Putting (2 2, 2 – 2), i.e., x = 2 2

and y = 2 – 2, we get

2 2 – 2 ( 2 – 2) = 4

� 2 2 – 2 2 + 4 = 4� 4 = 4 which is true.

��(i) The partition represents a straight line.To draw a straight line we must haveatleast two points on it.3x + 2y = 12

x 0 4 y 6 0

Thus the points are (0, 6) and (4, 0).Plotting these points on a graph paper andjoining them we get the required graph.(ii) Graph of a linear equation.(iii) Love for environment and Cooperat-ion i.e., Collective responsibility.

��

�� (D) Putting x = 1 in equation y = 2x – 1,we gety = 2 × 1 – 1 = 1 so, required solution is(1, 1).

�� The given point (0, – 6) lies on y-axisand a line parallel to x-axis at 6 unitsbelow of it, i.e., x = 0 or y = – 6.

�� Putting the points (2, 2), (– 2, – 1), etc.from the graph in given equations, wefind 8y – 6x = 4 is always true.

�� Yes. The given data satisfies the equationy = 2x – 1, as each point in the data is thesolution of the given equation.

� ��

�� The given equation is y = 3x. Let x = 0, then y = 3.0 = 0; (0, 0) Let x = 1, then y = 3.1 = 3; (1, 3) Let x = – 2, then y = 3.(– 2) = – 6; (– 2, –6) Thus, the table for values of x and y is

x 0 1 – 2

y 0 3 – 6

Plotting these points on the graph, we get

�� The given equation is2x – 3y + 11 = 0 can be written as

2x = 3y – 11 or x = 3 – 11

2y

Let y = 1, then

x = 3 1 11

2× −

= −3 112

= 8

2−

= – 4

So, ordered pair is (– 4, 1).Let y = 3, then

x = 3 3 11

2× −

= −9 112

= 2

2−

= – 1

So, ordered pair is (– 1, 3).Let y = 5, then

x = 3 5 11

2× −

= −15 112

= 42

= 2

So, ordered pair is (2, 5).Thus, the table for values of x and y is

x – 4 – 1 2

y 1 3 5

Plotting these ordered pairs on thegraph paper, we get

� Since the point (– 2, 4) lies on the graphof the equation y = mx + 5, then x = – 2,y = 4 is a solution of the given equation.Putting x = – 2 and y = 4, we get

4 = m (– 2) + 5or 4 = – 2 m + 5or 2 m = 5 – 4 = 1

or m = 12

.

� Consider the equations x + y = 16 and4x – y + 6 = 0.Both the above equations will besatisfied by x = 2, y = 14.Hence, these are the equations of twolines passing through (2, 14).We can write infinitely many such linesbecause infinitely many lines can bemade to pass through a given point.

�The graph for the values of x and ygiven in the table is as shown.

��

The graph is drawn by using theordered pairs (– 1, – 2) and (– 4, – 3).From the graph, we have when y = 1,x = 8 that means q = 8.Again when x = 5, y = 0 that meansp = 0.Thus, p = 0 and q = 8 are the requiredvalues.

�� Consider 2x + 5y = 12

� 5y = 12 – 2x � y = 12 – 25

x

Let x = 6, y= 12 – 2 × 65

= 12 – 125

= 05

= 0

Let x = – 4, y = 12 – 2× (–4)5

= 12 + 85

= 205

= 4

Let x = 11, y = 12 – 2 ×115

= 12 – 225

= –105

= – 2

Thus, we get a table as below:

x 6 – 4 11y 0 4 – 2

The graph of the equation is as shown.

(i) When x = 1, y = 2 (Using above graph)(ii) When y = 4, x = – 4

(Using above graph)

��

�� (D) On drawing the graph of givenequations, we see that they neverintersect each other at any point. So,they have no solution.

�� 3b + 1 = 2+ 2a – 3 = 5 (Subtracting)

– + – 3b – 2a + 4 = – 3

� 3b – 2a = – 7.�� At x = c, c being a real quantity,

y = mx gives y = mcHence, every distinct value of x gives acorresponding distinct value of y. Therefore,y = mx has infinitely many solution.

�� At y-axis, abscissa is always 0.So putting x = 0 in the given equationswe gety = 5 × 0 + 7 = 0 + 7 = 7Hence, solution is (0, 7).

�� Consider 3x + 2y = 12When x = 0, 3(0) + 2y = 12� 2y = 12� y = 6� x = 0, y = 6 is a solution.When y = 0, 3x + 2(0) = 12� 3x = 12� x = 4� x = 4, y = 0 is a solution.When x = 6, 3(6) + 2y = 12� 18 + 2y = 12� 2y = 12 – 18� 2y = – 6� y = – 3� x = 6, y = – 3 is a solution.These solutions can be put in the formof a table as shown below:

x 0 4 6

y 6 0 – 3

��

Points are: (0, 6), (4, 0), (6, – 3).The graph of the equation is:

��The point (2, 3) lies on the graph ofthe equation. Hence, x = 2, y = 3 is asolution of the equation.

�� The cost of 1 litre of petrol is � 50.Number of litres of petrol = x

Total cost of petrol = � y

� 50x = y

Table of values:

x 0 1 2

y 0 50 100

� Given equations are 3x + y + 1 = 0 and

2x – 3y + 8 = 0 which can be re-writtenrespectively as

y = – 1 – 3x ... (i)

and x = y3 82− ... (ii)

Table for values of x and y in equation (i)is as below:

x 0 1 – 2

y – 1 – 4 5

Table for values of x and y in equation (ii)is as below:

x – 4 – 1 2

y 0 2 4

From the graph, the point of intersec-tion is (– 1, 2).

� The two given equations are:2x – y = 3 ...(i)

and 3x + 2y = 1 ...(ii)

��

From equation (i), we have

y = 2x – 3

Table for values of equation (i) is givenbelow:

x 0 2 1

y – 3 1 – 1

From equation (ii), we have

2y = 1 – 3x �� y = 1 – 32

x

Table for values of equation (ii) is givenbelow:

x – 1 3 1

y 2 – 4 –1

With the help of these tables, we havedrawn two straight lines �1 and �2representing equations (i) and (ii)respectively as shown in figure.

From the graph, we find that the pointP(1, – 1) is the point of intersection ofthe two lines.

�� The two given equations are: 2x + 3y – 5 = 0 ...(i)

and 2x + 3y + 11 = 0 ...(ii)From equation (i), we have

x = y5 3

2−

Table for values of x and y in equation (i)is given below:

x 1 – 2 4

y 1 3 – 1

From equation (ii), we have

x = y11 3

2− −

Table of values for x and y in equation (ii)is given below:

x – 4 – 7 – 1

y – 1 1 – 3

Observing the graph, we infer that thelines are parallel.

��

�� (C) Putting x = 3 and y = 1 in eachequation, we see that only 3y = x issatisfied.

�� x = 6y – 12

�� The equation y = 23

x – 113

, is having

the coefficient of x as 23

and also satisfies

by the values x = 4, y = – 1.

�� x = 2y

+32

can be expressed as

2x = y + 3, i.e., y = 2x – 3.

So, a = 2.

��

�� Consider 3x + 2y = 12 When x = 0, 3(0) + 2y = 12� 2y = 12� y = 6� x = 0, y = 6 is a solution.When y = 0, 3x + 2(0) = 12� 3x = 12� x = 4� x = 4, y = 0 is a solution.When x = 6, 3(6) + 2y = 12� 18 + 2y = 12� 2y = 12 – 18� 2y = – 6� y = – 3� x = 6, y = – 3 is a solution.These solutions can be put in the formof a table as shown below:

x 0 4 6

y 6 0 – 3

Points are: (0, 6), (4, 0), (6, – 3).The graph of the equation is:

From the graph, the point (2, 3) lies onthe graph of the equation. Hence, x =2, y = 3 is a solution of the equation.

�� As given total distance = x kmand fare for first kilometre = � 8So, as per the condition, fare for next(x – 1) km = � (x – 1) × 5

Also given that total fare is � y.Hence, y = 8 + (x – 1) × 5� y = 8 + 5x – 5� 5x – y + 3 = 0. (Standard form)

� x + y = 100 is the linear equationsatisfying the given data.When x = 0, y = 100When x = 10, y = 90When x = 20, y = 80The table is

x 0 10 20

y 100 90 80

The points are (0, 100), (10, 90) and (20,80).Graphing and joining these points, weget the graph of x + y = 100 as shown.

� Let W = work done,F = constant force = 5 units,d = distance travelled

Then , W = FdHere, W = 5dFor d = 0, W = 0,For d = 1, W = 5,For d = 2, W = 10.

��

These values can be put into a tabularform as follows:

d 0 1 2

W 0 5 10

The points are (0, 0), (1, 5) and (2, 10).On taking d on x-axis and W on y-axis,the graph is:

(i) Let A represent x = 2 on the x-axis.From A, draw a line parallel to they-axis, meeting the graph at B. From B,draw a line perpendicular to y-axis,meeting the y-axis at C. The ordinateof C is 10.So, when the distance travelled is 2 units,the work done is 10 units.(ii) From the graph, when d = 0, W = 0,i.e., no work is done when distancetravelled is 0.

�� (i) F = 95

⎛ ⎞⎜ ⎟⎝ ⎠

C + 32

When C = 0, F = 0 + 32 = 32When C = 10, F = 18 + 32 = 50

When C = – 10, F = – 18 + 32 = 14The values can be tabulated as under:

x 0 10 – 10

y 32 50 14

Points are (0, 32), (10, 50) and (– 10, 14)Taking Celsius on the x-axis andFahrenheit on the y-axis, we draw graphusing the same scale on both the axes.The graph is

From the graph:(ii) When C = 30°, F = 86°Hence, 30°C = 86°F.(iii) When F = 95°, C = 35°Hence, 95°F = 35°C.

��

�� (A) Equation y = 3 represents a lineparallel to x-axis and 3 units away fromit in positive direction.

� ��

�� Consider y = 4 as an equation in onevariable so its geometrical representa-tion is a point at a distance of 4 units tothe right of the origin.

�� x = 4, y = 4, x = 0, y = 0

�� Equations of the sides OA and OB arey = 0 and x = 0 respectively.

�� We have 2x + 9 = 0.

This can be re-written as x = –92

The representation of the solution on the

number line is shown in figure. x = –92

is

treated as an equation in one variable.

�� y + 2 = 0 or y = – 2 is the line parallelto x-axis at a distance of 2 units belowthe x-axis.

� (i) origin (ii) the origin

(iii) (p, q)

� Equations x = 4 and x = – 4 representthe lines parallel to y-axis at thedistance of 4 units from both sides(right and left) of it. Similarly, y = 4and y = – 4 represent the parallels tothe x-axis at 4 units above and belowboth this axis.

Thus, a square ABCD is formed.

��

Equations of the sides of the rectangleABCD are

(i) AB is y = 2 (ii) BC is x = 6

(iii) CD is y = 4 (iv) AD is x = 2 .

Equations of the sides of the squarePQRS are

(i) PQ is y = – 6 (ii) QR is x = – 1

(iii) RS is y = – 3 (iv) SP is x = – 4.

��

��

�� (A) Anju had 8 pairs of shoes and Poonamborrowed x pairs. Then 6 pairs of shoeswere left with Anju. So, 8 – x = 6.

�� Consider 8x = 4 �� x = 48

= 12

and 2.4 x = 1.2 � x = .1 2.2 4

= 12

.

�� Yes�� By the definition of complementary

angles:

m�A + m�B = 90°� x + y = 90°� y = 90° – x.

�� (i) Given equation can be expressed as

2x + 3y – 9.35 = 0; on comparing withstandard equation, we get

a = 2, b = 3, c = – 9. 35.(ii) 2x + 5y + 0 = 0; a = 2, b = 5, c = 0.

�� Since the point (3, 4) lies on the graphof the equation 3y = ax + 7.� x = 3, y = 4 is a solution of the givenequation.� 3.4 = a.3 + 7

(Putting x = 3 and y = 4)� 12 = 3a + 7� 3a = 12 – 7 = 5

� a = 53

.

� All the equations can be written instandard form as an equation in twovariables, i.e., ax + by + c = 0(i) x = – 5 � 1.x + 0.y + 5 = 0

(ii) 3x + 5 = 0 � 3.x + 0.y + 5 = 0(iii) 5y = 2 � 0.x + 5.y – 2 = 0.

� The given equation 2x – 3y + 11 = 0 canbe written as:

2x = 3y – 11 � x = y3 11

2−

If y = 1, then

x = 3 1 11

2× −

= −3 112

= 8

2−

= – 4

So, the point is (– 4, 1).If y = 3, then

x = 3 3 11

2× −

= −9 112

= 2

2−

= – 1

So, the point is (– 1, 3).If y = 5, then

x = 3 5 11

2× −

= −15 112

= 42

= 2

So, the point is (2, 5).Thus, the table for values of x and y is

x – 4 – 1 2

y 1 3 5

Graph meets x-axis at (– 5.5, 0)

�� Distance covered = x km (x > 0)Total fare = � y ( y > 0)According to the question,y = (1 × 8) + (x – 1) × 5 (Fare for (Fare for first km) remaining kms)� y = 8 + 5x – 5

��

� y = 5x + 3� 5x – y + 3 = 0

(Standard form)When x = 1, y = 8When x = 2, y = 13When x = 3, y = 18These values can be put in the tabularform as follows:

x 1 2 3

y 8 13 18

Points are (1, 8), (2, 13) and (3, 18). Thegraph of the equation is shown in figure.

��

�� (B) Putting x = 1 and y = 1 in left sideof the equation 2x + 3y = 6, we get2 × 1 + 3 × 1 = 2 + 3 = 5 � 6.So, (1, 1) is not a solution of this equation.

�� x = 0�� No�� The equation y = 7x + 6 has infinitely

many solutions. As the graph of the givenequation is a straight line. A line has

infinite number of points lying on it andevery point on the line is a solution ofits equation.

�� (i) Putting x = 0 and y = 2 in x – 2y = 4,we get 0 – 2 × 2 = – 4 ��4Thus, (0, 2) does not satisfy the equationx – 2y = 4, so it is not a solution of theequation.

(ii) Putting x = 2 and y = 0 in the leftside of the equation, we get

x – 2y = 2 – 2 × 0 = 2 – 0= 2 � 4

Hence, (2, 0) is not a solution of theequation.

�� Putting x = 2 and y = – 3 in the givenequation, we get

– 3 = 2m + 5� 2m = – 8

m = – 4.� (i) Equation 2x – 3y = 5 can be re-written

as 2x – 3y – 5 = 0.Hence, a = 2, b = – 3, c = – 5.(ii) 7x = 3y can be expressed as7x – 3y + 0 = 0.Hence, a = 7, b = – 3, c = 0.(iii) Equation 9x = 4 – 3y can be expressedas 9x + 3y – 4 = 0.Hence, a = 9, b = 3, c = – 4.

� (i) We know that a linear equation in twovariables is of the form ax + by + c = 0.Putting a = 2, b = 3 and c = 4, we get

2x + 3y + 4 = 0.(ii) We know that a linear equation in twovariables is of the form ax + by + c = 0.Putting a = 5, b = – 1 and c = 0, we get

5x + (– 1)y + 0 = 0� 5x – y + 0 = 0.

Table of values is:

x 0 1 1–

2

y 0 5 5–

2

��

�� (i) Equations x = 6 and x = – 6 representthe lines parallel to y-axis at thedistance of 4 units from both sides(right and left) of it. Similarly y = 6and y = – 6 represent two lines parallelto the x-axis at 6 units above and belowx-axis. Thus a square ABCD is formed.

(ii) Square(iii) Rationality and Justice(iv) Ravi and Dinesh showed respect for

elders.

��

�� (B)�� Putting x = 3 and y = 4 in equation

3y = ax + 7, we get3. 4 = a . 3 + 7

� 12 = 3a + 7

� 3a = 12 – 7 = 5� � � � � � � � �� a = 53

.

�� Line IV.�� y = 5x + 3�� (i) 4n + 9 = 12 is a linear equation in

one variable.(ii) 9x – y = 3 – x is a linear equation intwo variables.

� �� Equations of AB is y = 2, CD is x = – 5

and GH is x = 4.

�� (i) Let the distance covered be x km andthe total fare be � y.Fare for first kilometre = � (1 × 40)Fare for (x – 1) kilometres

= � (x – 1) × 20So, fare for x kilometres will be givenby

y = 1 × 40 + (x – 1) × 20or y = 40 + 20x – 20or y = 20x + 20

(ii) Formation of linear equation.(iii) Self-reliance, i.e., value manual

labour, value self respect.

��

18 ��

��

QUADRILATERALS

��

�� (B) As all are closed figures formed bythree or more than three line segments.

�� False, as a sphere is a three-dimensionalstructure but a circle is two-dimensional.

�� Let reflex ∠ADC = y 30° + 48° + 40° + y = 360°

(Sum of interior angles)

⇒ y = 360° – 118° = 242°But x + y = 360° ⇒ x = 360° – 242° = 118°.

�� As CD || AB and BC is transversal,∠C + ∠B = 180°

(Sum of interiorangles on the sameside of transversal)

⇒ 95° + y = 180°⇒ y = 85°Similarly, 2x + 5° + x + 25° = 180°

⇒ x = 180º –5º – 25º50º

3� .

�� Since ABCD is a parallelogram, BC || AD

⇒ 5x = 25°(Alternate angles)

⇒ x = 5°Since AB || DC⇒ 60° = 6y

(Alternate angles)⇒ y = 10°

Hence, x = 5° and y = 10°.

�� Let x be the constant of proportionality.So, the angles of the quadrilateral canbe taken as 3x, 5x, 9x and 13xThen, 3x + 5x + 9x + 13x = 360°

(ASP of a quadrilateral)⇒ 30x = 360°⇒ x = 12°Thus, the angles of the quadrilateral are3x = 3 × 12° = 36°, 5x = 5 × 12° = 60°,9x = 9 × 12° = 108°, 13x = 13 × 12° = 156°.

� Let us take a quadrilateral ABCD.Join AC.In ΔABC, ∠1 + ∠2 + ∠B = 180° ... (i)

(ASP of a triangle)

In ΔACD,∠3 + ∠4 + ∠D = 180° ... (ii)

(ASP of a triangle)Adding equations (i) and (ii), we get∠1 + ∠2 + ∠3 + ∠4 + ∠B + ∠D

= 180° + 180°⇒ (∠1 + ∠4) + (∠2 + ∠3) + ∠B + ∠D

= 360°⇒ ∠A + ∠C + ∠B + ∠D = 360°.

� In parallelogram ABCD,AC = BD

(Given: Diagonals are equal)

19��

In ΔABC and ΔBAD,AB = BA (Common side)BC = AD

(Opposite sides of aparallelogram are equal)

AC = BD (Given)∴ ΔABC ≅ ΔBAD.

(Using SSS congruence rule)⇒ ∠ABC = ∠BAD (CPCT)We also have

∠A = ∠C and ∠B = ∠D(Opposite angles of a

parallelogram are equal)⇒ ∠A = ∠B = ∠C = ∠D = x (say)Also, ∠A + ∠B + ∠C + ∠D = 360°⇒ x + x + x + x = 360°⇒ 4x = 360°⇒ x = 90°⇒ ∠A = ∠B = ∠C = ∠D = 90°Hence, ABCD is a rectangle.

�� (i) ∠R = 80° (Given)SR �� PQ and RQ is atransversal∴ ∠R + ∠Q = 180°(Co-interior angles)⇒ ∠Q = 180° – 80° = 100°Similarly, ∠Q + ∠P = 180°⇒ ∠P = 180° – 100° = 80°and∠S + ∠R = 180°

⇒ ∠S = 180° – 80° = 100°Hence, ∠P = 80°, ∠Q = 100°,

∠R = 80°, ∠S = 100°(ii) Property of co-interior angles when apair of straight lines intersected byanother straight line (Geometry)

(iii) Diligence. i.e., dedication, deter-mination and Hard work.

��

�� (C)�� In quadrilateral ABCD, AB || CD

So, ABCD is a trapezium.

�� Let us represent thegiven figure by ABCD.Here, AB || CD andAD || BC.So, the given figurehas two sets of parallel sides.

�� In the given figure,∠A + ∠D = 90° + 90° = 180°

⇒ AB || DC (Cointerior anglesare supplementary, lines are parallel)

∴ ∠B + ∠C = 180°(Using ASP of quadrilateral)

But it is given that ∠B is an obtuseangle, therefore, ∠C must be an acuteangle.

�� Since opposite angles of a parallelogramare equal∴ 3x – 4 = 60 – x ⇒ 3x + x = 60 + 4

⇒ 4x = 64 ⇒ x = 644

⇒ x = 16.

Two adjacent angles are (3 × 16 – 4)° = 44°and 180° – 44° = 136°Hence, angles are 44°, 136°, 44°, 136°.

�� Let the smallest angle of the parallelogrambe x, then its adjacent angle = 2x – 24°⇒ x + 2x – 24° = 180°

(Adjacent angles of a parallelogramare supplementary)

⇒ 3x = 180° + 24°

⇒ x = 2043

� ⇒ x = 68°

So, 180° – 68° = 112°Hence, angles are 68°, 112°, 68°, 112°.

� ∠1 = ∠4 (Alternate angles) ... (i)In ΔACD,∠2 = ∠4 ... (ii)

(Angles opposite to equal sides of atriangle are equal)

20 ��

∠1 = ∠2 [From (i) and (ii)]Similarly, ∠3 = ∠4Therefore, AC bisects ∠A as well as ∠C.In the same way, we can also prove thatBD bisects ∠B as well as ∠D.

� We are given that diagonals AC and BDbisect each other at right angles at O.

In ΔAOB and ΔAOD, we haveAO = AO (Common)

∠AOB = ∠AOD = 90° (Given)OB = OD (Given)

Now, using SAS congruence criterion,we have

ΔAOB ≅ ΔAODNow, AB = AD (CPCT) ... (i)Similarly, we can prove that

AB = BC and AD = CD ... (ii)From equations (i) and (ii), we have AB = BC = CD = ADHenc, ABCD is a rhombus.

�� (i) Let the photo-frame be ABC suchthat BC = a, CA = b and AB = c and themid-points of AB, BC and CA arerespectively D, E and F.

We have to determine the perimeter ofΔDEF.

In ΔABC, DF is the line-segment joiningthe mid-points of sides AB and AC.So, DF is parallel to BC and half of it.

i.e., DF = BC2

= 2a

Similarly,

DE = AC2

= 2b

and EF = AB2

= 2c

∴ DF + DE + EF = 2a

+ 2b

+ 2c

= � �

2a b c

Hence, required perimeter is12

(a + b + c).s

(ii) Mid-point theorem (Geometry).(iii) Unity and cooperation.

��

�� (D) As SP || RQ and SQ is transversal,

∠QSP = ∠SQR = 60°

(Alternate angles)

In Δ PQS,

∠QSP + ∠SPQ + ∠SQP = 180° (ASP)

⇒ 60° + 75° + ∠SQP = 180°

⇒ ∠SQP = 45°.

�� In quadrilateral PQRS, PQ = SR andPQ || SR,∴ PS = QR and PS || QRTherefore, x = ∠RPQ

(Alternate angles) = 60°

Similarly, y = 30°.

�� Let the given parallelogram be ABCD.∠BAD + ∠ABC = 180°

(Sum of adjacent angles of a parallelogram)

∠1 + ∠2 + ∠3 + ∠4 = 180°

21��

But, ∠1 = ∠2 and ∠3 = ∠4∴ ∠1 + ∠1 + ∠3 + ∠3 = 180°⇒ ∠1 + ∠3 = 90°In Δ AOB,

∠1 + ∠3 + ∠AOB = 180° (ASP)⇒ 90° + ∠AOB = 180°⇒ ∠AOB = 90°.

�� In ΔPRS, 3a + 5a + 2a = 180° (ASP)⇒ a = 18°

SR || PQ (Opposite sidesof a parallelogram)

⇒ ∠1 = 2a(Alternate angles)

∴ ∠P = 3a + 2a = 5a = 5 × 18° = 90°.Similarly, ∠R = 2a + 3a = 90°,And ∠S = 5a = 90°,∠Q = 360° – (∠P + ∠R + ∠S)

= 360° – 270° = 90°.or

∠S = ∠Q = 90. (Opposite angles of a ||gm)

�� Let the given angles be 5x and 4x.Since, adjacent angles of a parallelogramare supplementary∴ 5x + 4x = 180°⇒ 9x = 180°⇒ x = 20°Therefore, angles are 100°, 80°, 100°,80°.

�� ABCD is a rectangle.⇒ ∠ABC = 90°⇒ ∠ABD + ∠DBC = ∠ABC⇒ 48° + ∠DBC = 90°⇒ ∠DBC = 90° – 48°

= 42°.

� Since ABCD is a rhombus. Therefore,∠AOB = ∠BOC = 90°.

In ΔABD, ∠BAD = 60° (Given)

∠ABD = ∠ADB(Angles opposite to equal

sides of rhombus are equal)Also, ∠1 + ∠2 + 60° = 180° (ASP)⇒ ∠1 + ∠1 = 180° – 60° = 120°⇒ 2∠1 = 120° ⇒ ∠1 = 60°In ΔAOB,

∠1 + ∠3 + ∠4 = 180°⇒ 60° + 90° + ∠4 = 180°⇒ ∠4 = 180° – 150° = 30°So, in ΔABO, angles are 90°, 30°, 60°.

� In ΔAPB and ΔCQD,∠1 = ∠2 = 90° (Given)∠3 = ∠4 (Alternate angles)

and AB = CD(Opposite sides of a parallelogram)

∴ ΔAPB ≅ ΔCQD(By AAS congruence criterion)

⇒ AP = CQ. (CPCT)

�� (i) In ΔAPD and ΔCQB,DP = BQ (Given)

∠ADP = ∠CBQ(Pair of alternate angles)

AD = BC(Opposite sides of a parallelogram)

∴ ΔAPD ≅ ΔCQB(SAS congruence criterion)

(ii) ⇒ AP = CQ (CPCT) (iii) Similarly, we can prove that

ΔAQB ≅ ΔCPD. (iv) ⇒ AQ = CP (CPCT)

22 ��

(v) Now, we have AP = CQand AQ = CP⇒ APCQ is a parallelogram.

��

�� (D) In ΔPQR, 30° + 60° + ∠PQR = 180° (ASP)

⇒ ∠PQR = 90°

As PQ || SR and PQ = SR

∴ PQRS is a ||gm

⇒ ∠PSR = ∠PQR(Opposite angles of a ||gm)

= 90°.

�� Sum of adjacent angles of a paralle-logram is 180°.∠A + ∠B = 180°

⇒ ∠B = 180° – ∠A= 180° – 100°= 80°.

�� No�� AB = 2AD ⇒ 2AP = 2AD

(� AP = BP)⇒ AP = AD ⇒ ∠2 = ∠3 ...(i)

Similarly, ∠4 = ∠5 ...(ii)

Now, (∠1 + ∠2 + ∠3) + (∠4 + ∠5 + ∠6)

= 180° + 180° (ASP)

⇒ ∠1 + ∠6 + 2∠3 + 2∠5 = 360°

[Using (i) and (ii)]

⇒ 2(∠3 + ∠5) = 360° – 180° = 180°

(� ∠1 + ∠6 = 180°)

⇒ ∠3 + ∠5 = 90° ... (iii)

Further, ∠3 + ∠5 + ∠7 = 180°(Angles at a point on the same

side of the line)

⇒ ∠7 = 180° – 90°[Using (iii)]

⇒ ∠CPD = 90°.

�� Since the diagonals of a parallelogrambisect each other. Therefore, O is themid-point of AC and BD.

∴ OC = 12

AC = 12

× 12.8 cm = 6.4 cm.

and OD = 12

BD = 12

× 7.6 cm = 3.8 cm.

�� Let line l || AB and passes through Eand meets BD in G.In ΔABD, E is the mid-point of AD andEG || AB.

∴ G is the mid-point of BD.(Converse of Mid-point theorem)

l || AB and AB || CD.∴ l || CD.In ΔBCD, G is the mid-point of BD andGF || DCF is the mid-point of BC.

� ABCD is a rectangle⇒ AC = BD (Diagonals of a rectangle

are equal and bisect each other)

⇒ 12

AC = 12

BD ⇒ OA = OD

⇒ ∠ODA = ∠OAD = x (say)In ΔOAD,

x + x + ∠AOD = 180°(ASP)

⇒ 2x + 40° = 180°(� ∠AOD = ∠BOC,

vertically opposite angles)

⇒ x = 180 – 40

2� �

⇒ x = 70°.i.e., ∠OAD = 70°.

� (i) In ΔDAC, S and R are the mid-pointsof the sides DA and DC respectively.

23��

⇒ SR || AC and SR = 12

AC

(Mid-point theorem)(ii) In ΔBAC, we have

PQ ||�AC and PQ = 12

AC.

We have PQ = SR � ��

1Each is AC

2

(iii) We have proved in (i) and (ii) thatPQ || AC, SR || AC

⇒ PQ || SR ...(i)(Each parallel to AC)

Also, PQ = 12

AC

SR = 12

AC

⇒ PQ = SR ...(ii)From (i) and (ii),⇒ PQRS is a parallelogram.

�� Join AC. In ΔDAC, S and R are the mid-points of the sides DA and DC respectively

⇒ SR || AC and SR = 12

AC ...(i)

(Mid-point theorem)Similarly, in ΔBAC, we have

PQ || AC and PQ = 12

AC ...(ii)

From equations (i) and (ii), we havePQ || SR and PQ = SR ...(iii)Similarly, we can prove thatPS || QR and PS = QR ...(iv)From relations (iii) and (iv), we obtainthat PQRS is a parallelogram.In ΔAPS, AP = AS⇒ ∠3 = ∠1

Similarly, ∠2 = ∠4 (�� BP = BQ)Also, ∠1 + ∠3 = 180° – ∠Ai.e., 2∠1 = 180° – ∠A ...(v)Similarly, 2∠2 = 180° – ∠B ...(vi)Adding (v) and (vi), we get

2(∠1 + ∠2) = 360° – (∠A + ∠B)⇒ 2(180° – ∠P) = 360° – 180°

(�� ∠A + ∠B = 180°)

⇒ ∠180° – ∠P = �180

2 = 90°

⇒ ∠P = 90°Therefore, PQRS is a rectangle.

��

�� (A) In a parallelogram, sum of a pair ofadjacent angles is 180°.∴ ∠A + ∠D = 180°⇒ ∠A = 180° – 120°

= 60°.

�� In ||gm ABCD, AC = BD (Given)⇒ ΔADB ≅ ΔBCA (By SSS)

[� AB = AB (Common)AD = BCBD = AC (Opposite sides of

parallelogram are equal)]⇒ ∠DAB = ∠CBA ...(i) (CPCT)

But, ∠DAB + ∠CBA = 180° ...(ii)⇒ ∠DAB = ∠CBA = 90° (Each 90°)From (i) and (ii)

24 ��

CM = AM ...(i) (CPCT)

Also, AM =12

AB ...(ii)

(Given: M is the mid-point of AB)From (i) and (ii),

⇒ CM = MA = 12

AB.

�� ∠DAB = ∠DCB(Opposite angles of a parallelogram)

⇒DAB2

�=�

DCB2

�

⇒ ∠FAE = ∠FCE(� AF and CE bisect ∠A

and ∠C respectively= ∠CEB

(� Alternate interior angle)⇒ Corresponding angles are equal

AF || CEAlso, AE || CF (� AB || CD)

⇒ AECF is a parallelogram⇒ AF || ECNow, in ΔDQC, F is the mid-point of DC.and FP || CQ (... AF || EC)⇒ P is mid-point of DQ⇒ DP = PQ ...(iii)Similarly, BQ = PQ ...(iv)From (iii) and (iv)

DP = PQ = BQHence, AF and EC trisect the diagonal BD.

��

�� (D) All four sides are equal so it is arhombus.Since parallelogram with one angle 90°is rectangle.Rhombus and rectangle both togetherform a square.

⇒ Parallelogram with one angle 90° isrectangle.

�� Let the required angles be 3x, 5x, 9x,13x.3x + 5x + 9x + 13x = 360°

(ASP of a quadrilateral)⇒ 30x = 360°⇒ x = 12°∴ 3x = 36°, 5x = 60°, 9x = 108°,

13x = 156°.

�� x + y + ∠BCD + ∠BAD = 360°(ASP for quadrilateral)

⇒ x + y + 180° – a + 180° – b = 360°⇒ x + y = a + b.

�� (i) True (ii) False�� (i) False (ii) False� (i) congruent (ii) rhombus

(iii) rhombus (iv) square(v) isosceles (vi) bisects

� (i) Through M, mid-point of AB, we drawline l || BC, l�intersects AC at D.⇒ D is the mid-point of AC.

(ii) As MD || BC and AC is transversal,∠ADM = ∠ACB = 90°

(Corresponding angles)⇒ MD ⊥ AC.

(iii) In ΔCMD and ΔAMD,CD = AD,

∠CDM = ∠ADM (Each = 90°)and MD = MDTherefore, ΔCMD ≅ ΔAMD (ASA)

25��

�� At least one pair of opposite sides of atrapezium is parallel.

�� No�� PQRS is a quadrilateral

OP = OR and OQ = OS

�� PQRS is a parallelogram�� In ΔPOQ and ΔROS,

OP = OR (Given)∠1 = ∠2

(Vertically opposite angles)OQ = OS (Given)

Using SAS congruence rule,ΔPOQ ≅ ΔROS

⇒ PQ = RS ...(i) (CPCT)Similarly, ΔQOR ≅ ΔSOP⇒ PS = QR ...(ii) (CPCT)From (i) and (ii),Quadrilateral PQRS is a parallelogram.

(� Quadrilateral with both pairs ofopposite sides is a parallelogram)

�� In ΔAPB and ΔCQD,∠1 = ∠2 = 90° (Given)∠3 = ∠4 (Alternate angles)

and AB = CD(Opposite sides of a parallelogram)

⇒ ΔAPB ≅ ΔCQD(By AAS congruence criterion)

⇒ AP = CQ. (CPCT)

�� (i) False (ii) True (iii) False (iv) True

� (i) ABCD is a parallelogram. DiagonalAC bisects ∠A.

∠BAC = ∠DAC (Given) ...(i)

Here, ∠BCA = ∠DAC ...(ii)(Pair of alternate angles)

From (i) and (ii), ∠BAC = ∠BCA.Also, ∠BAC = ∠DCA (Alternate angles)⇒ ∠BCA = ∠DCA⇒ AC bisects ∠C(ii) BC = AB.

(Sides opposite to equal anglesof a triangle are equal)

Further, AB = CDBC = AD

(Opposite sides of a parallelogram)

⇒ AB = BC = CD = AD⇒ ABCD is a rhombus.

� �� Let ABCD be a square.

Join AC and BD.∠A = ∠B = ∠C = ∠D = 90°

(Angles of a square)

⇒ ΔDAB is a right triangle.BD2 = AD2 + AB2

(Using Pythagoras Theorem)= AB2 + AB2 (... AD = AB)= 2AB2 ...(i)

(All sides of a square are equal)Similarly, AC2 = 2AB2 ...(ii)Thus BD2 = AC2

⇒ BD = AC ...(iii)[From (i) and (ii)]

As ABCD is a parallelogram, thediagonals AC and BD bisect each other,

OA = OC = 12

AC ...(iv)

OB = OD = 12

BD ...(v)

26 ��

From (iii), (iv) and (v), we getOA = OB = OC = OD ...(vi)

In ΔAOB and ΔAOD,AB = ADOA = OA (Common)OB = OD (Proved above)

⇒ ΔAOB ≅ ΔAOD(Using SSS congruence rule)

∠AOB = ∠AOD ...(vii) (CPCT)Also, ∠AOB + ∠AOD = 180° ...(viii)

(Linear pair)⇒ ∠AOB + ∠AOB = ∠AOD + ∠AOD

= 180°From (vii) and (viii), we get⇒ 2∠AOB = 2∠AOD = 180°⇒ ∠AOB = ∠AOD = 90° ...(ix)From (iii), (vi) and (ix), we obtain ACand BD are equal and bisect each otherat right angles.

��

�� (C) Let ∠Q = 2x, ∠R = 3x and ∠S = 7x;then

60° + 2x + 3x + 7x = 360°⇒ 12x = 300° ⇒ x = 25°∴ ∠S = 7x = 175°

�� In the given figure, RS || UT. And noother pair is parallel.Hence, only one pair isparallel.

�� The adjoining figure is noneof quadrilateral, parallelo-gram and rhombus.

�� Since opposite angles of a parallelogramare equal∴ 5x – 20 = 70 – 4x⇒ 9x = 90 ⇒ x = 10°So, the given angles are

(5 × 10 – 20)° = 30°,70°– 4 × 10° = 30°,

and 180° – 30° = 150°

Hence the required angles are 30°, 150°,30°, 150°.

�� (i) True (ii) True (iii) True(iv) False

�� Through A, draw a line parallel to CB

intersecting CD produced at E.

AB || DC (Given)

CB || EA (by const.) ∴ ABCE is a parallelogram.

Now, AE = BC(Opposite sides of a parallelogram)

AD = BC (Given)∴ AE = AD⇒ ∠2 = ∠3 ...(i)

(Angles opposite to equal sidesof a triangle are equal)

∠3 + ∠4 = 180° ... (ii)(Linear pair)

∠2 + ∠1 = 180° ...(iii)(Adjacent angles of a ||gm)

From (ii) and (iii),∠2 + ∠1 = ∠3 + ∠4 ...(iv)

[� ∠2 = ∠3]Hence, from (iv), ∠1 = ∠4.

� (i) Let the mid-points of AB, BC, CD andDA be P, Q, R and S respectively.Let AB = a and BC = b

∴ BP = AP = CR = DR = 2a

and BQ = CQ = AS = DS = 2b

27��

In right ΔBPQ,

PQ2 = BP2 + BQ2 = 2 2 2 2

4 4 4a b a b��

⇒ PQ = 12

2 2a b�Similarly, we can find

RQ = SR = PS = 12

2 2a b�

Therefore, PQ = QR = RS = SP⇒ PQRS is a rhombus(ii) Pythagoras theorem (Geometry).(iii) Self-reliance, Industrious.

�� (i) In ΔABC and ΔADC,

∠1 = ∠2 (AC bisects ∠A)AC = AC (Common side) ∠3 = ∠4 (AC bisects ∠C)

∴ ΔABC ≅ ΔADC (Using ASA)⇒ AB = AD (CPCT)

Also, AB = CD and AD = BC(�� Opposite sides of a rectangle)

⇒ AB = BC = CD = ADHence, ABCD is a square.(ii) Join BD. In ΔABD, AB = AD⇒ ∠5 = ∠7 ...(i)Also, ∠5 = ∠8 ...(ii)

(Alternate angles)and ∠7 = ∠6 (Alternate angles)⇒ ∠5 = ∠6 and ∠7 = ∠8Hence, BD bisects ∠B and ∠D.

��

�� (D)�� False

�� If both pairs of opposites sides of aquadrilateral are equal, then it must bea parallelogram. The quadrilateral maybe rectangle, rhombus or square.

�� A quadrilateral ABCD with∠A + ∠C = 180°�� ∠B + ∠D = 180°�� In quadrilateral ABCD,

∠A + ∠C + ∠B + ∠D = 360°(ASP of a quadrilateral)

⇒ 180° + ∠B + ∠D = 360°⇒ ∠B + ∠D = 180°.

�� Let given angles be 5x and 4x.Since adjacent angles of a parallelogramare supplementary∴ 5x + 4x = 180°⇒ 9x = 180° ⇒ x = 20°Therefore, angles are 100°, 80°, 100°,80°.

�� Since ABCD is a rectangle.

⇒ ∠ABC = 90°(Each angle of arectangle is 90°)

⇒ ∠ABD + ∠DBC = ∠ABC

⇒ ∠48° + ∠DBC = 90°⇒ ∠DBC = 90° – 48° = 42°.

� ��ABCD is a parallelogram.

∴ AD || BC

Now, AD || BC and AP is transversalAP intersects them at A and P.∠DAP = ∠APB ...(i) (Alternate angles)∠DAP = ∠PAB ...(ii) (AP bisects ∠A)From (i) and (ii), we get

∠PAB = ∠APB⇒ BP = AB

(Sides opposite to equal angles)

⇒12

BC = AB

⇒ BC = 2AB⇒ AD = 2AB. (� BC = AD)

� (i) Let x be the constant of proportionalitySo, the number of toffees on the eldestson to the youngest one as 6x, 5x, 4x and3x respectively.

28 ��

According to the given condition, we musthave 6x + 5x + 4x+ 3x = 360°Because sum of all interior angles of atrapezium (or quadrilateral) is 360°.

18x = 360° ⇒ x = 36018

��= 20°

∴ 6x = 6 × 20° = 120°5x = 5 × 20° = 100°4x = 4 × 20° = 80°3x = 3 × 20° = 60°

Therefore, the number of toffees on thesons were 120°, 100°, 80° and 60° in thegiven order.

(ii) Required ratio = 60

120��

= 12

i.e., 1 : 2.(iii) Angle sum property of a quadrilateral.(iv) Love for sons.

�� In rectangle ABCD,⇒ AB = CD

⇒12

AB = 12

CD

⇒ AP = DR ...(i)(P and R are the mid-points of AB

and CD respectively)

As S is the mid-point of AD,AS = SD ...(ii)

In ΔAPS and ΔDRS,AP = DR [From (i)]

∠PAS = ∠RDS (Each 90°)AS = SD [From (ii)]

⇒ ΔAPS ≅ ΔDRS(SAS criterion of congruence)

⇒ PS = RS ...(iii) (CPCT)Similarly, we can prove that

PQ = RQ, SP = QP and QR = SR...(iv)

Using equations (iii) and (iv), weconclude that

PQ= QR = RS = SP⇒ PQRS is a rhombus.

��

29��

��

AREAS OF PARALLELOGRAMS AND TRIANGLES

��

�� (C) DC = AB = 15 cm

AD × CF = DC × AE� AD × 10 = 15 × 8

� AD = 12010

= 12 cm.

�� Area of the parallelogram remains samefrom both the following manners.

AD × BN = AB × DM� AD × 11 = 12 × 9

� AD = 12 911�

= 10811

cm.

�� In option (a), �PDC and trapeziumABCD have a common base CD andtwo parallels AB and DC.In option (b), trapeziums APCD andABQD have a common base AD andtwo parallels AD and BQ.

�� Parallelogram PQRSand triangle BRS lieon the same base SRand between sameparallels PQ and SR.Also parallelogramPBCS and �EBC lie on the same base BCand between same parallels PS and BC.

�� AD × CF = CD × AE� AD × 10 = 16 × 8 (� AB = CD = 16 cm)� AD = 12.8 cm.

�� Let h be the length of altitude corre-sponding to AD.Area of a parallelogram= Base × corresponding

altitude

� AB × 8 = AD × h� 12 × 8 = 10 × h

� h = 12 8

10�

= 9.6 cm.

� Area of ��gm ABCD = AB × DM ... (i)Also, area of ��gm ABCD = BC × DN ... (ii)From (i) and (ii), we have

AB × DM = BC × DN� 8 × 6.6 = BC × 4.8

� BC =8 6.6

4.8�

= 11 cm.Now, perimeter = 2 (AB + BC)

= 2 (8 + 11)= 38 cm.

� Draw LM �� AB, passing through P.

(i) ar(�APB) + ar(�PCD)

= 12

ar(ALMB) +12

ar(LMCD)

(� �APB and parallelogram ABMLhave same base AB and are betweensame parallels AB �� LM. Also �PCD andparallelogram LMCD have same baseDC and are between same parallelsDC �� LM).

= 12

ar(ABCD) ... (i)

(ii) Similarly, ar(�APD) + ar(�PBC)

=12

ar(ABCD) ...(ii)

From (i) and (ii),ar(�APB) + ar(�PCD)

= ar(�APD) + ar(�PBC).

30 ��

�� From figure, transversal DB is intersect-ing a pair of lines DC and AB such that

�CDB = �ABD = 90°� DC �� AB

(A pair of alternateangles is equal, then

corresponding linesare parallel)

Also, DC = AB = 2.5 units.Therefore, quadrilateral ABCD is aparallelogram.

(If a pair of opposite sides of aquadrilateral is parallel and equal, then

the quadrilateral is a parallelogram)Now, area of parallelogram ABCD

= base × corresponding altitude= 2.5 × 4 = 10 sq. units.

��

�� (A) Draw AM � BC to intersect BC at M.Let BD = mx and DC = nx

Now, area of �ABD : area of �ACD

�12

× BD × AM : 12

× DC × AM

�12

× mx × AM : 12

× nx × AM

� m : n.�� We know that a median of a triangle

divides it into two triangles of equal areas.� ar(�PQS) = ar(�PRS)

��

�

arar

( PRS)( PQS)

= 1

Adding 1 to both sides,we get

�

�

arar

( PRS)( PQS)

+ 1 = 1 + 1

��

�

ar arar

( PRS) ( PQS)( PQS)

= 2

��

�

arar

( QRP)( PQS)

= 2

� ar(�PQS) : ar(�QRP) = 1 : 2.

�� Let height of the triangle be h inches.

Area of the trapezium= Area of the triangle

�12

× (AB + CD) × h = 12

× 18 × h

� AB + CD = 18 inches.

�� As parallelogramsABRS and PQRS areon the same base SRand between thesame parallels AQand SR,

ar(ABRS) = ar(PQRS) ... (i)As parallelogram ABRS and triangleAXS are on the same base AS andbetween the same parallels BR and AS,

ar(ABRS) = 2 ar(AXS) ... (ii)Using equations (i) and (ii), we get

ar(PQRS) : ar(AXS) : ar(ABRS)= 2 : 1 : 2.

�� ParallelogramsABCD and BPQRare on equalbases (AB = BP)and between thesame parallelsAP and CR.Therefore, ar(��gm ABCD) = ar(��gm BPQR).

�� The field is divided into three parts.The parts are �APS, �APQ and �AQR.

We have ar(�APQ) = 12

ar(PQRS)

31��

Therefore, ar(�APS) + ar(�AQR)

= ar(�APQ) = 12

ar(PQRS)

She can sow wheat in �APQ and pulsesin �APS and �AQR or vice versa.

� AD is the median of �ABC

� ar(�ABD) =12

ar(�ABC) ...(i)

BE is the median of the �ABD

� ar(�BED) = 12

ar(�ABD) ...(ii)

From (i) and (ii), we get

ar(�BED) =12

ar1

( ABC)2

⎛ ⎞Δ⎜ ⎟⎝ ⎠

= 14

ar(�ABC).

� ��

AD is the median of the �ABC.� ar(�ABD) = ar(�ACD) ...(i)and ar(�BDE) = ar(�CDE) ...(ii)

(� DE is median of �EBC)(Median of a triangle divides it into twotriangles of equal areas)

Subtracting (ii) from (i), we get ar(�ABD) – ar(�BDE)

= ar(�ACD) – ar(�CDE)� ar(�ABE) = ar(�ACE).

� �� Diagonals of rhombus are perpendicular

bisector of each other��AO � BD, CO � BD

ar(ABCD) = ar(�ABD) + ar(�BCD)

=12

BD × AO +12

× BD × CO

=12

BD (AO + CO) =12

× BD × AC

� ��

�� ar(�ABC) = ar(�ADC) ...(i)(A diagonal of a ��gm divides it into two

triangles of equal areas)O is the mid-point of AC and BD both.

(Diagonals of a parallelogrambisect each other)

BO is median of �ABC� ar(�AOB) = ar(�BOC)

= 12

ar(�ABC) ...(ii)

DO is median of �ADCand ar(�AOD) = ar(�COD)

= 12

ar(�ACD) ...(iii)

(Median of a �, divides it in twotriangles equal in area)

From (i), (ii) and (iii), we get ar(�AOB) = ar(�BOC)

= ar(�COD) = ar(�AOD).� ��

32 ��

��

�� (C) �ECD and squareABCD are on the samebase DC and between thesame parallels AB andCD.

� Area of �ECD =12

ar(square ABCD)

�=12

× x2 square units.

�� ar(PQRS) = PQ × SL� 156 = 13 × SL

� SL = 15613

= 12 cm.

�� ar(rectangle) = 10 × 18 = 180 cm2

Since �EDC andrectangle ABCDare on the samebase DC andbetween thesame parallelsAB and DC, so ar(�EDC) is halfar(rectangle ABCD).Now, ar(unshaded region)

= ar(rectangle ABCD) – ar(�EDC)= ar(rectangle ABCD)

– 12

ar(rectangle ABCD)

=12

ar(rectangle ABCD)

=12

× AB × BC

=12

× 180 = 90 cm2.

�� ar(rectangle ABCD)= DC × AD = 30 × 20 = 600 cm2

FD

A E B

C

G

DF = 12

× DC = 12

× 30 = 15 cm

Also, AE = DF = 15 cm

Similarly, DG = AG = 202

= 10 cm

ar(�DFG) =12

× DF × DG

=12

× 15 × 10 = 75 cm2

ar(�AGE) =12

× AE × AG

=12

× 15 × 10 = 75 cm2

Now, ar(GEBCF) = ar(rectangle ABCD)– ar(�DFG) – ar(�AGE)

= 600 – 75 – 75= 450 cm2.

�� CO = DO� AO is a median of �ACD� ar(�AOC) = ar(�AOD) ... (i)Similarly, ar(�BOC) = ar(�BOD) ... (ii)

Adding (i) and (ii), we havear(�AOC) + ar(�BOC)

= ar(�AOD) + ar(�BOD)� ar(�ABC) = ar(�ABD)

� ��

�� (i) Since �ACB and �ACF lie on thesame base AC and between the sameparallels AC and BF.

A E B

D C

10 cm

18 cm

33��

� ar(ACB) = ar(ACF)(Triangles on the same base and betweensame parallels are equal in area)(ii) Adding ar(ACDE) to both sides, we get

ar(ACDE) + ar(ACB)= ar(ACDE) + ar(ACF)

� ar(ABCDE) = ar(AEDF).� ��

� ar(�EBC) = 12

ar(��gm EBCY) ...(i)

and ar(�BCF) = 12

ar(��gm FCBX) ...(ii)

(Triangles on the same base and betweensame parallels are equal in area)But ��gm EBCYand ��gm FCBXare on the samebase BC andbetween thesame parallelsBC and EF� ar(��gm EBCY) = ar(��gm FCBX) ...(iii)From (i), (ii) and (iii), we get

ar(�EBC) = ar(�FCB).� ��

� (i) Here, FE �� BC and DE �� BA.� FE = BD = DC.

(Mid-point theorem)� FE �� BD and DE �� BF.� BDEF is a parallelogram.

(ii) �FBD, �EDC and �DEF are on equalbases and between the same parallelsbecause FE �� BC.� FE = BD = DC.� ar(�FBD) = ar(�EDC) = ar(�DEF)Similarly,

ar(�DCE) = ar(�FEA) = ar(�DEF)� ar(�DEF) = ar(�FBD)

= ar(�DCE) = ar(�AEF)

=14

ar(�ABC)

i.e., ar(�DEF) =14

ar(�ABC).

(iii) Now,ar(BDEF) = ar(�BDF) + ar(�DEF)

= 14

ar(�ABC) +14

ar(�ABC)

=12

ar(�ABC).

�� (i) We are given thatOB = OD ...(i)

and AB = CD ...(ii)We draw BL � OC and DM � OA.

In �OLB and �OMD, we have�LOB = �MOD

(Vertically opposite angles)�OLB = �OMD (Each 90°)

OB = OD [By (i)]�OLB � �OMD

(AAS congruence)� OL = OM ..(iii) (CPCT)and BL = DM ...(iv)In �ALB and �CMD, we have

�ALB = �CMD (Each 90°)AB = CD [By (ii)]BL = DM [By (iv)]

Then by RHS congruence criterion, wehave

�ALB ��CMD� AL = CM ...(v)Then from (iii) and (v), we have

OA = OC ...(vi)From (iv) and (vi),

OA × BL = OC × DM

�12

OA × BL= 12

OC × DM

34 ��

� ar(�AOB) = ar(�DOC)or ar(�DOC) = ar(�AOB).(ii) ar(�DOC) = ar(�AOB)� ar(�DOC) + ar(�COB)

= ar(�AOB) + ar(�COB)� ar(�DCB) = ar(�ACB).

� ��

��

�� (D)As parallelogramABCD and triangleABC are on the samebase AB and betweenthe same parallels ABand CD

ar(ABCD) = 2 ar(�ABC)

��

arar

(ABCD)( ABC)

= 21

� ar(ABCD) : ar(�ABC) = 2 : 1.

�� ar(�ACP) = ar(�ABC) ... (i)(�s are on samebase AC andbetween sameparallels ACand BP)

Now, ar(�ADP) = ar(�ADC) + ar(�ACP)= ar(�ADC) + ar(�ABC)

[From (i)]= ar(�ABCD).

�� In parallelogram ABCD,BP = PQ = QC

In quadrilateral ABPR, PR �� AB andBP �� AR� BP = ARSimilarly, PQ = RS and QC = SDConsequently, we have

ar(ABPR) = ar(RPQS) = ar(SQCD)(� Parallelograms with equal base

and between same parallels)

So, ar(RPQS) = 13

ar(ABCD)

Triangle APQ and parallelogram RPQSare on the same base PQ and betweenthe same parallels AD and BC.

� ar(APQ) = 12

ar(RPQS)

= 12

× 13

ar(ABCD)

= 16

ar(ABCD).

�� BD = 2 2AB + AD

= 2 24 + 3

= 16 + 9

= 25

= 5 units.Let BE be y� DF = y (� Opposite sides

of rectangle are equal)In right �DFC,

FC = y216 –

(Pythagoras theorem)In right �BCE,

CE = y29 –

(Pythagoras theorem)FE = BD = 5 units ...(i)

Also, FE = FC + CE

= y216 – + y29 – ...(ii)

y216 – + y29 – = 5

[From (i) and (ii)]

y216 – = 5 – y29 –

35��

Squaring both sides,

16 – y2 = 25 + 9 – y2 – 10 y29 –

(Using (a – b)2 = a2 + b2 – 2ab)

10 y29 – = 18 � 5 y29 – = 9

Squaring both sides, we get25(9 – y2) = 81225 – 81 = 25y2

144 = 25y2

y =125

�

= 125

(� Length cannot be negative)Area of DBEF= DB × BE

= 5 ×125

= 12 unit2.

�� DBC and �EBC have equal areas andsame base BC and vertices lie on sameside of base BC.� The two trianglesare between thesame parallels.��DE �� BC

�� ar(�ABD) = ar(�ABC). Why? ar(�ABD) – ar(�AOB)

= ar(�ABC) – ar(�AOB).

� We have, AP �� BQ (Given)Now, �ABQ and �PQB have same baseBQ and lie between same parallel linesAP and BQ.

� ar(�ABQ) = ar(�PQB) ... (i)Similarly, BQ �� CR and �CBQ and �RQBlie on same base BQ and between sameparallel lines BQ and CR�� ar(�CBQ) = ar(�RQB) ... (ii)Adding (i) and (ii), we have

ar(�ABQ) + ar(�CBQ)= ar(�PQB) + ar(�RQB)

� ar(�AQC) = ar(�PBR).

� Since, ar(�AOD) = ar(�BOC) (Given)

Adding ar(�COD) to both sides, we get ar(�AOD) + ar(�COD)

= ar(�BOC) + ar(�COD)� ar(�ADC) = ar(BCD)But these are the triangles lying on thesame base DC and are equal in areas.So, the line joining their vertices A andB is parallel to DC, i.e., AB �� DCHere, one of the pairs of opposite sides ofquadrilateral ABCD is parallel (AB �� DC)� Quadrilateral ABCD is a trapezium.

�� In figure, AB (produced) and AC (produced)meet XY at G and H respectively.Now, BGFC and BEHC are parallelograms.

� BC = GF and BC = EH� GF = EH� GF – EF = EH – EF� GE = FH� ar(�BGE)= ar(�CFH) ...(i)Also, we find that

ar(�AGE) = ar(�AHF) ...(ii)

36 ��

Subtracting (i) from (ii), we havear(�ABE) = ar(�ACF).

��

��

� (C) As �ABC and �AEC are on the samebase AC and between the same parallelsAC and BE,

� ar(�ABC) = ar(�ACE)� ar(�OAB) + ar(�AOC) = ar(�OCE)

+ ar(�AOC) � ar(�OAB) = ar(�OCE).

�� As AF = FB and AE = EC, FE �� BCor FE �� BD or FE �� DC

Consequently, we getar(�AFE) = ar(�BDF) = ar(CDE)

= ar(�DEF).

= 284

= 7 cm2

� ar(AEDF) = 14 cm2.�� ar(PQRS) = PQ × SL ...(i)

Also, ar(PQRS) = RQ × SM ...(ii)

From (i) and (ii), we getRQ × SM = PQ × SL

� 8 × SM = 12 × 6(� RQ = SP = 8 cm)

� SM = 9 cm.

�� CP is a median of �ABC,as P is the mid-point of the side AB.

� ar(�APC) = ar(�BCP) = 12

ar(�ABC)

As parallelogram ABCDand triangle ABC are onthe same base AB andbetween the sameparallels AB and CD,ar(ABCD) = 2 ar(�ABC)� ar(APCD) + ar(�PBC)

= 2 ar(�ABC)� ar(APCD) = 2ar(�ABC)

– 12

ar(�ABC)

� 36 =32

ar(�ABC)

� ar(�ABC) = 24 cm2.

�� ar(�ABD) =12

ar(�ABC).

(Median of a triangle divides itinto two triangles equal in areas)

� x =12

y

� 2x – y = 0.�� (i) parallels

(ii) corresponding altitude�� (i) equal (ii) area

(iii) one diagonal, other diagonal�� (i) We know that a median of a triangle

divides it into two triangles of equalareas. CD is the median of �ABC.� ar (�ACD) = ar (�BCD) = 400 m2

…(i)Also, M divides CD in 1 : 1� CM = DM� AM is the median of �ACD� ar(�AMD) = ar(�AMC)

= 12

ar(�ACD) = 12

× 400 = 200 m2

[Using (i)]� Area of Ramdin’s field = 200 m2

Area of Rohit Kapoor’s field = (400 + 200) m2 = 600 m2.

37��

(ii) A median of a triangle divides it intotwo triangles equal in areas (Geom-etry).

(iii) Honesty, Rationality i.e., able toreason and form judgements.

�� Draw AD � BCAD is perpendicularbisector of BC.

� BD = CD = 2a

and �ADB = �ADC = 90°

ar(ADB) = ar(ADC) = 12

ar(�ABC)

�12

BD × AD = 12

ar(�ABC)

In �ADB,AB2 = AD2 + BD2

(Using Pythagoras theorem)

� a2 = AD2 + 2

2a� �

� ��

� a2 –2

4a

= AD2

� AD2 =2 24 –4

a a =

234a

� AD =32

a

� ar(�ABC) = 2 ar(�ADB)

= 2 ×12

×2a

× 32

a

= 234

a sq. units.

��

�� (D)

�� The median of a triangle divides it intotwo triangles of equal areas.

�� Join AE

In �BDC, DE median

(� E mid-point of BC)

ar(�DEC) = ar(�BDE) = 12 cm ...(i)

In �AEC, ED is median[� D mid-point of AC (Given)]

ar(�AED) = ar(�DEC) ...(ii)From (i) and (ii),

ar(�AED) = ar(�BDE) = 12ar(�AEC) = ar(�ADE) + ar(�DEC)

= 12 + 12 = 24 cm2.�� ABCD is a rhombus with

AC = 16 cm, BD = 12 cm and P,Q,R,Sare the mid-points of the sides AB, BC,CD and DA respectively.�� area of the quadrilateralPQRS.In � ABD, S ismid-point of AD andP is mid-point of AB.

� SP �� BD

and SP =12

BD

(� Mid-point theorem) = 6 cm.

In � BCD,R is mid-point of DC and Q is mid-pointof BC.

� RQ �� BD and RQ =12

BD = 6 cm.

SP �� BD and RQ �� BD �� SP �� RQAlso, SP = RQ = 6 cm� SPQR is a parallelogram

Similarly, PQ = SR = 12

AC = 8 cm

Now, BD �� SP � SN �� OMAC �� RS � ON �� MS

� NOMS is a parallelogram�AOD = 90°(��Diagonals of parallelogram

bisect at right angle)� �MSN = 90°

(��NOMS is parallelogram)� SPQR is a rectangle� Area of SPQR = SP × QP

= 6 × 8 = 48 cm2.��

38 ��

�� (i) False (ii) True� Let side of a rhombus be a units.

So AC = aThe diagonals of arhombus bisect eachother at right angle

� OA = 12

a.

So, by Pythagoras theorem,

� OB = 2 2AB – OA

= aa

22 –

4=

32

a

� BD = 2 × 32

a

� BD : AC = 3 a : a or 3 : 1.� ar(ABCD) = 162 cm2 ... (i)

Let AP = x cm and BP = 2x cmLet altitude of the parallelogram corres-ponding to the base AB = h cmUsing equation (i), we have

3x × h = 162

Further, ar(�APD) =12

x × h

=12

x ×x

1623

= 27 cm2.

�� (i) In �ABD, BP = DP� AP is the median of �ABD� ar (�ABP) = ar (�ADP) …(i)Similarly in �BCD,

ar(�BCP) = ar(�DCP) …(ii)Adding (i) and (ii), we havear(�ABP) + ar(�BCP)

= ar(�ADP) + ar(�DCP)� ar(�ABC) = ar(�ADC).(ii) A median of a triangle divides it intotwo triangles of equal area (Geometry).(iii) Creative.

��

�� (C) As AQ �� SC and AS �� CQ,� AS = CQ ... (i)

Now, area of parallelogram PQRS= SR × CQ = SR × SA. [From (i)]

�� If a triangle and a parallelogram are onthe same base and between the sameparallels, then the area of the triangle ishalf the area of the parallelogram.� ar(triangle) : ar(parallelogram)= 1 : 2.

�� As AP = 4 cm, AS = 3 cm

� SP = 2 24 3� = 5 cm

Similarly, PQ = RQ = SR = 5 cmTherefore, PQRS is a rhombus.Now, SQ = DC = 8 cmAnd RP = DA = 6 cm

� ar(PQRS) = 12

× 8 × 6 = 24 cm2.

�� Perimeter of rectangle ABEF= 2 (AB + AF)

Perimeter of parallelogram ABCD= 2 (AB + AD)

In right triangle ADF, AD is thehypotenuse

39��

� AD > AF� AB + AD > AB + AF� 2(AB + AD) > 2(AB + AF)� Perimeter of ABCD

> Perimeter of ABEF�� (i) False (ii) True�� (i) True (ii) True�� (i) Let the proprietor

has the plot ABCD.Find a point E whichis the mid-point of AB.Join DE and CE. Nowjoin E to F, the mid-point of CD. Clearly,AEFD and EBCF are equalParallelograms as they have equal bases(AE = BE) and lie between same parallelsAB and CD.The �ADE and �BCE have equal bases(AE = BE) and lie between same parallelsAB and CD.

ar(�ADE) = ar(�BCE) …(i)EF is a median of �CDE� ar(�EDF) = ar(�ECF) …(ii)DE is a diagonal of �gm AEFD� �ADE � �EDF� ar(�ADE) = ar(�EDF) …(iii)From results (i), (ii) and (iii), we obtain

ar(�ADE) = ar(�EDF)= ar(�ECF) = ar(�BCE)

Hence the four equal triangular parts ofthe plot are ADE, EDF, ECF and BCE.(ii) Concepts are:

(a)Parallelograms on equal bases andlie between same parallels are equalin area.(b)Triangles on equal bases and liebetween same parallels are equal inarea.(c)A median of a triangle divides itinto two triangles equal in area.

(iii) Love for his country men, Collectiveresponsibility.

��

�� (B) We know that �ADF, �BDE, �EFCand �DEF are of equal areas.Then, ar(DECF)

= ar(DEF) + ar(EFC)

=14

ar(ABC) +14

ar(ABC)

=12

ar(ABC).

� Diagonal of parallelogram divides it intotwo triangles of equal areas.

� If two parallelograms are on equal basesand between the same parallels, thentheir areas are equal.� Ratio of their areas = 1 : 1.

�� 3a + b : a + 3b� �� Join BD to intersect EF at M.In �ABD and �EMD,

EMa

= 21

� EM = 2a

Similarly, FM = 2b

Area of a trapezium

=12

× sum of parallel sides

× distance between them.

�� (i) The base line

of trees remainssame before andafter the storm.Before the stormthey formed arectangular surface and after it theyform a parallelogram. The verticalheights of trees on June 2010 andJanuary 2012 are same. So their topslie on the straight line on these two

40 ��

days. Thus, the rectangle and theparallelogram are on the same base andlie between the same parallels.Hence required area = 108 m2.(ii) Parallelograms on the same baseand between the same parallels areequal in area.(iii) Love for environment and enviro-nmental cleanliness.

�� Let ABCD be the given plot of land inthe form of quadrilateral ABCD.Through D, draw DE �� CA, which meetsBA produced at E. Join EC.

We have to show thatar(quadrilateral ABCD) = ar(�EBC)Since �CAE and �CAD lie on the samebase AC and between the same parallelsCA and DE.� ar(CAE) = ar(CAD)� ar(CAE) – ar(AOC)

= ar(CAD) – ar(AOC)[Subtracting ar(AOC) from both sides]� ar(EOA) = ar(COD)Adding ar(ABCO) to both sides, we get

ar(EOA) + ar(ABCO)= ar(COD) + ar(ABCO)

� ar(EBC) = ar(ABCD).Clearly, land for Health Centre is

�COD.And new plot of land in triangular shape(�EBC) is equal in area with quadrila-teral ABCD.

��

41��

��

CIRCLES

��

�� (B) Let O be thecentre of the circle.Draw OM � AB.OM bisects AB, i.e.,AM = MB = 15 cm.In right triangle AMO,

OM = 2 2AO – AM = 2 217 – 15

= (17 +15)(17 – 15) = 64= 8 cm.

�� COD = �AOB= 70°

In �COD, CO = DO� �CDO = �DCO

= xFurther,

�CDO + �DCO + �COD = 180°� x + x + 70° = 180°� x = 55°.

�� AO = PO (Radii of same circle)= 5 cm

In right-angled �AMO,AO2 = AM2 + MO2

52 = AM2 + 42

� AM = 25 – 16 = 3 cm

We know that perpendicular, drawnfrom the centre of a circle to a chord,bisects the chord.� BM = AM = 3 cm� AB = AM + BM = 3 cm + 3 cm = 6 cm.

�� Centre O of the circlepassing through A, Band C, coincides withthe point of intersec-tion of the perpen-dicular bisectors ofAB and BC.

BM = 12

AB = 6 cm

OM = BN = 12

BC = 8 cm

In right triangle BMO,BO2 = BM2 + OM2

= 62 + 82 = 36 + 64= 100

� BO = 10 cmHence, radius of the circle is 10 cm.

�� Let AB be the chord,OC � AB.

� AC = CB (Perpendicular fromcentre bisects the chord)In� �ACO, AC2 = AO2 – OC2

(By Pythagoras Theorem)= 52 – 32 = 25 – 9 = 16

� AC = 4 cm.� AB = 4 × 2 cm = 8 cm.

�� In figure, we are given that�AOB = �PO�Q

Also,we have OA = O�P(... Radii of congruent circles)

and OB = O�Q(... Radii of congruent circles)

� �OAB ��O�PQ(SAS congruence rule)

Then by CPCT, AB = PQ.

� As the two circles with centres O andO� are congruent,

42 � � � � � ��

OA = O�Cand OB = O�D

AB = CD (Given)� �OAB ��O�CD

(SSS congruence rule)� �AOB = �CO�D (CPCT)

� (i) Let the butter-chords of the biscuit beAB and CD; and centreof the biscuit be O.Join each of A, B, C, Dto OIn �OAB and �OCD, AB = CD (Given)

OA = OC (Each equal to radius)OB = OD (Each equal to radius)

� �OAB � �OCD [SSS]� �AOB = �COD (CPCT)Therefore, the butter-chords subtendequal angles at the centre of the biscuit.(ii) We are given the length of eitherchord is greater than the radius and lessthan the diameter of the circle.Let length of either chord = l, radius = rand angled subtended by either butter-chord = Two cases arise:�� If l = rIn this case, the chord and corres-ponding radius form an equilateraltriangle with side r� = 60°�� If l = 2rIn this case, the butter chord passesthrough the centre.� = 180°Consequently, we arrive at the followinginequality:

60° < �< 180° (As r < l < 2r)

Thus, the required range is from 60° to180° excluding both.

(iii) (a) Congruence of triangles

(b) CPCT (Corresponding parts ofcongruent triangles are equal)

(c) Equilateral triangle and its angles.

(iv) Industrialist, thoughtfulness, self-confident, Rationality.

�� (i) OE, OD, OC (ii) CD

(iii) �DEC (iv) AB, CE, CD

(v) �CXE (vi) DOE

(vii) CXED (viii) <

��

�� (A) ABCD is a square, ABCis a right-angled trianglewith �ABC = 90°. AOB is anisosceles triangle withAO = BO.

�� Join AO. �� OM � AB � AM = BM

� AM = AB2

= 202

= 10 cm

In right-angled triangle AOM,AO2 = AM2 + OM2

= 102 + � �2

2 11

= 100 + 44 = 144� AO = 12 cm

Hence, radius = 12 cm.�� As OC is perpendicular to chord AB,

AC = BC = AB2

= 82

= 4 cmIn right-angled �AOC,

OC = 2 2AO – AC

= 2 25 – 4

= (5 – 4)(5 4)� = 3 cmAs AO and OD both are radii,

OD = AO

43��

� OC + CD = 5� 3 + CD = 5 [� OC = 3 cm]� CD = 2 cm.

�� Draw OM � ABProduce MO to inter-sect CD at N.As AB CD, ON � CDJoin CO.

AM = MB = AB2

=62

= 3 cm

In right-angled triangle AOM,

OM = 2 2AO – AM

= 2 25 – 3 = 25 – 9

= 16 = 4 cm

Similarly, in right-angled �CON,

ON = 2 25 – 4 = 3 cm

[� CO = AO = 5 cm]Now, MN = OM + ON

= 4 cm + 3 cm = 7 cm.As MN and PQ each is the perpendi-cular distance between the two parallellines AB and CD.

PQ = MN= 7 cm.

�� Let AB be the chord of the given circlewith centre O.Draw OM � AB.

AM = AB2

= 62

= 3 cm(Perpendicular drawn from the centreto a chord divides it into two equal parts)In right-angled �AOM,

OM = 2 2AO – AM = 2 26 – 3

= 36 – 9 = 27

= 5.20 cm (approximately).

�� (i) False (ii) True

� Let OE = x� OA = OC = OB = (x + 3) cm.Using Pythagoras Theorem in �OAE,we get

(3 + x)2 = x2 + 42

� 32 + x2 + 6x = x2 + 16� 6x = 16 – 9 = 7

� x = 76

= 116

cm

� OA = x + 3 = � �� 1

1 36

cm = 416

cm.

� O is the centre of the circle. Chords ABand CD of the circle are equal. P is thepoint of intersection of AB and CD. JoinOP. Draw OL � AB and OM � CD.Here, we find OL = OM ...(i)

(... AB = CD)In �OLP and �OMP,

�OLP = �OMP = 90°OP = OP (Common hypotenuse)OL = OM [By (i)]

Then we have �OLP � �OMP(RHS congruence rule)

By CPCT, PL = PM ...(ii)

Now, AL = BL = 12

AB

CM = DM = 12

CD

� AL = CM ...(iii)(... AB = CD)

� BL = DM ...(iv)Subtracting (ii) from (iii), we have

AL – PL = CM – PM� AP = CPAdding (ii) and (iv), we get

PL + BL = PM + DM� PB = PD.

�� Since AB = AC = 6 cm.So, D is the mid-point of BC.

(OD is right bisector of BC)Let OD = x

44 � � � � � ��

� 2 �CAB = 90°� �CAB = 45°.

�� If two chords of a circleare at equal distancesfrom the centre of thecircle, then the twochords are equal.

� CD = AB = 2 × 4.5 = 9 cm.

�� As OD � AB, AD = BD = 5 cm,In right triangle BOD,

BO = 2 2BD + OD

= 25 +16= 41 cm

� BC = 2 × BO = 2 41 cm.In right triangle ABC,

AC = 2 2BC – AB = 4 × 41 – 100

= 64 = 8 cm

In right triangle ADC,

DC = 2 2AC + AD = 64 + 25

= 89 cm.

�� AB = CD(Given)

� OL = OM(Equal chords areequidistant from

the centre)OP = OP (Common)

and �OLP = �OMP (Each 90°)� �OLP � �OMP� �OPL = �OPM. (CPCT)

�� Draw OL � AD.

BC is a chord ofthe smaller circle.P e r p e n d i c u l a rfrom the centre ofthe circle to thechord bisects the chord.� BL = LC ...(i)AD is a chord of the bigger circle� AL = LD ...(ii)

� AD = 5 – xIn �OCD, we have

OC2 = OD2 + CD2

� 52 = x2 + CD2

� 25 = x2 + CD2

� CD2 = 25 – x2 ...(i)In �ACD, we have

AC2 = AD2 + CD2

� 62 = (5 – x)2 + CD2

� 36 = 25 + x2 – 10x + CD2

� CD2 = 36 – 25 + 10x – x2

� CD2 = 11 + 10x – x2 ...(ii)From (i) and (ii), we get

11 + 10x – x2 = 25 – x2

� 10x = 14� x = 1.4 cmFrom (i), we get

CD2 = 25 – (1.4)2

� CD2 = 25 – 1.96� CD2 = 23.04 � CD = 4.8... BC = 2CD� BC = 2 × 4.8 = 9.6 cm.

��

�� (A) Given that �BAD = �ADC

� �BA + �AD = �AD + �DC

� �BA = �DC� BA = DC

(Chords corresponding to equalarcs are equal)

i.e., AB = CD.

�� We know that an angle made insemicircle is a right angle.

� �ACB = 90°

...(i)

In �ABC, as AC = BC� �ABC = �BAC ...(ii)Also,

�CAB + �ABC + �ACB = 180°� �CAB + �CAB + 90° = 180°

[Using (i) and (ii)]

45��

Subtracting (i) from (ii), we getAL – BL = LD – LC

� AB = CD.

� OD � chord AB (Given)� D is the mid-point of AB.

(Perpendicular drawn from thecentre of the circle to the chord

bisects the chord)As O is the centre of thecircle, O is the mid-pointof the diameter BC.In �ABC, D is the mid-point of AB and O is themid-point of BC.

� OD = 12

AC (Mid-point theorem)

� AC = 2OD.

� �AB � �CD (Given)

�AB + �BC � �BC + �CD

(Adding �BC to both sides)

� �AC � �BD� AC = BD

(Chords corresponding tocongruence arcs are congruent)

� �AOC = �BOD ...(i)(Equal chords subtend equal

angles at the centre)� OA = OB ...(ii)

OC = OD ...(iii)(Radii of the same circle)

From (i), (ii) and (iii),�AOC � �BOD(Using SAS congruence criterion)

� �A = �B. (CPCT)

�� R, S, M denote positions of Reshma,Salma and Mandeep on the circle withcentre O and radius 5 m such that RS =SM = 6 m.Join SO (and produce it if required) tointersect RM at N. We know that theline segment joining centre to thecommon point of equal chords bisectsthe angle between the two chords

� �RSN ��MSN by SAS� RN = MN (CPCT)� SO bisects RM

and �SNR = �SNM (CPCT)Also these angles form a linear pair.� �SNR = �SNM = 90°� ON ��RM�� Join OM,OS = OM = 5 m (... Radius of circle)Let ON = x m� SN = OS – ON = 5 – xIn right �ONM, NM2 = 52 – x2

= 25 – x2 ...(a)In right �SNM, NM2 = 62 – (5 – x)2

= 36 – 25 – x2 + 10x ...(b)� 25 – x2 = 36 – 25 – x2 + 10x

10x = 14 � x = 1.4�� Let ON = x� SN = OS + ON

= 5 + xIn right �ONM,

NM2 = 25 – x2

In right �SNM,NM2 = 62 – (5 + x)2

� 25 – x2 = 36 – 25 – x2 + 10x25 = 11 – 10x14 = – 10x

x = – 1.4 (Side cannot be –ve)� x = 1.4

NM2 = 25 – x2 = 25 – 1.96 = 23.04 NM = 4.8

RM = 2NM = 9.6 cm.

��

�� (C) �AB � �CD

� �AB + �BC � �BC + �CD

� �AC � �BD

46 � � � � � ��

� AC = BD

(Chords corresponding tocongruence arcs)

Now, in �OAC and �OBD,OA = OB, OC= OD, AC = BD� �OAC � �OBD� �2 = �3. (CPCT)

�� As angle made in semicircle is a rightangle

� �ACB = 90°In �ABC,�A + �B + �C = 180°

� x + 35° + 90° = 180°� x = 55°.

�� We know that the anglesubtended by an arc atthe centre is double theangle subtended by itat any point on theremaining part of thecircle.� �AOB = 2 �ACB� x = 2 × 35° = 70°.

�� Let us draw an angle BDC as shown inthe adjoining figure.

�BDC + �BAC = 180°(Opposite angles of cyclic

quadrilateral ABDC)� �BDC = 180° – 130° = 50°Now, x = 2 × �BDC

(Angles subtended at the centre istwice the angle subtended at any

point on the remaining partof the circle)

= 2 × 50° = 100°.

�� AD is the bisector of �BACmeeting the circle at D andpasses through the centreO of the circle.AD is a diameter

� �ABD = �ACD ...(i)Also, �BAD = �CAD (Given)

� �BD = �CD ...(ii)(Arc subtending equal angles are equal)Subtracting (ii) from (i), we get

� �ABD – �BD = �ACD – �CD

� �AB = �AC� AB = AC. ��

�� ADC = 12�AOC

(Angles subtended atthe centre is twice theangle subtended at anypoint on the remainingpart of the circle)

= 12

× (60° + 30°) = 45°.

� 150°, 30°�� AB = OA = OB� �AOB = 60°

�APB = 12

× reflex

��AOB.

� (i) Let the equalchords be AB and ACof the wheel (circle)with centre OJoin BC, BO, CO andAO. AO intersects BCat MWe have to determine the length of BC.In �AOB and �AOC,

AB = AC = 12 cmBO = CO = 10 cmAO = AO (Common)

� �AOB ��AOC (SSS axiom)� �BAO = �CAO …(i) (CPCT)In �ABM and �ACM,

AB = AC = 12 cm�BAM = �CAM [From (i)]

AM = AM (Common)� �ABM ��ACM (SAS axiom)

47��

� BM = CM (CPCT)� �BMO = �CMO …(ii) (CPCT)BMC is a straight line� �BMA + �CMA = 180° …(iii)From (ii) and (iii), �BMA = �CMA = 90°� �BMO = 90°In right triangle BMO,

BM2 = 102 – MO2 …(iv)In right triangle BMA,

BM2 = 122 – AM2 …(v)From (iv) and (v),

100 – MO2 = 144 – AM2

� 100 – (AO – AM)2 = 144 – AM2

� AM2 – (10 – AM)2 = 44 (� AO = 10 cm)

� (AM + 10 – AM) (AM – 10 + AM) = 44� 10 × (2AM – 10) = 44� 2AM = 10 + 4.4� AM = 7.2In �BMA,

BM = 2 212 7.2� = 92.16 = 9.6

BC = 2BM = 2 × 9.6 = 19.2Hence, the length of the required chordis 19.2 cm.(ii) SSS and SAS axioms of congruence oftriangles; and Pythagoras Theorem.(iii) Love for country, Patriotism

�� Join OA, OS and OD.�AOS = �SOD = �AOD = (say)

(... AS = SD = AD)� + + = 360°

(Angles at the centre of the circle)� 3�= 360° � = 120°

AO produced meets SD at M and the

circle at N. Here OM � SD and M ismid-point of SD. Put SM = x.We have �OSN in which �SON = 60°and OS = ON.

�OSN = �ONS = 60°�OSN is equilateral.

M is mid-point of ON.OM = 10 m.

In �OMD, by Pythagoras Theorem,x2 = (20)2 – (10)2 = 300

� x = 10 3 m.

Then SD = 2x = 20 3 m.

Thus, AS = AD = SD = 20 3 m.Hence, the length of the string of each

phone is 20 3 m.

��

�� (B) �AOC = 2 × �ABC(The angle subtendedby an arc at the cen-tre is twice the anglesubtended by it atany point on the re-maining part of thecircle) = 2 × 20° = 40°.

�� AOB = 2 × �ACB= 2 × 40°= 80°.

�� In �DEC, �CEB is an exterior angle.� �CEB = �ECD + �EDC� 130° = 20° + �EDC� �EDC = 110°

i.e.,�BDC = 110°�BAC = �BDC

(Angles in the same segmentsof a circle are equal)

= 110°.

48 � � � � � ��

�� Reflex �POR = 2 �PQR= 2 × 100° = 200°

�� Obtuse �POR = 360° – Reflex �POR= 360° – 200°= 160° ...(i)

In �POR, �OPR = �ORP ...(ii)(� OP = OR)

Also,�OPR + �ORP + �POR = 180°

� �OPR + �OPR + 160° = 180° � 2�OPR = 180° – 160°

[From (i) and (ii)]

� �OPR =180° – 160°

2= 10°.

�� Let the given parallelogram be ABCD� �B = �D ...(i)

(Opposite angles of a gm)Also,��B + �D = 180° ...(ii)

(Opposite angles of a cyclicquadrilateral)

From (i) and (ii), we get2�B = 180° � �B = 90°

Parallelogram with one right angle isrectangle.Hence, ABCD is a rectangle.

�� BAD + �BCD = 180° ...(i)(Opposite angles of cyclic

quadrilateral ABCD)�BCD + �BCE = 180°

(Linear pair axiom)� �BCD = 180° – �BCE ...(ii)

From equations (i) and (ii), we get�BAD + 180° –�BCE = 180°� �BCE = �BAD.

��

� AC is the diameter ofthe circle.��B = 90°and� � �D = 90°(Angles in a semicircle)BD is also a diameterof the circle.Similarly, �A = �C = 90°.Thus, �A = �B = �C = �D = 90°.Thus, ABCD is a rectangle.

� �DAC = �DBC = 70°(Angles in the same segment)

� �BAD = �BAC + �DAC= 30° + 70° = 100°

Now, �BCD + �BAD = 180°(Opposite angles of the cyclic

quadrilateral ABCD)� �BCD = 180°– 100° = 80°

Further, in �ABC,�BCA = �BAC

(� AB = BC)= 30°

Now, �BCD = �BCA + �ACD� 80° = 30° + �ACD� �ACD = 80° – 30° = 50°i.e., �ECD = 50°.

�� ABCD is a trapezium such that AB DCand AD = BC.Draw perpendiculars DL and CM on ABfrom D and C respectively.

49��

In �ALD and �BMC,�ALD = �BMC = 90°

AD = BC (Given)and DL = CM

(Perpendicular distances betweenparallels AB and CD)

� �ALD ��BMC(RHS congruence rule)

� �1 = �2� �1 + �LDC = �2 + �MCD

[��LDC = �MCD, both 90°]� �D = �C ...(i)Also, AB CD� �A + �D = 180° ...(ii)From (i) and (ii),

�A + �C = 180°Also, �A + �B + �C + �D = 360°� 180° + �B + �D = 360°� �B + �D = 180°Hence, the trapezium ABCD is a cyclicquadrilateral.

��

�� (B) As the sum of opposite angles of acyclic quadrilateral is 180°,

�A + �C = 180°

� �A = 180° – 110°

= 70°.

�� ADC = 180° – �ADX�= 180°– 80° = 100°

Also, y + �ADX = 180°

� y = 180° – 100° = 80°

Similarly, �BCD = 180° – 70° = 110°And x = 180° – 110° = 70°.

�� B + �D = 180°(Opposite angles of the cyclic quadrilateral are supplementary)� �D = 180° – 65°

= 115°�B + �C = 180° (Interior angles onthe same side of the transversal BC)� �C = 180° – 65° = 115°Similarly, we can find that

�A = 180° – 115° = 65°.

�� BDC = �BAC (Angles in the same segment)

= 45°In �BCD,

�BCD + �BDC + �DBC= 180°� �BCD + 45° + 60° = 180°� �BCD = 180° – 105°

= 75°.

�� ABC = �ADC = 90°

In quadrilateral ABCD,

�B + �D = 180°� ABCD is a cyclic quadrilateral.(See in figure we can draw circlethrough A, B, C, D)

� �CAD = �CBD.

(Angles in the same segment)�� Circles are drawn on AB and AC as

diameters. AP is their common chord.Point P is the second point of inter-section of circles other than A.

50 � � � � � ��

Now, �APB = 90°(Angle in a semicircle)

and �APC = 90°(Angle in a semicircle)

� �APB + �APC = 180°Also, these are adjacent angles.� �APB and �APC form a linear pair.� B, P and C are collinear.� P lies on the side BC of the �ABC.

� (i) �PQC + �PBC = 180°(Opposite angles of a cyclic

quadrilateral are supplementary)� �PQC + 80° = 180°� �PQC = 100° ...(i)But, �PQD + �PQC = 180°

[Linear pair axiom]� �PQD + 100° = 180°� �PQD = 80°.

(ii) Also, �PQD + �PAD = 180°(Opposite angles of a cyclic

quadrilateral)� 80° + �PAD = 180°� �PAD = 100°.Further, clearly �PAD + �PBC

= 100° + 80°i.e., �BAD + �ABC = 180°This is the sum of cointerior angles asAB is transversal for AD and BC.Hence, we conclude that AD BC.

� In cyclic quadrilateral ABCD,�A + �B + �C + �D = 360°

(Sum of the angles of a quadrilateralis 360°)

�12�A +

12�B +

12�C +

12�D

= 12

(360°)

� �1 + �2 + �3 + �4 = 180° ...(i)

(� AG, BG, CE and DE are bisectorof �A, �B, �C, �D respectively)

In �ABG, �1 + �2 + �6= 180° ...(ii)(ASP)

In �CDE, �3 + �4 + �5= 180° ...(iii)(ASP)

Adding (ii) and (iii), we get�1 + �2 + �3 + �4 + �5 + �6

= 180° + 180° = 360°� 180° + �5 + �6 = 360°

[From (i)]� �5 + �6 = 180°� �E + �G = 180° ...(iv)

Also,��E + �F + �G + �H = 360°(Sum of the angles of a quadrilateral)

� 180° + �F + �H = 360°� �F + �H = 180° ...(v)From (iv) and (v),Quadrilateral EFGH is a cyclicquadrilateral.

�� Produce QR to meet ABat S. QR meets CD at L.�� In triangles QDL and QBS,

�3 = �4(... QR is the bisector of �Q)

and �QDL = �QBS(... An exterior angle of a cyclic quadri-

lateral is equal to interior opp. angle)� �3 + �QDL = �4 + �QBS� 180° – �DLQ = 180° – �QSB

(... �3 + �QDL + �DLQ = 180°)� �DLQ = �QSB� �PLR = �QSB

[... �PLR = �DLQ,(Vertically opposite angles)]

� �PLR = �QSP = �RSP = �PSR...(i)

51��

In triangles PRL and PRS,��2 = �1

(... PR is the bisector of �P)and �PLR = �PSR [From (i)]On adding, �2 + �PLR = �1 + �PSR� 180° – �PRL = 180° – �PRS

(... �2 + �PLR + �PRL = 180°and �1 + �PSR + �PRS = 180°)

� �PRL = �PRS� �PRL + �PRS = 180°

(SL is a straight line)� 2�PRL = 180°� �PRL = 90°� �PRQ = 90°Hence, �PRQ = �PRS = 90°.

��

�� (B) �AOB = 2 × �ACB(Angle subtended byan arc at the centre ofa circle is double theangle subtended by itat any point on the re-maining part of thecircle)� �AOB = 2 × 50° = 100°Now, x + �AOB = 360°� x = 360° – 100° = 260°.

�� AOC = 180° – 140° = 40°

�ADC = 12�AOC

= 12

× 40°

= 20°� � x = 20°.

�� AOB + �AOC + �BOC

= 360°� 100° + 140° + �BOC

= 360°� �BOC = 120°But �BOC = 2 × �BAC

� �BAC = 120

2�

= 60°.

�� In �ABD,�DAB +�ABD +�ADB = 180°

(Angle sum property)

� 60° + 50° + �ADB = 180°� �ADB = 180° – 110°

= 70°Now, �ACB = �ADB

(Angles in the same segment)� �ACB = 70°.

�� (i) radius (ii) segment

�� AOC = �AOB + �BOC= 60° + 30° = 90°

�ADC = 12�AOC

(Angle subtended by an arc at thecentre is double the angle subtended

by it at any point on the remainingpart of the circle)

= 12

× 90° = 45°.

� � AB = 12 cm� AD = BD = 6 cm

OD = 2 2AO – AD

= 2 210 – 6

= 64 = 8 cm.

52 � � � � � ��

� (i) �� A circleC(O, r) and chord AB= chord AC. AD isbisector of �CAB.

To Prove: Centre O lieson the bisector of �BAC.

Construction: Join BC, meetingbisector AD of �BAC, at M.

Proof: In triangles BAM and CAM,AB = AC [Given]

�BAM = �CAM [Given]

and AM = AM [Common]� �BAM ≅ ��CAM [SAS]� BM = CMand �BMA = �CMA

As ��BMA + �CMA = 180°[Linear pair]� �BMA = �CMA = 90°� AM is the perpendicular bisector ofthe chord BC.� AM passes through the centre O.[... Perpendicular bisector of chord of acircle passes through the centre of thecircle]

Hence, the centre of the park lies onthe angle bisector of �BAC.(ii) Congruence of triangles by SASaxiom (Geometry)(iii) Cleanliness, respect for labour.

�� Draw OM � AB and ON � CD

� AM = 12

AB = 12

× 8 = 4 cm

and CN = 12

× CD = 12

× 6 = 3 cm.

MN = 1 cm (Given)Let OM = x cmIn �OMA,

OA2 = AM2 + OM2

= (4)2 + x2 ...(i)In �ONC,

OC2 = CN2 + ON2

= (3)2 + (x + 1)2 ...(ii)

Now, OA = OC(Radii of the same circle)

� OA2 = OC2

�� 42 + x2 = 32 + (x + 1)2

� 16 + x2 = 9 + x2 + 1 + 2x� 16 – 10 = 2x ��2x = 6�� x = 3 cm.�Radius of the circle

= OA = 2 2OM AM�

= 2 23 4�

= 25 = 5 cm.

!"��

��

�� (B) Let us draw �BDC with D on thecircle other than the arc BAC

�� BDC = 2t

As ABDC is a cyclic quadrilateral,

x +2t = 180°

i.e., 2x + t = 360° ...(i)In �BOC,

y + y + t = 180°(Angle sum property)

2y + t = 180° ...(ii)Subtracting equation (ii) from equation (i),we get

2x – 2y = 180°i.e., x – y = 90°.

�� BCO = �CBO(... OB = OC)= y

�BOC = 2 × �BAC= 2x

Now, in �BOC,�BOC + �BCO + �CBO = 180°

(Angle sum property)� 2x + y + y = 180°

x + y = 90°.

53��

�� In �ABO, AO = BO(Radii of same circle)

� �ABO = �BAO = 60°� �AOB = 180° – 60° – 60°

= 60°� �AOC = 180° – �AOB

= 180° – 60°= 120°

� �ADC = 12

× �AOC

= 12

× 120° = 60°.

�� As �B and �D are the opposite anglesof cyclic quadrilateral ABCD,�B + �D = 180°

� �B = 180° – 140° = 40° ...(i)Let us join AC.Since, �ACB isthe angle in thesemicircle� ACB = 90° ...(ii)Now, in �ABC,

�BAC + �ABC + �ACB = 180°(Angle sum property)

� �BAC + 40° + 90° = 180°[From (i) and (ii)]

� �BAC = 50°.

�� A =��D = 40°(Angles in the same

segment)65° + 40° + x = 180°

(ASP)� x = 75°.

�� (i) 360° (ii) two� (i) False (ii) False (iii) False� In cyclic quadrilateral PQRS,

�R + �P = 180° (Opposite angles)x + 3x = 180°

� 4x = 180°� x = 45°� 3x = 135°Similarly, y = 30°

And 5y = 150°Hence, �P = 135°, �Q = 30°, �R = 45°,

�S = 150°.

!"

We are given a circle. We have to find outits centre.

We will use concept: Perpendicularbisector of chord always passes throughcentre. ...(i)#�� (i) Choose threedistinct points A, Band C on the circle.(ii) Join AB and BC.(iii) Draw the per-pendicular bisectorsof AB and BC.� Perpindicular bisectors of AB andBC both pass through centre using (i)(iv) The perpendicular bisectors drawnin step (iii) intersect each other at O(say). which is the required centre ofthe given circle.

�� y is an exterior angle of �DFB� �y = �1 + �2

(Exterior angle theorem)� �2 = �y – �1 ...(i)Similarly, �3 is an exterior angle of �AEB� �3 = �x + �1 ...(ii)

(Exterior angle theorem)

But �2 = �3 ...(iii)(Angles with the same segment)

Also, �z = 2�2 (Angle subtendedby a chord at the centre of the circle istwice the angle subtended by it at anyother point on the alternate segment)

= �2 + �2= �2 + �3 [From (iii)]

54 � � � � � ��

= (�y – �1) + (�1 + �x)[From (i) and (ii)]

= �y + �xHence, �x + �y = �z.

!"Let the twogiven circleswith centres C1and C2 inter-sect each otherat A and B.Join C1C2, AB,C1A, C2A, C1B and C2B.In �C1AC2 and �C1BC2,

C1A = C1B(Radii of the same circle)


C1C2 = C1C2 (Common)Using SSS congruence rule, we have

�C1AC2 � �C1BC2� �AC1C2 = �BC1C2 ...(i)

(CPCT)and �AC2C1 = �BC2C1 ...(ii)

(CPCT)Now, in �C1AM and �C1BM,


�AC1M = �BC1M [From (i)]C1M = C1M (Common)

� �C1AM ��C1BM(SAS congruence)

Now, by CPCT, we haveAM = BM

and �C1MA = �C1MBAlso, �C1MA + �C1MB = 180°� AM = BMand �C1MA = �C1MB = 90°� C1M is right bisector of chord ABand similarly, C2M is right bisector ofchord AB.� C1C2 is perpendicular bisector of thecommon chord AB.

��

�� (C) �BAD = 180° – 30° – 70° (ASP)= 80°

�BCD + �BAD = 180°(Opposite angles of a cyclic quadrilateralare supplementary )� �BCD = 180° – 80°

= 100°.

�� CD AB� �CDE = �ABE = 75°

(� Corresponding angles)�DCE = �ABE = 75°

(Exterior angle of a cyclic quadrilateralis always equal to its interior

opposite angle)

Now, �CDE + �DCE + �CED= 180° (ASP)

� 75° + 75° + �CED = 180°� �CED = 30°i.e., �AEB = 30°.

�� ACB =12

× �AOB

= 12

× 90°

= 45°

(Angle subtended at the centre byan arc of a circle is double the anglesubtended by the same arc at anypoint on the remaining part of thecircle)

�CAB + �ABC + �BCA = 180°(ASP)

55��

� �CAB + 30° + 45° = 180°� �CAB = 105°� �CAO + �BAO = 105°

In �OAB, OA = OB� �BAO = �ABO� �CAO + 45° = 105°� �CAO = 60°

�� ABC + �ADC = 180°(Sum of opposite angles of

a cyclic quadrilateral)��ABC = 180° – 85° = 95°

Since, �ABC is an exterior

angle of �BCP,

� 40° + �BCP = 95°

� �BCP = 55°

� �DCQ = 55°(Vertical opposite angles)

Further, �CDQ = 180° – �ADC(Linear pair)

= 180° – 85° = 95°Now, in �CDQ,

�CQD + �CDQ + �DCQ = 180°(ASP)

� �CQD + 95° + 55° = 180°� �CQD = 30°

Now, �BCD = 360 – DCQ – BCP

2� � �

= 360 – 55 – 55

2� � �

= 125°

� �BAD = 180° – �BCD= 180° – 125° = 55°.

��

� � � � � � � � �!"��

�� (i) Join OA, OS and OD.

�AOS = �SOD = �AOD = (say)(... AS = SD = AD)

� + + = 360°(Angles at the centre of the circle)

� 3�= 360° � = 120°

AO produced meets SD at M and thecircle at N. Here OM � SD and M ismid-point of SD. Put SM = x.We have �OSN in which �SON = 60°and OS = ON.

�OSN = �ONS = 60°�OSN is equilateral.

M is mid-point of ON.OM = 10 m.

In �OMD, by Pythagoras Theorem,x2 = (20)2 – (10)2 = 300

� x = 10 3 m.

Then SD = 2x = 20 3 m.

Thus, AS = AD = SD = 20 3 m.Hence, the length of the string of each

phone is 20 3 m.(ii) Bisector of an angle made by tworadii at the centre of a circle, bisects thecorresponding chord at right angle.(iii) Inventive and Creative thoughts.

��

��

56 ��

��

CONSTRUCTIONS

��

�� (B)12

× 60° = 30°.

��( )120 – 60 °

60° +2

= 60° + 30° = 90°.

�� Since, sum of any two sides of a triangleis greater than the third side, therefore,in �ABC,

AB + AC > BC.�� To construct triangle ABC, given AB – AC

must be non-zero and non-negative.Therefore, AB > AC.We know that difference of any two sidesof a triangle is less than the third side.Hence, AB – AC < BC, where AB > AC.

�� We will construct an angle of 90° at theinitial point A of the ray AB.

�EAB = 90°

��

�BAD = 30°

� We will construct a 45° angle at theinitial point A of the ray AB.

�� 1. First of all, we construct �BAC = 60°.2. We draw bisector AD of �BAC.

Such that, �BAD = �DAC = 30°.

3. Now, we draw bisector AE of �CAD.Such that,�DAE = �CAE = 15°.

�BAE = �BAD + �DAE= 30° + 15° = 45°.

�BAE is the required angle of 45°. Alternately,

Make an angle of 90° and bisect it toget an angle of 45°.

�� 1. We draw a line segment BC = 4 cm.2. We make an angle CBX of measure

45° at B.

3. We cut a line segment BD = 1.8 cmfrom the ray BX.

57� ��

4. We join DC and draw perpendicularbisector of it, which meets BX at A.

5. We join AC.Thus obtained �ABC is the required triangle.

�� 1. We draw a line segment QR = 6 cm.2. We make an angle RQX of measure

60° at Q.

3. We produce XQ and cut QS = 2 cmfrom it. (PQ < PR)

4. We join RS and draw perpendicularbisector of it, which meets QX at P.

5. We join PR.Thus obtained �PQR is the requiredtriangle.

�� 1. We draw a line segment PQ of length

13 cm.

2. We make �XPQ = 60° and �YQP = 75°at P and Q respectively.

3. We bisect the angles obtained in step 2.Suppose these bisectors intersecteach other at A.

4. We draw perpendicular bisectors ofsegments AP and AQ. Suppose thesebisectors meet PQ at B and Crespectively.

5. We join AB and AC.Thus obtained �ABC is the requiredtriangle.

58 ��

�� We will construct an angle of 105° atthe initial point A of the ray AB.

�BAC = 105°��

Line PMQ is the required right bisector.

� Let the given side is of length 2.7 cm.�� 1. Draw BC = 2.7 cm

2. Draw BX��

such that �CBX = 60°.

3. Draw CY��

such that �BCY = 60°.

4. Let BX��

and CY��

intersect at A.5. �ABC is required triangle.

Using Angle sum property, we have�A + �B + �C = 180°

� �A + 60° + 60° = 180°� �A = 60°Hence, the �ABC is equilateral triangle.

��

�� (D) �XYZ = 90°� �XYL + �LYZ = 90°

� �XYL = 902�

= 45° ...(i)

(� �LYZ = �XYL)Since, �LYZ is bisected by YM

� �LYM = 12

�LYZ

= 12

× 45°

= 1

222� ...(ii)

Adding equations (i) and (ii), we get

�XYL + �LYM = 45° + 1

222�

� � �XYM = �167

2.

�� As YP = PR, �XYR = 60° But PQ = QR

� �XYZ = 12�XYR

= 12

× 60°

= 30°.

�� Given, BD = AB + AC� AB + AD = AB + AC� AD = AC.

�� In the figure ofconstruction of�PQR, PL isthe perpen-dicular bi-sector ofSR.

Therefore, any point of PL is equidistantfrom S and R.� PS = PR.

59� ��

��

1. We draw a line segment BC = 7 cm.2. We make an angle of measure 75° at

the point B such that �CBX = 75°.

3. We cut BD = 13 cm from BX��

.4. We join CD and draw right bisector

of it, which intersect BD at A.5. We join AC.Thus obtained �ABC is the requiredtriangle.

�� 1. We draw a line segment GH = 11 cm.

2. We draw angles of 30° and 90° atthe points G and H respectively. Thenwe draw bisectors of these angles.

3. These bisectors obtained in step 2intersect each other at X.

4. We draw right bisectors of GX andHX, which meet GH at Y and Zrespectively.

5. We join XY and XZ.Such obtained �XYZ is the requiredtriangle.

60 ��

��

�� (C) �BAX = 60°

As AY��

is the bisector of �LM,�BAY = 90°�XAY = �BAY – �BAX

= 90° – 60° = 30°

As AC��

is the bisector of �XAY,

�CAX = 12�XAY

= 12

× 30°

= 15°Now, �BAC= �BAX + �CAX

= 60° + 15° = 75°.�� BAY = 120°

�XAY = 60°

�ZAY = 12�XAY

(� AZ bisects �XAY)

= 12

× 60° = 30°

Now, �CAY = 12�ZAY

(� �CAZ = �CAY)

= 12

× 30° = 15°.

Hence, �BAC= �BAY + �CAY= 120° + 15° = 135°.

�� As BM = LM = CY = XY = R,�CBA = �BCA = 60°

Now, in �ABC,�A + �B + �C = 180° (ASP)

� �A + 60° + 60° = 180°� �A = 60°

Therefore, AB = BC = CA� �ABC is an equilateral triangle.

�� Y is point on the perpendicular bisector of GX� GY= XY (Any point on the perpen-

dicular bisector of a linesegment is equidistant

from its end points)� �YGX = �YXG = 15°

(Angles opposite to equalsides of a triangle are equal)

�XYZ =�YGX + �YXG(Exterior angle of a triangle is equalto the sum of its interior oppositeangles)

= 15° + 15° = 30°.

61� ��

Similarly, Z is a point on the perpendi-cular bisector of XH.Thus, ZX = ZH� �ZXH = �ZHX = 45°� �XZY = �ZXH + �ZHX = 90°.

�� We will construct an angle of 22 12

° at

the initial point A of the ray AB.

�BAC = 22 12

°.

�� We will construct an angle of 135° atthe initial point A of the ray AB.

�BAC = 135°

� ��

1. We draw a line segment AB of length5.8 cm.

2. We make �ABX = 60° at B.3. We cut BD = 7 cm from BX.4. We join AD and draw perpendicular

bisector of it, which meets BD at C.5. We join AC.Thus obtained �ABC is the requiredtriangle.

�� 1. We draw a line segment QR = 6 cm.2. We make an �SQR = 60° at Q.

3. We produce SQ and cut QT = 2 cmfrom it.

4. We join TR and draw perpendi-cular bisector of it, which meetsQS at P.

5. We join PR.Thus obtained �PQR is the requiredtriangle.

�� 1. We draw a line segment GH of length

13 cm.2. We make �XGH = 60° and �YHG = 75°

at G and H respectively.3. We bisect the angles obtained in step

2. These bisectors intersect each otherat A.

62 ��

4. We draw perpendicular bisectors of GA and HA, which meet GH in B and Crespectively.

5. We join AB and AC.

Thus obtained �ABC is the required triangle.

��

63��

��

SURFACE AREAS AND VOLUMES

��

�� (D) Let the length of the side of the cube= a cm

Volume of the cube= Volume of the sheet

� a3 = 27 × 8 × 1� a = 3 × 2 × 1

= 6 cmSurface area of the sheet

= 2 (27 × 8 + 8 × 1 + 1 × 27)= 2 × 251= 502 cm2

Surface area of the cube= 6a2

= 6 × 62

= 216 cm2

Now, difference of the two surfaces= 502 – 216 = 286 cm2.

�� Let a be the length of the side of thecube. Then area of one face = a2

and area of all six faces = 6a2

� Required ratio = a2 : 6a2 = 1 : 6.

�� We know that the longest rod that canbe fitted in a cube is equal to itsdiagonal.� Required length of the rod

= 4 3 m = 4 × 1.73 m

= 6.92 m.

�� Let the actual side of the cube = a� Initial surface area of the cube = 6a2

After increment, the side of the cube

= a ×150100

= 1.5a

After increment, the surface area of thecube = 6 × (1.5a)2

= 13.5a2

Now, percentage increase in the surfacearea

=2 2

213.5 – 6

6a a

a× 100%

=2

27.56

aa

× 100% = 125%

�� Edge of the cube = 6.5 cm

Length of the diagonal = 3 edge.

= 3 × 6.5 cm

= 1.7 × 6.5 cm

(Given: 3 = 1.7)

= 11.05 cm.

�� Obviously the longest rod can be placedalong either of the diagonal of the room.� Length of the longest rod

= l b h2 2 2+ + = 2 2 2(12) (9) (8)+ + m

= 144 81 64+ + m = 289 m = 17 m.

� (i) l = 1.5 m, b = 1.25 m,

h = 65 cm = 65

100m = 0.65 m.

Area of the sheet required for makingthe box which is open at the top

= l × b + 2 (l + b)h= 1.5 × 1.25 + 2(1.5 + 1.25) × 0.65 m2

= (1.875 + 3.575) m2

= 5.45 m2

(ii) Cost = � 20 × 5.45 = � 109.

�Let height of the hall be h metres.

Area of the 4 walls= 2(l + b) × h= Perimeter of floor × h[As perimeter = 2(l + b)]= 250 × h m2

64 � � � � � � ��

� Cost of painting on 1 m2 = � 10� Cost of painting on 250 × h m2

= � (250 × h) × 10But given total cost of painting

= � 15000.� (250 × h) × 10 = 15000� 250 × h = 1500

� h = 1500250

� h = 6 m.

�� The length, breadth and height ofbigger box are 25 cm, 20 cm and 5 cmrespectively.� Surface area of a bigger box = 2( lb + bh + hl)

= 2(25 × 20 + 20 × 5 + 5 × 25)= 2(500 + 100 + 125) = 2 × 725= 1450 cm2

� Area of the cardboard required for250 bigger boxes

= 250 × 1450 = 362500 cm2

The dimensions of the smaller box are15 cm, 12 cm and 5 cm respectively.Surface area of a smaller box

= 2 (lb + bh + hl) = 2 (15 × 12 + 12 × 5 + 5 × 15) cm2

= 2 (180 + 60 + 75) = 630 cm2

� Area of the cardboard required for 250smaller boxes = 250 × 630 = 157500 cm2

Total area of the cardboard required= 362500 + 157500 = 520000 cm2

Cardboard (5%) required for overlapping

= ×520000 5

100 = 26000 cm2

Total area of the sheet includingoverlapping

= 520000 + 26000 = 546000 cm2

Cost of cardboard at the rate of � 4 for1000 cm2

= � 546000 4

1000×

= � 2184.

� We are given that, the length and thebreadth of the hall as 20 m and 16 mrespectively.� l = 20 m and b = 16 mLet the height of the hall be h m.Now, the sum of the areas of the floorand the flat roof of the hall = 2(l × b)

= 2(20 × 16) m2 = 640 m2

The sum of the areas of the four wallsof the hall

= 2(l + b) × h = 2(20 + 16) × h m2

= 72h m2

It is given that, 72h = 640

� h = 64072

m = 809

m = 8.888 m.

So, the height of the hall is about 8.89 m.

��

�� (A) A closed right circular cylinder hastwo circular surfaces and one curvedsurface.

�� r =52

m, h = 14 m

Curved surface area = 2 �rh

= 2 × 227

×52

× 14

= 220 m2

Now, cost of white washing= 220 × 50 paise= � 110.

�� Let the length of the cloth = a mArea of the cloth = a × 5Slant height of the cone,

l = 2 2r h�

= 2 27 24�

= 25 mCurved surface area = �rl

= 227

× 7 × 25

= 22 × 25

65��

According to the given condition,a × 5 = 22 × 25

� a =22 25

5�

= 110 m.

�� Radius = r (say)

Slant height = l (say)Curved surface area = �rl

��Radius = r (say)

Slant height = l + l ×100

p

= l100

100p ��

� ��

Curved surface area = �rl100

100p��

� ��

Hence, percentage increase in thecurved surface area

=

100–

100p

rl rl

rl

��

× 100%

=

⎛ ⎞⎡ ⎤π ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦π

prl

rl

+100–1

100×100%

= 100– 1

100p��

� �� × 100%

= ��

� �� 100 – 100100

p× 100%

= p %.

�� r = 0.7 m S.A. = 4.4 m2

� 2�r × h = 4.4

� 2 ×227

×7

10h =

4410

� h = 4410

×7 10

2 22 7�

� �

= 1 m.

�� h = 1 m, diameter (d) = 140 cm

� r = 2d

= 140

2 = 70 cm

= 70

100m = 0.7 m

Sheet required = 2�r (r + h)

= 2 ×227

× (0.7)

× (0.7 + 1) m2

= 4.4 × 1.7 m2 = 7.48 m2.

� The radius of the cylindrical blockof wood = r = 70 cm and the length= h = 200 cm.

The total surface area of the cylindricalblock

= 2�r(r + h)

= 2 ×227

× 70 × (70 + 200) cm2

= 440 × 270 cm2

= 118800 cm2

� Cost of painting on 100 cm2

= � 1.25 = � 54


= �5

400


= � 5

400 × 118800

= � 54

× 1188 = � 1485.

Hence, the cost of painting the blockof wood is � 1485.

�

(i) Diameter, d = 3.5 m =3510

m

� r = 2d

= 3520

=74

m, h = 10 m.

66 � � � � � � ��

Inner curved surface area of the well= 2�rh

= 2 ×227

×74

× 10 m2

= 110 m2.

(ii) Cost of plastering = � 40 × 110 = � 4400.

�� h = 120 cm, Diameter = d = 84 cm

� r = 2d =

842

= 42 cm.

Area leveled in 1 revolution= C.S.A. of the roller= 2�rh= 2� × 42 × 120 cm2

= � × 84 × 120 cm2

Area levelled in 500 revolutions= 500 × Area levelled in one

revolution= 500 × � × 84 × 120

= 500 ×227

× 84 × 120 cm2

=500 22 1440

10000� �

m2

= 1584 m2.

�

r = 3 cm, h = 10.5 cmCardboard required for one penholder

= �r2 + 2�rh

= �r × (r + 2h)

=227

× 3 × (3 + 2 × 10.5)

=227

× 3 × (3 + 21)

=227

× 3 × 24 cm2

Cardboard required for 35 competitors

=227

× 3 × 24 × 35 cm2

= 7920 cm2.

� (i) Diameter (d) = 4.2 m = 4210

m

� r = d2

=42

2 10×m =

2110

m, h =4510

m

The lateral surface area = 2�rh

= 2 × 227

× 2110

× 4510

m2

= 59.4 m2.(ii) Total surface area = 2�r2 + 2�rh

= 2 × 227

× 2110

× 2110

m2 + 59.4 m2

= 27.72 m2 + 59.4 m2 = 87.12 m2

Let the required area of the steelactually used be x m2

Now, x – x × 112

= 87.12

�1112

x = 87.12

� x = 87.12 × 1211

= 95.04 m2

Hence, the steel actually used was95.04 m2.

��

�� (B) r = 3 cm, h = 11 cmTotal surface area = 2�r (h + r)

= 2 ×227

× 3 (11 + 3)

= 2 ×227

× 3 × 14

= 264 cm2.�� r = 5 cm, l = 13 cm

Total surface area of the cone= �r (l + r)= � × 5 × (13 + 5)= 90� cm2.

�� Radius (R) = 2r

, slant height (L) = 2 l

Total surface area =��R (L + R)

67��

= � ×2r

× ⎛ ⎞⎜ ⎟⎝ ⎠2 +

2r

l

= �r4r

l� �� .

�� Let the required length = x mArea of floor = 346.5 m2

� �r2 = 346.5

� r = 346.57

22� = 10.5 m

Slant height, l = 2 2r h�

= 2 2(10.5) 14� = 17.5 m

Area of canvas required= lateral surface area of the conical tent + area of floor

� x × 1.2 = �rl + 346.5

� x × 1.2 = 227

× 10.5 × 17.5 + 346.5

� x × 1.2 = 577.5 + 346.5

� x = 9241.2

= 770 m.

�� Slant height of the right circular cone,l = 25 cm.

Radius of the base of the cone,r = 7 cm.

Curved surface of a right circular cone= �rl

= 227

× 7 × 25 cm2

= 550 cm2.

�� Let radius of the base = r mHeight = 16 m

Circumference of the base = 33 m� 2�r = 33

�� r = 332�

= 33

2 22 / 7�

= 33 744� =

23144

=214

m

l = 2 2h r� = 2

2 2116

4� ��

= 441256

16� = 4096 441

16�

= 453716

= 283.5625

= 16.84 m (approx.)

Surface area of the cone= � r (r + l)

= 227

× 214

⎛ ⎞⎜ ⎟⎝ ⎠

21+ 16.84

4

= 332

× 22.09

= 364.49 m2 (approx.).

� r = 6 m , h = 8 m,

l = 2 2r h� = 2 26 8�

= 36 64� = 100 = 10 m

C.S.A. = �rl = 3.14 × 6 × 10 m2

= 188.4 m2

Area of tarpaulin required= C.S.A. = 188.4 m2

Now, length of tarpaulin required to

make tent =188.4

3= 62.8 m.

Total length of tarpaulin requiredincluding margins and wastage

= 62.8 + 0.20 = 63 m.

�� l = 25 m, r = 7 mCurved surface = �rl

= 227

× 7 × 25 m2

= 550 m2

� Cost of white washing on 100 m2

= � 210�� Cost of white washing on 1 m2

= 210

100�

68 � � � � � � ��

� Cost of white washing on 550 m2

= �210100

× 550

= � 1155.

� r = 7 cm, h = 24 cm� l2 = h2 + r2 = (24)2 + 72

= 576 + 49 = 625� l = 625

� l = 25 cm.Sheet required for one cap

=227

× 7 × 25 cm2 = 550 cm2

Sheet required for 10 such caps= 10 × 550 cm2 = 5500 cm2

� r = 20 cm = 20

100m = 0.2 m and h = 1 m

l2 = h2 + r2 = 1 + 0.04 = 1.04

� l = 1.04 = 1.02 m

Curved surface area of 50 cones= 50 × �rl= 50 × (3.14 × 0.2 × 1.02) m2

= 32.028 m2

Total cost on painting = � 12 × 32.028= � 384.34 (approximately.)

��

�� (D)Volume of 27 balls = Volume of thebig sphere

� 27 ×43�r3 =

43

�R3

� R = 3rThe ratio of surface area (S1) of the bigsphere to the surface area (S2) of a ballis given by

π= = =π

2 21

2 22

S 4 R (3 ) 9S 14

rr r

i.e., S1 : S2 = 9 : 1.

�� Let R1 = r, R2 = 2r

V1 =43

�r3, V2 =43� (2r)3 =

323

�r3

Now, percentage increase in the volume

= 2 1

1

V – V×100%

V

= π π

π

3 3

3

32 4–

3 343

r r

r × 100% = 700%.

�� When radius R1 = 6 cm,surface area S1 = 4�R1

2 = 4�(6)2.When radius R2 = 12 cm,surface area S2 = 4� R2

2 = 4� (12)2.

Now, 1

2

SS

= 4 × 6 × 6

4 ×12 ×12�

�

� S1 : S2 = 1 : 4.

�� Let the initial radius = r. Then new

radius = r + r ×100

p

= r 100

100p��

� ��

Initial surface area, S1 = 4�r2

New surface area, S2 = 4�r2 2100

100p��

� ��

Now, percentage increase in the surfacearea

= 1

2 1S – SS

× 100%

=

+⎛ ⎞π π⎜ ⎟⎝ ⎠π

22 2

2

1004 – 4

1004

pr r

r× 100%

= ⎡ ⎤+⎛ ⎞⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

2100– 1

100p

× 100%

= ⎛ ⎞+ +⎜ ⎟⎜ ⎟⎝ ⎠

210000 200– 100 %

100p p

= 2200

100p p�

% = ⎛ ⎞

+⎜ ⎟⎝ ⎠

2

2 %100p

p .

69��

�� Here, r = 10 cm.Total surface area of hemisphere = 3�r2

= 3 × (3.14)(10)2 = 942 cm2.

�� Let r be the radius of the sphere. Then,4�r2 = 154

� 4 × 227

× r2 = 154

� r2 = 154 74 22

�

�

= 7 7

4�

= 7 72 2�

�

� r = 72

= 3.5 cm.

� (i) Radius of the sphere= rThen surface area of the

sphere = 4�r2. (ii) Radius of the sphere = r

Height of the cylinder = 2rCurved surface area of the cylinder

= 2� × (R) × (H)= 2� × r × 2r= 4�r2.

(iii) Required ratio = 2

2

44

r

r

�

�

= 11

, i.e., the

ratio is 1 : 1.

�� Here d = 10.5 cm

� r = 2d

= 10.52

= 5.25 cm

Inner surface area of the bowl

= 2 �r2

= 2 ×227

× 5.25 × 5.25 cm2

Cost of tin plating

= � 16

100× 2 ×

227

×525100

×525100

= � 2 22 63100

� � = � 27.72.

�

Inner radius, r = 5 cmThickness of steel sheet

= 0.25 cmOuter radius, R = 5 cm + 0.25 cm

= 5.25 cm =525100

cm

Outer curved surface area of the bowl= 2�R2

= 2 ×227

× 525100

× 525100

cm2

= 173.25 cm2.

� Let the diameter of the moon = 2r(r is the radius)

and let the diameter of the earth = 2R(R is the radius)

Given that, 2r =14

× 2R, i.e., R = 4r

Ratio of the surface areas of the moonand earth = 2 24 : 4 Rr� �

= r2 : R2 = r2 : (4r)2 = 1 : 16

Thus, the required ratio is 1 : 16.

��

�� (B) Let the number of planks be xVolume of x planks = volume of the pit

� x × 4 ×50

100× 20 = 40 × 12 × 160

� x × 40 = 40 × 12 × 160� x = 1920.

�� Let the side of the cube = a m� Lateral surface area = 4a2

� 256 = 4a2

� a2 = 64� a = � 8 m�� Side length cannot be negative.� a = 8 m

Now, volume a3 = 83 = 512 m3.�� Let the side of the cube be a cm, then

6a2 = 96 � a2 = 16 � a = � 4 cmBut side length cannot be negative.� a = 4Now, volume = a3 = 43 = 64 cm3.

�� Inner dimensions are:l1 = 115 cm, b1 = 75 cm, h1 = 35 cm

70 � � � � � � ��

Outer dimensions are:l2 = 115 + 2.5 + 2.5 = 120 cmb2 = 75 + 2.5 + 2.5 = 80 cmh2 = 35 + 2.5 + 2.5 = 40 cm

Now, volume of the wood= Outer volume – Inner volume= l2 b2 h2 – l1 b1 h1= 120 × 80 × 40 – 115 × 75 × 35= 384000 – 301875= 82125 cm3.

�� Internal volume of the pit

= 8 × 6 × 3 m3

Cost of digging it at the rate of� 30 per m3

= � 8 × 6 × 3 × 30 = � 4320.

�� Maximum number of wooden crates

= Capacity ( volume)of the GodownVolume of one wooden crate

i.e.

=60 × 25×10

1.5×1.25× 0.5 =

150003 5 1

× ×2 4 2

= 15000 ×16

15 = 16000.

� Let the side of the new cube be a cm.Volume of 8 new cubes

= Volume of the original cube� 8 × (a)3 = (12)3

� a3 =12 ×12 ×12

8� a3 = 6 × 6 × 6� a = 6Hence, the length of the side of thenew cube is 6 cm.

Required ratio = 2

26(12)6(6)

=41

, i.e., 4 : 1.

�� Length of water flowing in 1 hour= 2 km = 2000 m

Length of water flowing in 1 min.

= 200060

m

Volume of water flowing in 1 minute

=200060

× 40 × 3 m3 = 4000 m3.

�

Capacity of tank = 50000 litres

=500001000

m3 = 50 m3

Breadth of the tank=Volume

×l h

=50

2.5 × 10= 2.

Hence, breadth is 2 m.

� Let the edges of the cubes be 3a cm,4a cm and 5a cm.

� (3a)3 + (4a)3 + (5a)3 = 3

12 3

3

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

(... Diagonal of a cube = 3 edge )

� 27a3 + 64a3 + 125a3 =12 × 12 × 12� (27 + 64 + 125)a3 = 1728

� a3 = 1728216

= 8

� a = 13(8) = 2.

Hence, edges of the three cubes are3 × 2 cm, 4 × 2 cm and 5 × 2 cm, i.e.,6 cm, 8 cm and 10 cm.

��

�� (A) r = 3.5 cm, h = 10 cmVolume, V = �r2h

=227

× 3.5 × 3.5 × 10

= 385 cm3.

�� Capacity of the vessel

= Inner volume of the cylinder

= �r2h =227

× 7 × 7 × 30

= 4620 cm3 = 46201000

l = 4.62 l.

71��

�� 1

2

rr =

23

, 1

2

hh

= 53

�1

2

VV

= 2

1 12

2 2

r h

r h

�

�

= ⎛ ⎞⎜ ⎟⎝ ⎠

21

2

rr ×

⎛ ⎞⎜ ⎟⎝ ⎠

1

2

hh

=22

3� ��

× 53

= 2027

i.e., V1 : V2 = 20 : 27.

�� Let h1 = h, h2 = 3h

Given that V1 = V2

� �r12 h1 = �r2

2 h2

� r12 × h = r2

2 × 3h

�

21

22

rr =

31

i.e., r1 : r2 = 3 : 1.

�� Diameter (d) = 7 cm

� r = 72

cm, h = 4 cm

Volume of soup filled in one bowl

= 227

×27

2� �� × 4 cm3

= 227

×494

× 4 cm3

= 154 cm3

Soup required for 250 patients= 250 × 154 cm2

= 38500 cm3 = 385001000

litres

= 38.5 litres.

�� r (inner radius) = 12 cm,R (outer radius) = 14 cm,

h = 35 cmVolume of wood = �(R2 – r2) × h

=227

(142 – 122) × 35

= 227

(14 – 12) × (14 + 12) × 35

= 22 × 2 × 26 × 5

= 220 × 26 = 5720 cm3

Mass = 5720 × 0.6 g = 3432 g= 3.432 kg.

� (i) h = 10 m Inner curved surface area (in m2)

= 2Total cost

Cost per m =

220020

m2

= 110 m2.

(ii) 2� × r × h = 110

� 2 ×227

× r × 10 = 110

� r =74

= 1.75 m.

(iii) Capacity = �r2h

=227

×27

4� ��

× 10 m3

= 96.25 m3 = 96.25 kl.

�� (i) Let us first consider horizontal partof the article.

l1 = length = 22 cm,

b1 = breadth = 8 + 3 = 11 cmh1 = height = 3 cm

Volume of horizontal part = l1b1h1

= 22 × 11 × 3 = 726 cm3

Now, we will consider the vertical part.

l2 = length = 22 cm

b2 = breadth = 3 cm

h2 = height = 5 cm

� Volume of vertical part (segment)

= l2b2h2 = 22 × 3 × 5 = 330 cm3

Thus volume of the article

= 726 + 330 = 1056 cm3.

(ii) Volume of a cuboid.

(iii) Independent, Constructive, Dedica-tion, Hardworking, Understanding.

� Diameter of pencil = 7 mm

� r1 = 72

mm = 7

20cm

72 � � � � � � ��

Volume of the pencil

= � ×27

20� ��

× 14 cm3

Diameter of graphite= 1 mm

� r2 =12

mm = 1

20 cm.

Volume of the graphite

= � ×21

20� ��

× 14 cm3

=227

×1

400× 14 cm3

=44400

cm3 = 0.11 cm3

Volume of wood= Volume of pencil – Volume of graphite

=2 27 1

14 – 1420 20

� �� cm3

= ��× 14 49 1

–400 400

� ��

cm3

= 227

× 14 49 – 1400

� ��

cm3

= 44 ×48

400 cm3 = 5.28 cm3.

��

�� (B) r = 82

= 4 cm, l = 5 cm,

h = 2 2–l r = 25 – 16 = 3 cm

Volume =13�r2h =

13� × 42 × 3

= 16 � cm3.

�� Let radius of the cone = rLet initial height (h1) = hThen new height (h2) = 2h

Initially, the volume (V1) of the cone

= 13�r2h1 =

13�r2h

The new volume (V2) of the cone

= 13�r2h2 =

13�r2 × (2h)

= 23�r2h

Percentage increase in the volume

= 2 1

1

V – VV

× 100%

=

2 2

2

2 1–

3 313

r h r h

r h

� �

�

× 100%

= 100%.

�� After revolution, we obtain a shape ofcone.For such cone,

r = 6 cmand h = 10 cm

� Volume = 13�r2h

= 13� × 36 × 10

= 120 � cm3.

�� Volume of the conical glass

=13�r2h

=13

×227

(3.5)2 × 10

=385

3cm3

Required quantity= 30 × Volume of the glass

= 30 ×3853

= 3850 cm3

= 3.85 l.

�� Diameter, d = 3.5 m = 3510

m = 72

m

� r = 12

× d = 74

m, h = 12 m

73��

Now, capacity = 13

×227

×27

4� ��

× 12

= 13

×227

× 74

×74

×121

= 772

m3 = 38.5 m3

= 38.5 kl.

�� Solid obtained is a cone withr = 5 cm, h = 12 cm,l = 13 cm

Volume =13�r2h

=13

(�) × (5)2 × 12 cm2

= 100 � cm3.

� h = 9 cm

Volume = 48� cm3

�13�r2 × h = 48�

or13

r2 × 9 = 48 � r2 = 483

= 16

� r = � 4 but r = 4 as r can't be –ve� Diameter = 2 r = 2 × 4 = 8 cm.

�

In this case, r = 12 cm, h = 5 cm, l = 13 cm.

Volume = 13� × (12)2 × 5

=13� × 144 × 5

= � × 48 × 5 cm3

= 240� cm3

Required ratio = 100� : 240�,= 5 : 12.

�� Diameter d = 10.5 m

� r = 2d

= 5.25 m

h = 3 mVolume of wheat = Volume of cone

=13

�r2h

=13

× 227

× 5.25 × 5.25 × 3 m3

=227

×214

×214

m3

= 86.625 m3.Area of canvas = �rl

= 227

×214

× 2

221(3)

4� � ��

m2

= 332

× 6.046 m2 = 33 × 3.023 m2

= 99.759 m2

� (i) Diameter = 28 cm� r = 14 cm

Volume = 9856 cm3,

�13�r2h = 9856

�13

× 227

× (14)2 × h = 9856

� h = 9856 3 722 14 14

� �

� �

� h = 48 cm.

(ii) l = 2 2r h� = 2 2(14) (48)�

= 196 2304� = 2500 = 50 cm.

(iii) Curved surface area = �rl

= 227

× 14 × 50

= 22 × 2 × 50= 2200 cm2.

��

�� (A) Volume =43�r3

=43

×227

× 3.5 × 3.5 × 3.5

Density = 0.9 g /cm3

Mass = Volume × Density

74 � � � � � � ��

= 4 223.5 3.5 3.5

3 7� ��

× (0.9)= 161.7 g.

�� Volume of the sphere = 34(radius)

3�

= 34(2 )

3r�

= 323��r3.

�� Let radius of the hemisphere = r cmVolume of the hemisphere = 19404 cm3

�23�r3 = 19404

�23

×227

× r3 = 19404

� r3 = 19404 3 7

2 22� �

�

� r3 = 9261� r = 21 cmNow, total surface area of the hemisphere

= 3�r2

= 3 ×227

× 21 × 21

= 4158 cm2.�� r = 2.1 cm

Volume of the hemispherical bowl

=23�r3

=23

×227

× (2.1)3

= 19.404 cm3

Total capacity of 20 bowls= 20 × Volume of the bowl= 20 × 19.404 = 388.08 cm3.� 388 cm3 approximately.

�� Radius of the hemisphere

= 72

cm = 3.5 cm = 3510

cm

Volume of the hemisphere =23�r3

= 23

×227

×3510

×3510

×3510

= 89.8 cm3.�� Diameter, d = 4.2 cm

� r = 2d

= 4.22

= 2.1 cm

Volume of the metallic ball

= 43�r3 =

43

×227

× (2.1)3 cm3

= 43

× 227

× 2.1 × 2.1 × 2.1

= 88 4411000� cm3 = 38.808 cm3

Mass of the ball = 38.808 × 8.9 g

= 345.39 g (approx.)

� Let r be the radius of the moon and Rbe the radius of the earth.

Diameter of the moon

= 14

(diameter of the earth)

(Given)

� 2r = 14

(2R)

(� Diameter = 2 × radius)

� 2r = 12

R � R = 4r

Volume of the moonVolume of the earth

=3

3

43

4(R)

3

r�

�

=3

3(R)r

=3

3(4 )rr

= 3

364r

r

=164

.

� Volume of the moon =164

volume of

the earth.

Therefore, the required fraction is 164

.

�� For the cone, we haver = 2 cm and h = 8 cm

75��

� Volume of the cone = 13�r2h

= 13� (2)2 8 =

323�

cm3

Let the radius of the sphere be R

Then volume of the sphere =43�R3.

According to given condition,

323�

cm3 = 43�R3

� R3 = 324

� R3 = 23 � R = 2� Diameter of the sphere = 2R

= 2 × 2 = 4 cm.

� (i) Radius (r) = 72

,�Height (h) = 8 cm

Volume of glass of type C = �r2h

= 227

�× 72

× 72

× 8 = 308 cm3

(ii) Volume of hemisphere

= ��

7 7 7 223 2 2 2 7

= 89.83 cm3

� Volume of glass of type A = 308 – 89.83 = 218.17 cm3

Volume of cone

= � � �1 22 7 73 7 2 2

× 8

= 12.83 cm3

� Volume of glass of type B = 308 – 12.83 = 295.17 cm3

Thus, the glass of type A hasminimum capacity.

(iii) Volume of solid figures (Mensura-tion)

(iv) Honesty.

��

�� (C) Radius of sphere (r1) = 62

= 3 cm

Radius of wire (r2) = 22

= 1 mm

= 0.1 cmLet the length of the wire = h.

Volume of the wire = Volume of thesphere

� � r22 h =

43�r1

3

� (0.1)2 h = 43

(3)3

� h = 4 90.01�

= 3600 cm

� h = 36 m.

�� 2�r2 = 2772

� ��r2 = 1386Total surface area

= Curved surface area + Area of the cross-section= 2�r2 + �r2

= 2772 + 1386 = 4158 cm2.�� Let the given radius and height be r

and h respectively.

Volume (cylinder)Volume (cone)

= 2

213

r h

r h

�

�

= 31

Thus, Volume (cylinder) : Volume (cone)= 3 : 1.

�� Let the required height be h cm.

Here, 1

2

rr =

1h

� r2 = hr1 ...(i)

Also, 1

Rr =

21

� R = 2r1 ...(ii)

Volume of water in upper cone+ Volume of water in lower cone= Volume of either cone

76 � � � � � � ��

� r2 = 15400 722 100

�

�

� r2 = 7 × 7 ��r = 7 cm

Area of the metal sheet

= Total surface area of thecylinder

= 2�r(r + h)

= 2 × 227

× 7 × (7 + 100) cm2

= 44 × 107 cm2 = 4708 cm2

= 4708

10000 m2 = 0.4708 m2.

�� Volume of the earth to be dug out

= volume of the well

= 227

72

72

22 5× × ×��

��. m3

= 866.25 m3

Area of the inner curved surface = 2�rh

= 2227

72

22 5� � ��

��. m2= 495 m2

� Cost of plastering the inner curvedsurface = � (495 � 3) = � 1485.The quality adopted by the farmer Itwariis a social work.

� Radius of the hemispherical bowl,

R = 9 cm.Radius of each cylindrical bottle,

r = 32

cm.

Height of each cylindrical bottle, h = 4 cm.

Volume of water = Volume of the hemi-spherical bowl

= 23�R3

= 23� (9)3 cm3

= 23� × 9 × 9 × 9 cm3

= 486 ��cm3

�13�r1

2 × 1 + 13�r2

2 × h

=13�R2 × 2

� r12 + r2

2 h = 2R2

� r12 + h2 r1

2 × h = 8r12

[From (i) and (ii)]

� 1 + h3 = 8

� h3 = 7

� h = 3 7 cm.

�� Here, d = 21 cm � r = 212

cm

Surface area of the sphere

= 4�r2

= 4 ×227

× 221

2� ��

= 4 ×227

× 212

×212

= 1386 cm2.

�� Diameter of spherical ball = 21 cm.

� Radius of spherical ball, r = 212

cm.

Volume of ball = 43�r3

= 43

×227

× 212

×212

× 212

= 4851 cm3. Side of cube = 1 cm.� Volume of the cube = 1 × 1 × 1 cm3

= 1 cm3.

Now, number of cubes = 4851

1= 4851.

� h = 1 m = 100 cm.Capacity = 15.4 litres

= 15.4 × 1000 cm3

= 15400 cm3

� �r2h = 15400

�227

× r2 × 100 = 15400

77��

Volume of a cylindrical bottle

= �r2h = �32

⎛ ⎞⎜ ⎟⎝ ⎠

32

⎛ ⎞⎜ ⎟⎝ ⎠

× 4

= 9��cm3

� Number of bottles

= Volume of hemispherical bowlVolume of a cylindrical bottle

= 4869

ππ

= 54.

��Let the radius of the base of ice cream cone

= x cm.Then, height of the conical ice cream

= 2 × diameter = 2 (2x) = 4x cm.Volume of ice cream in 1 cone

= Vol. of conical portion + Vol. of hemispherical portion

= 13�x2 (4x) +

23�x3

= 34

3x� +

23�x3

= 36

3x� = 2�x3 cm3

Diameter of cylindrical container

= 12 cm

Its height h = 15 cm.

� Volume of cylindrical container

= �r2h = �(6)2 15 = 540 �

Number of children

= Volume of cylindrical container

Volume of ice cream cone in 1 cone

� 10 = 3

5402 x

�

�

� 10x3 = 270

� x3 = 27010

= 27

� x3 = 27 � x3 = 33

� x = 3� Diameter of ice cream cone

= 2x = 2(3) = 6 cm.

��

�� (B) Area of the base of the prism= area of a right triangle

=12

× 8 × 15 = 60 cm2

Volume of the prism= area of the base × height= 60 × 20 = 1200 cm3

�� Let the required number of cubes = xVolume of x small cubes

= Volume of the large cube � x × 53 = 203

� x =20 × 20 × 20

5×5×5 = 64.

�� Number of bullets

=Volume of rectangular solid

Volume of one bullet

= 3

9 11 12

0.643 2

� �

� ��

=9 11 12 3 7 2 2 2

4 22 0.6 0.6 0.6� � � � � � �

� � � �

= 10500.

�� Let the height of the cylinder = HLet the common radius = RVolume of the whole solid

= 3 × Volume of the cone

�

13� R2 h + �R2 H = 3 ×

13�R2 h

� �R2 H =23�R2 h

� H =23h

Hence, the height of the cylinder is 23h

.

� r = 0.7 mC.S.A. = 4.4 m2

78 � � � � � � ��

� 2�r × h = 4.4

� 2 ×227

×7

10× h =

4410

�

4410

× h = 4410

� h = 1 m.

� ��

�� Internal radius (r)of the vessel

= 242

= 12 cm

External radius (R) of the vessel

= 252

= 12.5 cm

� The surface area of the vessel= Outer curved surface area+ Inner C.S.A + Area of ring

= 2�r2 + 2�R2 + �(R2 – r2)= 2�(r2 + R2) + �(R + r)(R – r)= 2�(122 + (12.5)2) + �(12.5

+ 12)(12.5 – 12)= 2�(144 + 156.25) + �(24.5) × (0.5)= �[2(300.25) + (12.25)]= �[600.50 + 12.25] = �[612.75]

= 227

× 612.75 = 1925.79 cm2

Cost of painting on 1 cm2 = � 0.05� Cost of painting on 1925.79 cm2

= � 1925.79 × 0.05 = � 96.29.��

Let the length, breadth and height ofthe cube be l cm, b cm and h cmrespectively.Then, l + b + h = 19 ... (i)Now, diagonal = 11 cm.

� 2 2 2l b h� �= 11

� 2 2 2l b h� � = 121 ... (ii)

Now, l + b + h = 19 (Given)� (l + b + h)2 = (19)2

(Squaring)� l2 + b2 + h2 + 2(lb + bh + lh)

= 361

� 121 + 2(lb + bh + lh) = 361� 2(lb + bh + lh) = 361 – 121

= 240.Hence, the surface area of the cuboidis 240 cm2.

�� Here AG = 22 cm, GH = 2 cm, HI = 5 cm,IJ = 8 cm and JK = 3 cm.Now, volume of upper cuboid GHICPAB

= AG × GH × HI= 22 × 2 × 5 cu. cm= 220 cu. cm ...(i)

Volume of lower cuboid= PD × CI × JK= (PC + CD) × AG × JK= (GH + IJ) × 22 × 3= (2 + 8) × 22 × 3 cu. cm= 660 cu. cm ...(ii)

Thus total volume of the piece= (220 + 660) cu. cm= 880 cu. cm. [Using (i) and (ii)]

� Length of cylinder = 48 cm.Diameter of copper wire = 4 mm

= 0.4 cm.So, one round of wire will cover thesurface 0.4 cm thick.� Number of rounds to cover 48 cm

= 480.4

= 120

Diameter of cylinder = 10 cm.

Length of the wire in one round

= Circumference of the base of thecylinder

79��

= � × 10 cm = 10� cm.

� Length of the wire to cover the wholesurface

= Length of wire in 120 rounds

= 10 � × 120 = 1200 � cm.

Radius of the copper wire

= 42

mm = 0.2 cm

� Volume of the wire

= �(0.2)2 × 1200� cu. cm

= 48�2 cu. cm

Mass of the wire= Volume × specific gravity

= 48�2 × 8.80 g = 422.4�2 g.

��

�� (A) Radius (r) of the sphere

= 12

× side of the cube

= 12

× 5 cm.

Volume of the sphere

= 43�r3 = 4

3� ×

352

� ��

= 1256� cm3.

�� Let the side of the cube = a unitsVolume = Surface area

� a3 = 6a2

� a3 – 6a2 = 0� a2 (a – 6) = 0� a = 0 or a = 6 units

but a cannot be zero.� a = 6 units� Volume = a3 = 216 cu. units.

�� Number of cones

=Volume of the cylinder

Volume of a cone

=2

2

R H1 R H3

�

�

= 3.

�� The bullet is in the shape of sphere.Number of bullets

=Volume of the cube

Volume of one bullet

=22 22 22

4 22 1 1 13 7

� �

� � � �

=22 22 22 3 7

4 22� � � �

�

= 2541.

� Volume of a cone = 13�r2h

Let r1 and r2 be the radii of the twocones.

Then, 1

2

rr =

35

�

21

22

rr

= 925

� Ratio of volumes =

21

22

1313

r h

r h

� �

� �

=2

12

2

rr

= 925

.

Hence, ratio of volumes of the two conesis 9 : 25.

� �� Volume of a cuboid = (18 × 12 × 9) m3

Volume of each cube of 3 metre edge= (3)3 = 27 m3

Let the required number of cubes = n,then

n × 27 = 18 × 12 × 9

� n = 18 12 9

27� �

= 1944

27= 72.

�� r = 0.25 m, h = 3.5 mCurved surface of the pillar

= 2 �rh

= 2 ×227

× 25100

×3510

m2

= 112

m2 = 5.5 m2

80 � � � � � � ��

Cost of painting = � 12.50 × 5.5= � 68.75.

� Radius, r = 10.5 = 212

cm

Slant height, l = �r = 227

× 212

= 33 cm

Curved surface area of the capS = �rl = l × l = l2

= 33 × 33 cm2

Volume of mixture V = 54.45 cm3

(i) The painting on the cap is doneuniformely, so the thickness of the painton the cap must be uniform (identical)

Let it be h.

� S × h = V� 33 × 33 × h = 54.45

� h = 54.4533 33�

= 5445

33 33 100� �

=1653300

= 0.05 cm

= 0.5 mmThus, the thickness of the paint on thecap is 0.5 mm.(ii) Let the required volumes be x, 2xand 3x.� x + 2x + 3x = 54.45

� x = 54.45

6 = 9.075

� 2x = 2 × 9.075 = 18.15and 3x = 27.225So, the required volumes are 9.075 ml,18.15 ml and 27.225 ml.(iii) (a) Volume of a uniform solid isequal to product of its base area andheight.(b) Curved surface area of a right circularcone.(iv) Constructive thoughtful, Creatorand Entertainer.

��

81��

��

PROBABILITY

��

�� (B) Total number of elements in thegiven set = 10The even numbers in the set are 2, 4, 6,8 and 10Number of even numbers = 5Probability of choosing an even number

= 5

10=

12

.

�� In a single throw of a die, multiple of 3are 3 and 6.Required probability

= Number of favourable eventsNumber of all possible events

= 26

= 13

.

�� After removing the cards, number ofcards of clubs = 13 – 3 = 10After removing the cards, number of allcards = 52 – 3 = 49

� Required probability = 1049

.

�� Number of all possible outcomes= 6 × 6 = 36

Favourable outcomes are {(2, 3), (2, 6),(3, 2), (3, 4), (3, 6), (4, 3), (4, 6), (6, 2),(6, 3), (6, 4), (6, 6)� Number of favourable outcomes = 11Hence, required probability

= Number of favourable outcomesNumber of all possible outcomes

= 1136

.

�� Total number of trails or chances = 30Number of chances when the boundaryis not hit = 30 – 6 = 24.

P(the boundary is not hit) = 2430

= 45

.

�� Total number of students = 40.Number of students born in August = 3.P (A student of the class was born in August)

= 3

40.

� Total number of trials (tosses) = 500No. of heads = 280

� No. of tails = 500 – 280 = 220(i) Probability of getting a head

= No. of heads

Total no. of tosses

� P (head) = 280500 =

1425 .

(ii) P (tail) = No. of tails

Total no. of tosses

= 220500 =

1125 .

� (i) There are 6 possible ways in which adie can fall, i.e., 1, 2, 3, 4, 5, 6.There is only one way of throwing 3.Total number of outcomes = 6� The required probability

= Number of favourable outcomes

Total number of possible outcomes

= 16

.

(ii) Number of outcomes of falling evennumber, i.e., 2, 4, 6 = 3.

The required probability = 36

= 12

.

�� Here S = {1, 2, 3, 4, 5, ......, 25}� n(S) = 25The prime numbers from 1 to 25 are:

2, 3, 5, 7, 11, 13, 17, 19, 23.

82 ��

Let E = event of getting a prime number� n(E) = 9

Now, P(E) = (E)(S)

nn

= 925

Required probability that the selectednumber is not a prime number

= 1 – the probability that the number selected is a prime number

= 1 – 925

= 25 – 925

= 1625

.

��

�� (C) Total number of trials (tosses) = 500No. of heads = 280

� No. of tails = 500 – 280 = 220(i) Probability of getting a head

= No. of heads

Total no. of tosses

� P (head) = 280500 =

1425 .

(ii) P (tail) = No. of tails

Total no. of tosses

= 220500

= 1125

.

�� P (the marble drawn is green)

= Number of green marblesSum of all blue and green marbles

= 4

6 4�

= 4

10 =

25

.

�� No. of aces in the deck = 4

Required probability = 452

= 1

13.

�� Total number of outcomes = 52There are 26 red cards having 2 queens.Other two queens are black.Therefore, number of cards having neithera red card nor a queen = 52 – 26 – 2 = 24Thus, number of favourable outcomes = 24

Now, required probability = 2452

= 613

.

�� Total number of bags = 11Number of bags containing more than5 kg of flour = 7.

Therefore, probability of choosing a bagcontaining more than 5 kg of flour

=

Number of bags containingmore than 5 kg flour

Total number of bags = 7

11.

�� Total number of trials = 250.Number of chances favouring 2 heads

= 72

� Required probability=72250

= 36

125.

� Total number of students = 90(i) Number of students obtained lessthan 20% marks = 7


(ii) Number of students obtained 60 marksor more = 15 + 8 = 23


.

� (i) Let area of the wheel be A.(a) Area of the region “play”

=30360

��

× A = A12

� Required probability =A / 12

A=

112

(b) Area of the region “school”

= 75360

��

× A = 5 A24

� Required probability =5A/24

A=

124

(c) Area of the region “others”

= 60360

��

× A = A6

� Required probability = A/6A

= 16

(ii) (a) Time spent in playing

= 30360

��

× 24 hours = 2 hours

(b) Time devoted in homework

= 75360

��

× 24 hours = 5 hours

83��

No. of favourable outcomes = 3

� P (getting at least one head) = 34

.

�� Number of outcomes in the samplespace = 36Possible doublets are:(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)Number of these doublets = 6

� P (getting a doublet) = 636

=16

.

�� From the given data, the frequencydistribution table is as below:

Blood Tally Frequency orgroup Marks number of students

A IIII �� 9B IIII � 6O IIII IIII �� 12

AB �� 3

Total number of students = 30

The number of students having theirblood group AB = 3.

�� The probability that a student,selected at random, has his blood group

AB = 3

30 =

110

.

� Outcomes in the sample space are givenby {HHH, HHT, HTH, HTT, THH, THT,TTH, TTT}� Number of outcomes in the samplespace = 8All possible outcomes having at leastone head are{HHH, HHT, HTH, HTT, THH, THT, TTH}� Number of all possible outcomes

having at least one head = 7

� P (getting at least one head) = 78

.

� Total number of tickets = 100Multiples of 3 from 1 to 100 are:

3, 6, 9, ......, 99

1

) (3 3310099

� 33 multiples of 3

(iii) Rashmi gave sufficient time to eachactivity. The time duration of playing isonly two hours a day is a good thinking.I think her plan is ‘very good’(iv) Accountability, Rationality

�� (i) Total number of families = 1500Number of families having 2 girls

= 475

P(a family having 2 girls) = 4751500

= 1960

(ii) Number of families having 1 girl= 814

P(a family having 1 girl) = 8141500

= 407750

(iii) Number of families having no girl= 211

P (a family having no girl) = 2111500

.

Sum of the probabilities

= 4751500

+ 8141500

+ 2111500

= 15001500

= 1.

��

�� (C) Total number of numbers = 100Prime numbers between 1 and 20 are:

2, 3, 5, 7, 11, 13, 17, 19.� P(card is a prime less than 20)

= 8

100 =

225

.

�� P(the boundary is not hit) = 30 – 630

= 2430

= 45

.

�� All possible outcomes are 1, 2, 3, 4, 5, 6.Numbers greater than 3 are 4, 5, 6.P (getting a number greater than 3)

= 36

= 12

.

�� All possible outcomes are:{HH, HT, TT, TH}No. of all possible outcomes = 4Favourable outcomes are: {HH, HT, TH}

84 ��

) (5 20100100× � 20 multiples of 5

) (15 610090

10� 6 multiples of 15

Multiple of 3 or 5 are 33 + 20 – 6 = 47.Hence, probability of drawing a ticket,whose marked number is a multiple of

3 or 5 = 47

100.

�� Number of days in leap year = 366.Now, 366 days = 52 weeks and 2 days.The remaining two days can be one ofthe following seven cases:

(i) Sunday and Monday(ii) Monday and Tuesday

(iii) Tuesday and Wednesday(iv) Wednesday and Thursday(v) Thursday and Friday

(vi) Friday and Saturday(vii) Saturday and Sunday.For a leap year to contain 53 Sundays,last two days are either Sunday andMonday or Saturday and Sunday.� Number of such favourable outcomes

= 2.

Total number of possible outcomes = 7.� P(a leap year containing 53 Sundays)

= 27

.

��

�� (B)�� There are 4 kings out of 52 playing

cards.� Total no. of all possible outcomes = 52And no. of favourable outcomes = 4� P (getting a king from a well shuffleddeck of 52 cards)

= 4

52 =

113

.

�� We know that the sum of the proba-bilities of occurring and non-occurringan event is one. So, according to thequestion,

2x

+ 23

= 1

�

2x = 1 – 2

3

� x = 23

.

�� As |x|< 2, x = – 1, 0, 1

Total number of outcomes = 7

Number of favourable outcomes = 3

� P (|x|< 2) =37

.

�� From the given data, the frequency

distribution table is made as below:

Concentration ofSO2 (in ppm)

Frequency

0.00-0.04 40.04-0.08 90.08-0.12 90.12-0.16 20.16-0.20 40.20-0.24 2

Total = 30

Number of days when the concentrationof level of sulphur dioxide is in theinterval 0.12-0.16 = 2.� The required probability, i.e.,

P(0.12-0.16) = 230

= 115

.

� Let H denotes head and T denotes tail.So, on tossing two coins simultaneously,all the possible outcomes are:

HH, HT, TH, TT.

(i) The probability of getting two heads= P(HH)

85��

=

No. of outcomes of occurringtwo heads

Total number of possibleoutcomes

= 14

.

(ii) The probability of getting at leastone head

= P(HT or TH or HH)

=

No. of outcomes of occurring at least one head


= 34

.

(iii) The probability of getting no head= P(TT)

=

No. of outcomes of occurring no head


= 14

.

� Randomly drawing of blocks ensuresequally likely outcomes.(i) The number of outcomes representing

A = 2Total number of possible outcomes = 6P(A)

=Number of favourable outcomes


= 26

= 13

.

(ii)Number of outcomes representing D = 1Total number of possible outcomes = 6

� P(D) = 16

.

(iii) Letters B, C and E have the sameprobability as the letter D.

�� Total number of students = 200.(i) The number of students who likeStatistics = 135.

� The probability of a student, chosen

at random, likes Statistics = 135200

= 2740

.

(ii) Number of students who dislikeStatistics = 65.� The probability that a student,chosen at random, dislikes Statistics

=65

200=

1340

.

��

�� (C) No. of red balls = 4� No. of favourable outcomes = 4

Total no. of all balls = 4 + 2 = 6� No. of all possible outcomes = 6P (getting a red ball)

= No. of favourable outcomesNo. of all possible outcomes

= 46

= 23

.

�� On throwing a die, the sample space isgiven by {1, 2, 3, 4, 5, 6}The favourable outcomes are: 2, 4, 6.P (coming up an even number)

=36

= 12

.

�� Let H denotes the head and T denotesthe tail. On tossing two coins, thesample space is given by {HH, HT, TH,TT}

The outcome of getting "head-head" isonly HH.

� P (getting HH) = 14

.

�� The total number of ladies = 200

The number of ladies who dislike coffee= 118

� P (a randomly chosen lady dislikes coffee)

= 118200

=59

100.

86 ��

�� Total number of trials = 200.Number of chances favouring 2 heads

= 72Probability of getting 2 heads coming up

= 72200

= 925

.

�� All possible outcomes are: 1, 2, 3, 4, 5, 6Favourable outcomes are: 1, 2, 3, 4Probability of getting a number less than

5 in a single throw of a die = 46

= 23

.

� Number of all possible outcomes= 6 × 6 = 36

(i) Favourable outcomes for 9 are(3, 6), (4, 5), (5, 4) and (6, 3)

� Number of favourable outcomes = 4

Now, P(getting a sum of 9) = 436

= 19

.

� (i) There are 6 possible ways in whicha die can fall, i.e., 1, 2, 3, 4, 5, 6There is only one way of throwing 3.Total number of outcomes = 6� The required probability

=Number of favourable outcomes


=16

.

(ii) Number of outcomes of falling evennumber, i.e., 2, 4, 6 = 3.

The required probability = 36

= 12

.

�� (i) Marks 98.11% and 98.89% areassociated with the months February andMay respectively. So the number offavourable outcomes = 2.

Number of all possible outcomes

= Total number of marksheets = 6

� Required probability = �2 16 3�

(ii) There is no marksheet associated withthe month of July.


(iii) The probability of an impossible eventis zero.

(iv) Brilliant student.

��

87��

��

STATISTICS

��

�� (C)�� Frequency�� 15.00 hrs, 19.00 hrs.

�� (i) Decrease in percentage

= 99 – 98 ×100%99

= 1.01%.

(ii) Increase in percentage

= 104 – 100 ×100%100

= 4%.

�� (i) variate(ii) 12Range = maximum value of the variable

– minimum value of the variable= 15 – 3 = 12.

�� Class size = Difference between classmarks of two adjacent classes

= 115 – 105 = 10.We need classes of size 10 with classmarks 105, 115, 125,.....,175 . Thus, class

limits for the first class are 105 – 12

(10)

and 105 + 12

(10), i.e., 100 and 110.

Therefore, the first class is 100-110.Hence classes are 100-110, 110-120,120-130, 130-140, 140-150, 150-160,160-170, 170-180.

� Blood group Tally marks Number ofstudents

A IIII�� 9B IIII� � 6O IIII IIII�� 12

AB �� 3

Total 30

The most common blood group is ‘O’as it has maximum frequency and therarest blood group is ‘AB’ withfrequency 3.

� (i) Relative Tally Frequencyhumidity marks

84-86 � 186-88 � 188-90 �� 290-92 �� 2

92-94 IIII �� 7

94-96 IIII � 696-98 IIII �� 798-100 �� 4

Total 30

(ii) The data appears to be taken inthe rainy season as the relativehumidity is high.(iii) Range = 99.2 – 84.9 = 14.3%.

�� Arranging the marks in ascendingorder, we get

11, 12, 15, 16, 18, 19, 20, 22, 22, 22,23, 27, 27, 27, 29, 29, 29, 31, 31, 33,35, 37, 40, 43, 47, 49, 50, 51, 56, 58.

(i) � �� Frequency distribution of marks:

MarksTally Marks Frequency(Class intervals)

11-20 IIII �� 7

21-30 IIII IIII 10

31-40 IIII � 6

41-50 �� 4

51-60 �� 3

Total 30

88 ��

(ii) The most frequently occurring digitsare 3 and 9. The least occurring is 0.

�� Since the class size is the differencebetween the any two consecutive classmarks, so the class size = 52 – 47 = 5.� The lower limit of the first class

= 47 – � ��

52

= 44.5

and the upper limit of the first class

= 47 + � ��

52

= 49.5

Therefore, the class limits are44.5-49.5, 49.5-54.5, 54.5-59.5,59.5-64.5, 64.5-69.5, 69.5-74.5,74.5-79.5, 79.5-84.5, 84.5-89.5,89.5-94.5, 94.5-99.5, 99.5-104.5.This being the exclusive form offrequency distribution. True class limitsare the same as the class limits.

� (i) Concentration of sulphur Frequencydioxide (in ppm)

0.00-0.04 4

0.04-0.08 9

0.08-0.12 9

0.12-0.16 2

0.16-0.20 4

0.20-0.24 2

Total 30

(ii) The concentration of sulphur dioxidewas more than 0.11 ppm for (2 + 4 + 2 =)8 days.

� � Range = maximum value of the variable – minimum value of the variable = 38 – 10 = 28

Number of class intervals = RangeClass size

= 284

= 7

We can take 10-13 as first classinterval.

(ii)�� Frequency distribution of marks:

MarksTally Marks Frequency(Class intervals)

10.5-20.5 IIII� �� 720.5-30.5 IIII IIII 1030.5-40.5 IIII � 640.5-50.5 �� 450.5-60.5 �� 3

Total 30

��

�� (D) First class interval is 0-5, secondclass interval is 5-10, third class intervalis 10-15, etc.

�� The variables in the class interval 15-20 are 15 and 17 only.

�� C.I. f cf

10-20 7 720-30 11 1830-40 20 3840-50 46 8450-60 57 14160-70 37 17870-80 22 200

From the table, the cumulative frequencyof the class interval 50-60 is 141.

�� Limits for the first class are 52 – 52

and

52 + 52

, i.e., 49.5 and 54.5.

So, first class interval is 49.5-54.5.Similarly, the second class interval is54.5-59.5.

�� (i) Digits Frequency

0 21 52 53 84 45 56 47 48 59 8

Total 50

89��

�� !� �� "

MarksTally marks Frequency(Class intervals)

10-13 IIII 514-17 IIII �� 818-21 IIII �� 822-25 IIII �� 726-29 IIII 530-33 �� 434-37 �� 238-41 � 1

Total 40

�� Arranging the electricity bills in ascendingorder, we get5, 8, 10, 12, 14, 15, 16, 18, 20, 22, 22, 24, 24,25, 25, 27, 30, 30, 36, 42, 44, 45, 47, 50, 56.

Maximum value of the bill = � 56.

Minimum value of the bill = � 5.

� Range = 56 – 5 = � 51.

Let us suppose that we want to make 7class intervals.

Now, Range � Number of class

= 517

= 727

� We take the size of each class = 8.

We now make two frequency tables.

(i) � ��"

Class interval Tally Marks Frequency

5-12 �� 4

13-20 IIII 521-28 IIII �� 729-36 �� 337-44 �� 245-52 �� 353-60 � 1

Total 25

(ii) ��"In this case, we start with 5 andmake 6 classes each of size = 10.

Class interval Tally Marks Frequency

5-15 IIII 515-25 IIII �� 825-35 IIII 535-45 �� 345-55 �� 355-65 � 1

Total 25

��

�� (D)�� 8�� 3 : 4 : 6 : 2 : 5�� Colony A has the tallest bar as it contains

maximum number of pupils, i.e., 400.Colony D has the shortest bar as it con-tains minimum number of pupils, i.e., 50.

��

�� Let the amount spent (in �) onfootball, hockey, cricket, basketball andvolleyball be 3x, 4x, 6x, 2x and 5xrespectively.

� 3x + 4x + 6x + 2x + 5x = 200000

� x = 10000

(i) Amount spent on hockey = 4x

= � 40000.

(ii) Amount spent on cricket = 6x

= � 60000.

90 ��

�

� (i) The histogram is shown here.

(ii) Required number of lamps

= 74 + 62 + 48

= 184.

�� As the given distribution is discontinu-ous, we shall first change it to a con-tinuous distribution. To do this, we takehalf of the difference of the lower limitof second class and the upper limit of

first class. Here it is 0.5 � ��

51 – 502

.

Proceeding this way, we get the fol-lowing distribution.

Daily earningsNo. of workers(in rupees)

0.5-50.5 3 50.5-100.5 7100.5-150.5 4150.5-200.5 5200.5-250.5 4250.5-300.5 3300.5-350.5 2350.5-400.5 2

The required histogram is shown below.

��

�� (A) On x-axis, the y-coordinate of eachpoint is zero.

�� 0 to 6

�� In a continuous class interval the lowerlimit is included but the upper limit isnot included in the interval.

�� We will take class intervals along y-axisand frequency along x-axis.

�� (i) Expenditure on wheat = 35%Expenditure on pulses = 20%

Excess of expenditure on wheat

than that on pulses = 35 – 20 ×100%20

= 75%.

91��

(ii) Total percentage expenditure onpulses and ghee = (20 + 15)% = 35%.

�� # �$�� $�� %�� $��&� "(a) Consider a histogram of given data(say) with six adjacent rectangles.(b) We join the mid-points of the uppersides of the adjacent rectangles of thehistogram by means of line segments.Let us call the mid-points in step (b) beB, C, D, E, F and G. We obtain thefigure BCDEFG.(c) To complete the polygon, we join Bto A and G to H. Here, A and H are themid-points of class intervals withfrequency zero before the class intervalcorresponding to the point B and afterthe class interval corresponding to thepoint G respectively.

ABCDEFGH is the frequency polygoncorresponding to the given data.

� We represent the names of educations(variable) on the horizontal axis (thex-axis) and the expenditure (value) onthe vertical axis (the y-axis).

� Daily pocket Class Frequencyexpenses of a mark (no. of students)student (in �)

0-5 2.5 105-10 7.5 1610-15 12.5 3015-20 17.5 4220-25 22.5 5025-30 27.5 3030-35 32.5 1635-40 37.5 12

Now, we plot the ordered pairs (2.5, 10),(7.5, 16), (12.5, 30), (17.5, 42), (22.5,50), (27.5, 30), (32.5, 16), (37.5, 12).

We also plot the ordered pairs (–2.5, 0)and (42.5, 0). On joining all these pointsby line segments, we obtain the requiredfrequency polygon which is shown above.

�� The class mark of the classes 20-29,30-39, 40-49, 50-59, 60-69, 70-79 are 24.5,34.5, 44.5, 54.5, 64.5, 74.5 respectively.The corresponding frequencies for clubsA and B are 5, 10, 15, 10, 20, 5 and 10,15, 10, 5, 10, 15 respectively.For the club A, the frequency polygon isdrawn by joining the points: (14.5, 0),(24.5, 5), (34.5, 10), (44.5, 15), (54.5,10), (64.5, 20), (74.5, 5), (84.5, 0).

92 ��

For the club B, the frequency polygon isdrawn by joining the points: (14.5, 0),(24.5, 10), (34.5, 15), (44.5, 10), (54.5,5), (64.5, 10), (74.5, 15), (84.5, 0).

��

�� (C) As 17 is the mean of 10, q, 16, 26,12 and 20

�10+ +16+26+12+ 20

6q

= 17

� 84 + q = 17 × 6

� q = 18.

�� First 'n' even natural numbers are2, 4, 6, 8, ....., 2n

Mean = 2+ 4 +6+8+.........+ 2nn

=2(1+ 2+3+ 4 +.......+ )n

n

=2 ( +1)

2n n

= n(n + 1)

= n2 + n.

�� Required mean = 2 +3+5+7 +115

= 285

= 5.6.

�� Average wage paid per day

= 18×12+10×13.5+5×25+ 2×4218 +10 + 5+ 2

= 56035

= � 16.

�� The positive factors of 12 are 1, 2, 3, 4,6, 12.

� Mean = x = � ixn

= 1+ 2+3+4 +6+126

= 286

= 143

or 423

Hence, the mean of all positive factors

of 12 is 423

.

'(Since, on replacing the observation 50by 50, the sum of the observations willremain same. Hence the mean will alsobe remain same, that is, 50.

�� x =� ix

n

= +( +2)+( +4)+( +6)+( +8)5

x x x x x

or 24 = 5 + 205

x [Mean = 24 (given)]

or 120= 5x + 20or 5x = 100 or x = 20Hence, the value of x is 20.

� The given table shows the demand ofbiscuits on different number of days.To find the average, we prepare thefollowing table:

Daily demand Frequencyfixi(xi) ( fi)

12 36 43217 160 272021 92 193227 56 151230 26 780

if∑ = 370 i if x∑ = 7376

x = i i

i

f x

f∑∑

= 7376370

= 19.94

Hence, average (or mean) number ofbiscuits demanded = 19.94.

93��

� Let us rearrange the given data as below:

x fi fixi

10 3 3015 10 15020 a 20a25 7 17535 5 175

if∑ = 25 + a i if x∑ = 530 + 20a

� Mean = i i

i

f x

f∑∑

� 20.6 = a

a530 20

25++

[... Mean = 20.6 (given)]� 20.6 (25 + a) = 530 + 20a

� 530 + 20a = 515 + 20.6a� 530 – 515 = 20.6a – 20a� 15 = 0.6a � a = 25.

�� (i) Required number of students= 8 + 32 = 40

(ii) Here, we notice that classes arecontinuous but class-size is not the samefor all the classes. We notice minimumclass-size is of class 45-50, i.e., 5. We willfirst find proportionate length ofrectangle (adjusted frequency) for eachclass.Length of rectangle (adjusted frequency)

= Frequency of class

Width of class× Minimum class-size

Therefore, the value of the 13th entryis 29.

Marks Number of students Width of class Length of rectangle(C.I.) (f) (class-size)

0-10 8 10 810

× 5 = 4

10-30 32 203220

× 5 = 8

30-45 18 151815

× 5 = 6

45-50 10 5105

× 5 = 10

Now, we construct rectangles with respective class-intervals as widths and adjustedfrequencies as heights.Histogram representing marks obtained by students in unit test of Mathematics.

(iii) Hardwork, Dilligence.

94 ��

��

�� (B) Let the 40 items be x1, x2, x3,....., x40.

As � � � �x x x x1 2 3 40.........

40= 35 ...(i)

� New mean

= � � �ax ax ax1 2 40.........

40

� � � �a x x x1 2 40( ......... )

40= a × 35 [Using (i)]Hence, the new mean will be 35a.

�� According to the given conditions,x1 + x2 + x3 + ............ + xn = A × n

� (x + a) (x1 + x2 + x3 + .....+ xn) = (x + a) An

�� nx a x x a x x a x x a x

n1 2 3( ) ( ) ( ) ..... ( )

= (x + a)AThus, the new average is (x + a) A.

�� Sum of 10 observations having themean as 20 = 10 × 20 = 200

Sum of another 15 observations havingthe mean as 16 = 15 × 16 = 240

Thus the sum of these 25 observations= 440

Hence, the required mean = 44025

= 17.6.

�� We know that

x =� ix

n �� xi =�n × x

Hence, n = 40, x = 60� � xi = 40 × 60 = 2400But this value of � xi was incorrect as48 was misread as 84.� Correct value of � xi = 2400 – 84 + 48

= 2364

Hence, correct mean is 236440

= 59.1.

�� Let the numbers be x1, x2, x3, ......, x16

Given that � � � �x x x x1 2 3 16......16

= 8

or x1 + x2 + x3 + ....... + x16 = 128

If 2 is added to each number, then theobservations become

x1 + 2, x2 + 2, x3 + 2, ......... , x16 + 2� New mean

= x x x x1 2 3 162 2 2 ....... 2

16+ + + + + + + +

= x x x1 2 16........ 32

16+ + + +

= 128 32

16+

= 16016

= 10

Thus, the new mean is 10.

�� Let the observations be x1, x2, x3, ......., x21.

Mean x =x x x x1 2 3 21......

21+ + + +

= 15

� x1 + x2 + x3 + ....... + x21 = 315If each observation is multiplied by 2,then the observations become

2x1, 2x2, 2x3, ...... 2x21

New mean = x x x x1 2 3 212 2 2 ...... 2

21+ + + +

= x x x x1 2 3 212( ...... )

21+ + + +

= 2 × 31521

= 2 × 15 = 30

Thus, the new mean is 30.

� Mean score of nine innings = 58� Total score of nine innings = 58 × 9

= 522Mean score of 10 innings = 61

� Total runs of ten innings = 61 × 10= 610

Difference = 610 – 522 = 88Hence, 88 runs to be scored in the tenthinnings.

'(

We know that

x = ∑ ixn

� ix∑ = n × x

Here, n = 25 and x = 78.4

� ix∑ = 78.4 × 25 = 1960

But this value, that is, 1960 of ix∑was incorrect as 96 was misread as 69.

95��

� Correct value of ix∑ = (1960 – 69 + 96)

= 1987

Hence, correct mean =198725

= 79.48.

� (i) Let us converge the given data intofrequency distribution table.

Hours (xi) Frequency (fi) fixi

5 3 15

18 10 18020 6 120

12 28 336

40 3 120

��fi = 50 ��fixi = 771

Mean x = i i

i

f xf

��

= 77150

= 15.42

Thus the required mean time is 15.42hours per week.(ii) Mean of observations is the sum ofproduct of all the observations and theircorresponding frequencies divided bysum of frequencies.(iii) Social work.

�� Rewriting the given data, we have

x f fx

2 3 64 2 86 3 1810 1 10

p + 5 2 2p + 10

if∑ = 11 i if x∑ = 52 + 2p

Mean = i i

i

f x

f∑∑ =

p52 211+

� 6 =p52 2

11+

(Given mean = 6)

� 66 = 52 + 2por 2p = 66 – 52 = 14� p = 7.

��

�� (B) Rearranging the data in ascendingorder, we have

9, 13, 15, 21

Median = 13+152

= 282

= 14.

�� Rearranging the data in descendingorder, we have

27, 25, 24, 23, 21

Median = Value of middle term = Value of 3rd term = 24.

�� Number of observations = 11So, the middle term is the sixth termwhich is x.� x = 63.

�� Squares of first five natural numbers is1, 4, 9, 16, 25

Median = Value of middle term = Value of 3rd term = 9.

�� Arranging the given observations inascending order, we get

3, 5, 7, 8, 12, 14, 17.Here n = 7, which is odd.

� Median = � ��

th+12

n item = � ��

th7 +12

item

= 4th item = 8.

'(

Arranging the given observations inascending order, we get

3, 4, 6, 9, 10, 11, 18, 22.Here n = 8, which is even.

� Median = Mean of � ��

th

2n and � �

� �th

+12n

items.

= Mean of � ��

th82

and � ��

th8 +12

items

= Mean of 4th and 5th items

= 9 +102

= 192

= 9.5.

96 ��

�� Here n = 16, which is even.

� Median is the mean of � ��

th

2n and

� ��

th+1

2n term i.e.,

162

= 8th and 9th term

Required median = Average of 8th and9th observations

= 25 272+ = 26.

'(Here, n = 9 (odd)

So, median is the value of th

9 +12

� ��

obser-

vation, i.e., 5th observation� x + 2 = 24� x = 22.

� First ten prime numbers in ascendingorder are 2, 3, 5, 7, 11, 13, 17, 19, 23,29, Here n = 10, which is an evennumber

Median = � ��

� �th th

item + +1 item2 2

2

n n

= th th5 item +6 item

2= 11+13

2

= 242

= 12.

� Here n = 10 which is even.

)%��" Median

= 12

� � � ��

n

n

th

th

Value of item Value of2

1 item2

= 12

� � � ��

th

th

10Value of item Value of

2

101 item

2

= 12

[Value of 5th item + Value of 6th item]

� Median = 12

[30 + x] � 35 = � x302

� 35 × 2 = 30 + x �� 70 = 30 + x� x = 70 – 30 = 40.

)%��" If 48 is changed to 28, the newincreasing order is:

18, 20, 25, 26, 28, 30, 37, 38, 39, 40.� New median

= 12

� � ��

� ��

n

n

th

th

value of item2

value of 1 item2

= 12

� � ��

� ��

th

th

10value of item

2

10value of 1 item

2

= 12

[value of 5th item + value of 6th item]

= 12

(28 + 30) = 582

= 29

Hence, x = 40 and the new median = 29.�� Rearranging the given data in ascending

order, we have19, 25, 30, 31, 32, 35, 48, 51, 59

Here, number of observations, i.e., n = 9,which is odd

� Median = Value of th

9 +12

� ��

observation= Value of 5th observation= 32

If 25 is replaced by 52, then the newarrangement in ascending order will be

19, 30, 31, 32, 35, 48, 51, 52, 59Now, the new median

= Value of 5th observation= 35.

97��

��

�� (C) Rearranging the given data, we have1, 2, 2, 3, 3, 4, 4, 5, 5, 7 and PThis data has the mode 3. It is possibleonly when 3 has maximum frequency.Hence, P = 3.

�� mode = 3 median – 2 mean�� We have, 3 median= mode + 2 mean

� Median = 21+ 2× 243

= 23.

�� Number of all observations = 32Sum of all observations = 111

Mean = 11132

= 3.47

Rearranging the given observations inascending order, we have1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 6, 7.

Median = 12

(value of 16th observation

+ value of 17th observation)

= 12

(3 + 3) = 12

× 6 = 3

Mode = Value of the digit having maximum frequency

= 3.�� The given data, except x, can be repre-

sented in the form of a table as below:

Observation 13 14 15 16 17

Frequency 1 1 3 3 3

Frequency of 15, 16 as well as 17 is 3.But the mode of the given data = 17.Frequency of 17 should be the maxi-mum. That is, more than 3, which ispossible only if x = 17.Hence, x = 17.

�� We make frequency distribution tablefor the given data as below:

xi Tally Marks Frequency

6 �� 38 � 1

9 IIII 5

14 IIII 515 �� 321 IIII �� 825 �� 427 �� 429 �� 2

Total = 35

Here, the observation 21 has the maxi-mum frequency. Therefore, the mode ofthe data is 21.

� Mean = Sum of all observationsNumber of observations

= ixn�

� 14 = 11+15+17+( +1)+19+( – 2)+37

x x

� 98 = 64 + 2x� 2x = 98 – 64 = 34� x = 17� x + 1 = 17 + 1 = 18and x – 2 = 17 – 2 = 15Therefore, given observations are 11, 15,17, 18, 19, 15, 3. The number 15 occursmaximum number of times, i.e., 2 times.� Mode of the given data = 15Hence, x = 17 and mode = 15.

� The observations are 42, 43, 44, 44,(2x + 3), 45, 45, 46, 47.... The number of observations is 9.Here n = 9 (odd)

� Median = Value of n 1

2+⎛ ⎞

⎜ ⎟⎝ ⎠th term

= Value of 9 1

2+⎛ ⎞

⎜ ⎟⎝ ⎠th term

= Value of 5th term� Median = 2x + 3� 45 = 2x + 3 � 2x = 45 – 3 = 42� 2x = 42 � x = 21.� The observations are 42, 43, 44, 44,

45, 45, 45, 46, 47.The number 45 occurs maximum (3) times.� Mode of the data = 45.Hence, x = 21 and mode = 45.

98 ��

�� (i) The most often observation of thegiven data is called mode.(ii) 8 (iii) 4 (iv) Self-reliance.

*##�##+�,-�#.��-/��

�� (A)�� 0�� Required mean

= Total number of marksTotal number of students

= 10×75+12×60 + 8 × 40 + 3 × 3010 +12+8 + 3

= 750 +720 + 320 + 9033

= 188033

= 56.97 =57 marks (approximately).�� Let number of boys and that of girls be

x and y respectively.� x + y = 60 ... (i)Sum of weights of all 60 students

= 60 × 40 = 2400 kgSum of weights of all x boys = x × 50Sum of weights of all y girls = y × 30Now, 50x + 30y = 2400

5x + 3y = 240 ... (ii)Solving equations (i) and (ii), we get

x = 30, y = 30.

�� (i) variate (ii) 12

�� (i) 35, 46 (ii) 65

� x = 1 2 6+ +.......+

6

x x x

or 18 =

+ ( + 3)+ ( + 6)+ ( +9) + ( +12)+ ( +15)

6

x x x xx x

or 18 = 6 + 456

x

or 108 = 6x + 45or 6x = 108 – 45 or 6x = 63

� x = 636

= 10.5

� First 4 observations are x, x + 3, x + 6, x + 9

i.e., 10.5, 10.5 + 3, 10.5 + 6, 10.5 + 9i.e., 10.5, 13.5, 16.5, 19.5

Mean of the first 4 observations

= 10.5 + 13.5 + 16.5 + 19.54

= 60.04

= 15Hence, mean of the first four observa-tions is 15.

'(

�� (i) The graph represents production of

rice (in lakh tons) during each year fromthe year 1978 to 1983.

(ii) The bar for 1980-1981 has length of55 cm.

Therefore, the production of rice in theyear 1980-1981 is 550 lakh tons.

(iii) In the year 1981-1982, the produc-tion of rice is maximum and is equal to650 lakh tons.

In the year 1979-80, the production ofrice is minimum and is equal to 250 lakhtons. The difference between themaximum and the minimum production

= 650 lakh tons – 250 lakh tons= 400 lakh tons.

99��

�� The given data has 10 values. Arrangingthe values of the given data in ascendingorder, we have 0, 1, 2, 3, 3, 3, 3, 4, 4, 5.

Mean = Sum of the observationsNo. of the observations

=0 +1 + 2 + 3 + 3 + 3 + 3 + 4 + 4 + 5

10= 28

10 = 2.8 goals

Since there are 10 valuesi.e., n = 10 (even).

So the median is the mean of � ��

n th

2term

and � �� n th

12

term

= mean of � ��

th102

term and � ��

th101

2term

= mean of 5th term and 6th term

� Median = 12

(3 + 3) = 62

= 3 goals.

Making discrete frequency table, we have

No. of goals Frequency

0 11 12 13 44 25 1

Since, the value 3 has the maximumfrequency, therefore, 3 goals is themode of the data.

'(

��

��

�� (A) As 17 is repeated 6 times, thefrequency of 17 is 6.

�� Primary data

�� Sum of the nine numbers = 9 × 8 = 72Let the tenth number = x

Now, � x72

10 = 9 �� x = 18.

�� We have, average = x = � ixn

Here, x = 15 and n = 12

� ix� = 15 × 12 = 180

But 18 is taken at the place of 0

� ix� = 180 is incorrect

Correct ix� = 180 – 18 + 0 = 162

� Correct average = 16212

= 13.5.

�� Let the numbers be x1, x2, x3, ......., x21.

Mean x = x x x x1 2 3 21......

21+ + + +

= 15

� x1 + x2 + x3 + ....... + x21 = 315

If each observation is multiplied by 2,then the observations become

2x1, 2x2, 2x3, ...... 2x21

New mean = x x x x1 2 3 212 2 2 ...... 2

21+ + + +

= x x x x1 2 3 212( ...... )

21+ + + +

= 2 × 31521

= 2 × 15

= 30Thus, the new mean is 30.

�� MarksTally Marks Frequency

(Class intervals)

0-10 � 110-20 �� 420-30 �� 3

30-40 IIII � 640-50 IIII �� 850-60 IIII �� 760-70 � 1

'((i) Looking at the numbers, we find that56 occurs maximum number of times,i.e., 3 times.� Modal age = 56 years.

100 ��

(ii) Re-writing the values after replacingany one 56 by 65, we get48, 42, 47, 48, 65, 56, 65, 56, 65, 60.� Modal age = 65 years.

�� Salary No. of workersfixi(in �) (xi) ( fi)

3000 16 480004000 12 480005000 10 500006000 8 480007000 6 420008000 4 320009000 3 27000

10000 1 10000

if∑ = 60 i if x∑ = 305000

Now, x = i i

i

f x

f∑∑

= 30500060

= � 5083.33.

'(

x = 1 2 3 4 5+ + + +5

x x x x x

� x1 + x2 + x3 + x4 + x5 = 5 xNow,

� x1 – x + x2 – x + x3 – x + x4

– x + x5 – x = 0

� (x1 – x ) + (x2 – x ) + (x3 – x ) + (x4 – x )

+ (x5 – x ) = 0..� ��$��

� Mean = ix

n∑

=

� � � � � �

� � �

17 16 25 23 22 23 2825 25 23

10

=22710

= 22.7

Arranging the observations in ascendingorder, we get16, 17, 22, 23, 23, 23, 25, 25, 25, 28.

Here n = 10 (even)

� Median = 12

� � ��

th10value of term

2

��

th10value of 1 term

2

= 12

(value of 5th term

+ value of 6th term)

= 23 23

2+

= 23

Mode = 3 median – 2 mean= 3 × 23 – 2 × 22.7= 69 – 45.4 = 23.6.

��

��

�� (B) Average speed during 5.00 hrs– 7.00 hrs

= 40 + 502

= 902

= 45 km/hr.

�� The speed 60 km /hr occurs maximumnumber of times, that is, 3 times. So,the modal speed of the car is 60 km /hr.

�� The average speed between 5 hrs and9 hrs

= 40 + 602

= 1002

= 50 km / hr.

�� Growth in plant 1 = 17 – 12 = 5 cmGrowth in plant 2 = 19.5 – 15.5 = 4 cmGrowth in plant 3 = 17.5 – 12.5 = 5 cmGrowth in plant 4 = 22 – 18 = 4 cmGrowth in plant 5 = 23 – 17.5 = 5.5 cmNow, mean growth of the plants

= 5 + 4 + 5 + 4 + 5.55

= 23.55

= 4.7 cm.

��

101��

�� The observations are 42, 43, 44, 44,(2x + 3), 45, 45, 46, 47.Since, the number of observations is 9(odd).

� Median = value of ��

n th12

term

= value of ��

th9 12

term

= value of 5th term� Median = 2x + 3� 45 = 2x + 3� 2x = 45 – 3 = 42� 2x = 42� x = 21.So, the observations are42, 43, 44, 44, 45, 45, 45, 46, 47.The number 45 occurs maximum (3)times.� Mode of the data = 45.Hence, x = 21 and mode = 45.

� Mean = Sum of all observationsNumber of observations

= ixn�

� 14 = 11+15+17+( +1)+19+( – 2)+37

x x

� 98 = 64 + 2x

� 2x = 98 – 64 = 34� x = 17� x + 1 = 17 + 1 = 18 and x – 2 = 17 – 2 = 15Therefore, given observations are 11, 15,17, 18, 19, 15, 3. The number 15 occursmaximum number of times, i.e., 2 times.� Mode of the given data = 15Hence, x = 17 and mode = 15.

� ��

�� (i) The required frequency distribution

Table is given below:

Date Tally Marks Frequency

0-5 �� 35-10 �� 1110-15 � 115-20 �� 420-25 �� 625-30 �� 2

(ii) Range = Highest observation– Least observation

= 29 – 1 = 28.(iii) Blood donor, Helpful, Thankful,

Grateful.

��

��

Practice Paper –1

SECTION-A

1. Infinite number of solutions as graph ofa linear equation in two variables is astraight line, there are infinite numberof points on a line and every point onthe line is a solution of the correspondingequation.

2. Let AD be the median of the ΔABC⇒ BD = CD ...(i)Draw AE ⊥ BC.

ar(ΔADB) =12

BD × AE ...(ii)

⇒ ar(ΔACD) =12

CD × AE ...(iii)

From equations (i), (ii) and (iii), we get

ar(ΔADB) = ar(ΔACD)

Thus, the median of a triangle divides itinto two triangles of equal area.

3. ��

� � � � � � � � � � � � � � � �!��

�!��

�

4. "� #��$� �� %�� &��

∴ '&��%��

!� ( ) �

!�� * ��+��

!��,��

SECTION-B

5. Since area of a triangle

= 12

base × corresponding altitude

= 12

× 8 × 5 cm2 = 20 cm2.

�� %��%��$��

�� -�� -��&��

�� . * /0 ! �/. * �1⇒ �.*�2 ! �� * 3��4 ! /. ! �� ⇒ �. !��3� ��

8. '��%��5��

= Maximum height – Minimum height

= 160 cm – 138 cm = 22 cm.

9. 6�� %��&��!��7�+�∴� ��!��!��%��5�&��⇒ ��!��7�+��!��!�� ⇒ �� !��7�+ ⇒��!��8�+�!��

⇒ � !�53�

10. 9��5��

:�� x !x x x x1 2 3 21......

21+ + + +

!��

⇒ �� 7�� 7��7� ��7��!��;%� � �� 5�� -� ��5��

��

�$��!�x x x x1 2 3 212 2 2 ...... 2

21+ + + +

��

103��

!�x x x x1 2 3 212( ...... )

21+ + + +

!� ��*� 31521

�!��*��!��2

<��$��2�

SECTION-C

11. � � � � � � � � � � ��!��!��22� ��/�� - !��

!�� * �222 ��

! ��22 ��

⇒ π� �� !��22

⇒ ��

� �� * �22 !��22

⇒ �� !�=>(?? @AA =??

�

�

⇒ �� !�+ * +⇒ � !�+� ��%��

!�<��%� ��%�� -��!��π�� 7 �

!�� * ��

* + * �+ 7 �22 ��

!�� * �2+� ��!��+23� ��

!�(@?B=????

��!�2��+23��

12. (i) From the table it is clear thatnumber of persons ‘aged 60’

= 16090

and number of persons ‘aged 61’

= 11490

∴ Number of persons of ‘aged 60’ whowill die within a year

= 16090 – 11490= 4600

Therefore, P(a person ‘aged 60’ of dyingwithin a year)

= 460016090

= 4601609

.

(ii) Number of persons ‘aged 61’ = 11490

Number of persons ‘aged 65’ = 2320

P(a person ‘aged 61’ will live for 4 years)

= 2320

11490

= 232

1149.

OR

Total number of girls = 250

Number of girls who like coffee = 115

Number of girls who dislike coffee = 135

(i) P (selected girl likes coffee)

=Number of girls who like coffee

Total number of girls

= =1150.46

250

(ii) P (selected girl dislikes coffee)

=Number of girls who dislike coffee

Total number of girls

= =1350.54

250.

13. Fare for the first km = � 10

Remaining distance = (x – 1) km

Fare for subsequent distance

= � 6 per km

Fare for subsequent (x – 1) km

= ��(x – 1) × 6

= ��(6x – 6)

Total fare (y) = 10 + 6x – 6

= 6x + 4

or 6x – y + 4 = 0

Let us form a solution table.

x 1 2 4 5

y 10 16 28 34

��

Distance-Fare Graph

14. Radius of the sphere = 5 cm

∴ Its surface area = 4π(radius)2

= 4 × π × (5)2 cm2

Let slant height and height of the conebe l and h respectively.

Its radius = 4 cm

∴ Its C.S.A. = πrl

= π × 4 × l cm2

According to the question,

4π(5)2 = 5π × 4 × l

⇒ l = 5 cm

l2 = h2 + r2

⇒ (5)2 = h2 + 42

⇒ 52 – 42 = h2

⇒ h2 = 9

⇒ h = 3 cm

Thus, height of the cone = 3 cm

Volume of the cone =13

πr2h

= 13

×227

× 4 × 4 × 3 cm3

= ×22 167

cm3 = 3527

cm3

= 50.29 cm3 (approximately).

ORLet radii of two spheres be r1 and r2respectively. Let the volumes of twospheres be V1 and V2 respectively, then

1

2

VV

= 6427

=π

π

31

32

4r

34

r3

⇒ �3

132

6427

r

r=

( )( )

3

3

4

3

⇒ �1

2

43

rr

⇒ r1 =24

3r

... (i)

Now, r1 + r2 = 42

⇒ 22

43r

r� = 42 [From (i)]

⇒ 273r

= 42

⇒ r2 = 18 cm and r1 = 24 cm

Hence, the radii of two spheres are24 cm and 18 cm.

15. ∠C4/ 7 ��2D�! �32D

(Interior angles on same side oftransversal as PQ || CB and QB is

transversal)

⇒ ∠QBC = 60°

∠ABC + y = 180°

(Cointerior angles)

⇒ x + ∠QBC + y = 180°

⇒ x + y = 120°.

105��

OR

16.

��

!�

��

�!��

��

!�164

�

⇒� � �E��%�� !164

5��%

��

<��%��&��%�� 164

�

20. Let the angles of the quadrilateral be 3x,4x, 4x and 7x. Using angle sum propertyof a quadrilateral, we have

3x + 4x + 4x + 7x = 360° ⇒ 18x = 360°

∴ x = °360

18 = 20°

Therefore, four angles of the quadrilat-eral are

3x = 3 × 20° = 60°4x = 4 × 20° = 80°4x = 4 × 20° = 80°

and 7x = 7 × 20° = 140°.

SECTION-D

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∠�94 !�∠/:. �1� ��,2D �4 !�/. G4-�� H49 !�.: G4-�� H

<��-�'I6� �� $��5Δ�94 ≅ Δ/:.

17. Given:

arc �A X B = 12

arc �BYC

⇒ � �( )AXB !�12

m �( )BYC

⇒ ∠AOB =12

∠BOC

Also, AOC is diameter,

⇒ m(arc �ABC) = 180°

⇒ ∠�F4�7�∠4F/ !��32D

⇒�12

∠4F/�7�∠4F/ !��32D

⇒32

∠4F/ !��32D

∴ ∠4F/ !�23

�*��32D�!��2D

�� :��!� ��

�!� ��

⇒��!��

� �

⇒ ,3 !��7��⇒ �� !�,3�8��!��⇒��!��+∴ � 7 � ! �+ 7 �! �3�� 8� ! �+ 8 � !��<��%�� 5�� 5�� +�� 3�� ,�� <�� J��%��∴ :��%��5��!��I� ��!��+��!��

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⇒ ��Δ�F4 !��Δ.F/ �� Δ.F/ !��Δ�F4 �� Δ.F/ !��Δ�F4 ⇒ ��Δ.F/ �7��Δ/F4

!��Δ�F4 �7��Δ/F4 ⇒ ��Δ./4 !��Δ�/4 �

��

22.

l(AB) of protractor = π r = 227

× 14

= 44 cmCone is formed using semicircularsheet.∴ Let radius of cone = R

Base ring of cone = l(AB)2πR = 44

R =×

×44 72 22

= 7 cm

l of cone = r of protractor = 14 cm

Depth = h = −2 2l R = −2 214 7

= −2 27 (2 1) = 7 × 3 cm

Volume = π 213

r h

= 13

× 227

× 7 × 7 × 7 3

=1078 3

3cm3 = 622.16 cm3.

23. 9��4/.��&�� /��4.�∠��!�∠4�!�∠/�!�∠.�!�,2D

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⇒ Δ.�4��

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�� Total outcomes = 8(i) Try yourself(ii) At least 2 heads: HHT, HTH, THH,HHH

∴ Probability = (B

= =A

.

(iii) At most 2 heads: TTT, HTT, THT,TTH, HHT, HTH, THH.

∴ Probability = @B

.

(iv) Try yourself

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27. Join PR and QS.

Now PQ || RS and PQ = RS (Given)

⇒ PQSR is a parallelogram

⇒ PR || QS and PR = QS ...(i)

PN || QM

⇒ ∠1 + ∠2 + ∠5 = 180°

PR || QS

⇒ ∠6 + ∠5 + ∠1 = 180°

∴ ∠1 + ∠2 + ∠5 = ∠6 + ∠5 + ∠1

⇒ ∠2 = ∠6

Similarly, ∠3 = ∠4

Also PR = QS [From (i)]

⇒ ΔPNR ≅ ΔQMS (By ASA)

⇒ PN = QM and RN = SM (CPCT)

Now PN || QM (Given)

PN = QM

⇒ PNMQ is a parallelogram.

⇒ PQ = MN and PQ || MN.Hence proved.

28. Let us suppose that the two circlesintersect at three different points say atP, Q and R. This means P, Q and R arenot collinear.

We know that through three non-collinear points we can draw one andonly one circle.

⇒ There cannot be two different circlespassing through the points P, Q and R.

Thus, two circles cannot intersect atmore than two different points.

ORSuppose ∠1 represents ∠COD and ∠2represents ∠ABD.

We know that an angle subtended by anarc of a circle at the centre is double theangle subtended by the same arc at apoint on the remaining part of the circle.

∴ ∠AOD = 2∠2 ...(i)

(Corresponding arc is minor arc AD)

And ∠1 = 2y ...(ii)

(Corresponding arc is minor arc CD)

∠AOD = 90° (Given)

But from equation (i),

2∠2 = 90°

⇒ ∠2 = 45° ...(iii)

Since AM is perpendicular to both ODand BC, so OD is parallel to BC.

Now, OD || BC and OC is transversal

∴ ∠1 = 30° ...(iv)

(Alternative interior angles)

From equations (ii) and (iv),

2y = 30°

⇒ y = 15° ...(v)

In ΔABM,x + ∠2 + y + 90° = 180°

(Angle sum property of a triangle)

⇒ x + 45° + 15° + 90° = 180°[Using (iii) and (v)]

⇒ x = 180° – 150° ⇒ x = 30°

Hence, x = 30°, y = 15°.

29. Rain water falls on a flat rectangularsurface whose

length (l) = 6 m = 600 cm and breadth (b) = 4 m = 400 cm

Level of rain fall (h) = 1 cmWater collected on this surface forms acuboid.

��

∴ Volume of water collected = l × b × h = 600 × 400 × 1 cm3 = 240000 cm3

This water is transferred into cylindricalvessel.Radius of cylindrical vessel = r = 20 cm.Let height of water collected in thecylinder be h cm.∴ Volume of water in cylindrical vessel

= πr2h= 3.14 × (20)2 × h cm3

Now, 3.14 × 400 × h = 240000

⇒ h =×

2400003.14 400

=6003.14

= 191 cm (approx.).

30. Given frequency distribution has the classintervals of unequal width. Therefore,we will have to make changes in the

lengths of the rectangles in the histo-gram so that areas of the rectangles areproportional to the frequencies. Thus,we have the following table:

Marks Frequency Width Length

100-150 60 50 5050

× 60

= 60

150-200 100 505050

× 100

= 100

200-300 100 100 50100

× 100

= 50

300-500 80 20050200

× 80

= 20

500-800 180 300 50300

× 180= 30

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Practice Paper – 2

SECTION-A

1. Putting x = 2 and y = 0 in 2x + 3y = k, wehave

2 × 2 + 3 × 0 = k ⇒ k = 4.

2. �� S�� $��5

,��3��+��+��

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��$��!��

!� ��

��

��! �� !��

3. As length of the longest pole

= diagonal of the room

= + +2 2 2l b h = + +2 2 210 10 5

= + +100 100 25

= 225 = 15 m.

4. In a sample study, total number of people = 642.

The number of persons who have high school certificate = 514

∴ Required probability = 514642

= 0.8.

SECTION-B

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6. Lower limit of a class

= Mid value – 12

class size

= 10 – 12

× 6 = 10 – 3 = 7.

7. Let the quadrilateral formed by joiningthe mid-points of the given rectangleABCD be PQRS such that

AB = 8 cm, BC = 6 cm

PQRS is a rhombus with side 5 cm, asAPS and DRS are right-angled triangles.

⇒ PS = 5 cm = PQ = SR = RC.

Also PR = BC = 6 cm and SQ = AB = 8 cm

(� Opposite sides of parallelogram

are equal)

ar(rhombus PQRS) =12

× PR × SQ

=12

× 6 × 8 = 24 cm2.

OR

In right ΔPQR,

PR = 2 217 – 8 = 289 – 64 = 15 cm

In right ΔPSR,

SR = 2 215 – 9 = 225 – 81 = 12 cm

ar(quadrilateral PQRS)= ar(ΔPQR) + ar(ΔPSR)

= 1 1QR PR PS SR

2 2� � � � �

=1 1

8 15 9 122 2� � � � �

= 60 + 54 = 114 cm2.

��

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9. Since diagonals of a rhombus are per-pendicular bisectors of each other.

Therefore, AO =12

AC =12

× 24 = 12 cm

And BO = 12

BD = 12

× 10 cm = 5 cm

ΔAOB is a right triangle.

⇒ AB2 = OA2 + OB2

= 122 + 52 = 169

⇒ AB2 = 13 cm

∴ Side of the rhombus is 13 cm.

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SECTION-C

11. Steps of construction:

1. Draw line segment BC = 5 cm andmake ∠CBX = 60°.

2. Mark the point D on the ray BX suchthat BD = 7.5 cm.

3. Join CD and draw perpendicularbisector of it to meet BD at A.

4. Join AC.

5. Such formed ΔABC is the requiredtriangle.

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� ��

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13. Number of observations = 10 which isan even number.∴ Median

=th th1 10 10

observation 12 2 2

observation

��

⇒ 63 =12

(5th observation + 6th

observation)

113��

63 =12

[x + x + 2]

⇒ 126 = 2x + 2

⇒ 2x = 124 ⇒ x = 62.

OR

Sum of given marks = Σxi =

80 + 22 + 30 + 90 + 90 + 53 + 50 + 56 + 60+ 72 + 90 + 88 + 88 + 72 + 62 + 60 + 46

+ 40 + 36 + 40 + 80 + 33 + 69 + 40 + 66

+ 92 + 50 + 48 + 56 + 92 = 1851

Number of students = 30

Now, mean = x = 1851

30 = 61.7.

14.

Yes, lines AP and BQ are parallel toeach other.

Verification:As AP ⊥ AB and BQ ⊥ AB,

∴ ∠PAB = 90° and ∠QBA = 90°

⇒ ∠PAB + ∠QBA = 180°

Also AB is a transversal and cointeriorangles∴ AP || BQ are supplementary.

15. In ΔPQR, PQ = PR (Given)∠PRQ = ∠PQR = 35°

(Angles opposite to equalsides of a triangle are equal)

⇒ ∠QPR = 180° – (∠PQR + ∠PRQ)

(ASP of a triangle)

⇒ ∠QPR = 180° – (35° + 35°)

= 180° – 70° = 110°.

∠QPR and ∠QSR are in the samesegment of the circle. Since, angles inthe same segment of a circle are equal.

∴ ∠QSR = ∠QPR = 110°.

OR

We are given a circle with centre O,radius OA = 13 cm, chord AB and thedistance of AB from O is 12 cm, i.e.,OM = 12 cm. M divides AB in the ratio

AM : MB = 1 : 1.

∴ AB = 2 AM ... (i)

Since OM ⊥ AB, therefore, ΔAOM is aright-angled triangle.

∴ OA2 = AM2 + OM2

(Using Pythagoras theorem)

⇒ 132 = AM2 + 122

⇒ AM2 = 132 – 122 = 169 – 144 = 25

⇒ AM = 5 cm ... (ii)

From equations (i) and (ii), we have

AB = 2 × 5, i.e., AB = 10 cm

Hence, the length of the chord is 10 cm.

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17. Given: ABCD is a parallelogram.

To prove: ar(ΔAPB) = ar(ΔBQC)Proof: In figure, ΔAPB and �gm ABCDhave same base AB and lie between sameparallels AB and DC.

∴ ar(ΔAPB) = 12

ar(�gm ABCD) …(i)

Again, ΔBQC and �gm ABCD have samebase BC and lie between same parallelsAD and BC.

∴ ar(ΔBQC) = 12

ar(�gm ABCD) …(ii)

From equations (i) and (ii), we get

ar(ΔAPB) = ar(ΔBQC). Proved

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20. Total number of tossing the two coins= 100

Number of getting two heads = 25

115��

Number of getting one head = 40 Number of getting no head = 35

(i) P(two heads) = 25

100 =

14

or 0.25

(ii) P(one head) = =40 2100 5

or 0.4

(iii) P(no head) = =35 7100 20

or 0.35.

SECTION-D

21. Let the ratio constant be x then theradius = 3x and the height = 4x

Volume = π 213

r h

⇒ 301.44 =13

× 3.14 × (3x)2 × 4x

⇒ ×× ×

301.44 33.14 9 4

= x3 ⇒ x3 = 8

⇒ x = 2 cm. ∴ Radius = 3x = 6 cmHeight = 4x = 8 cm

Slant height = +2 26 8 = +36 64

= 100 = 10 cm.Curved surface area = πrl= 3.14 × 6 × 10 cm2 = 188.4 cm2.

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23. x f fx

10 6 6015 8 12020 p 20p25 10 25030 6 180

Σf = 30 + p Σf(x) = 610 + 20p

Mean = 20.2 = fxf

∑∑

⇒610 20

30p

p++

= 20.2

⇒ 610 + 20p = 20.2 (30 + p)⇒ 610 + 20p = 606.0 + 20.2p⇒ 4 = 20.2p – 20p⇒ 4 = 0.2p

⇒ 40.2

= p ⇒ p = 20.

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26. To draw graph of a line, we need at leasttwo solutions of its corresponding equa-tion. The equation x = 3 is satisfied byany value of y. So, two solutions of thisequation are:

x = 3, y = 2 and x = 3, y = 4

And for the equation y = – 2,

x = 0, y = – 2 and x = 4, y = – 2.

Each solution of the equation y = 2x ison the way that the value of y is twicethe value of x. So, any two of thesolutions of y = 2x are:

x = 2, y = 4 and x = 0, y = 0.

Let us draw the graph of lines x = 3,

117��

y = – 2 and y = 2x on the same set ofaxes.

These lines form a triangle ABC.

Coordinates of vertices are A(– 1, – 2),B(3, – 2) and C(3, 6).

27. (i) We know that triangles on the samebase and between the same parallelsare equal in area.

In the given figure, ΔABD and ΔABCare on the same base AB and between

the same parallels AB and DC, so theymust have equal area.

i.e., ar(ΔABD) = ar(ΔABC) ...(i)

(ii) Now, subtracting ar(ΔAOB) fromboth the sides of (i), we get

⇒ ar(ΔABD) – ar(ΔAOB)

= ar(ΔABC) – ar(ΔAOB)

⇒ ar(ΔAOD) = ar(ΔBOC).

Hence proved.

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��

29. Radius of the well = r =�82

= 4 m

Depth of the well at which the earth istaken out h = 14 m.

Clearly, the earth is taken out in theform of cylinder.∴ Volume of the earth dug out

= πr2h= π × 42 × 14= 16 × 14 × π m3

Width of the embankment = 3 m Area of the embankment = π (72 – 42)

= 33π m2

Let height of the embankment be h metres.Then, volume of the embankment

= 33πh m3.The volume of the embankment must besame as the earth dug out.∴ 33πh = 16 × 14π

⇒ h =��

�� ∴ h = 6.79 m

Hence, the height of embankment is 6.79 m.OR

Internal diameter of the hemisphericalwater tank = 14 m

∴ Internal radius (r) =142

= 7 m

Volume of the tank =23

πr3

= 23

×227

× (7)3 = 44 49

3�

= 718.67 m3.

∴ Capacity of the water tank = 718.67 kl(... 1 m3 = 1000 l or 1 kl)

Since the tank contains 50 kl of water.Water is pumped into the tank to fill toits capacity is (718.67 – 50 =) 668.67 kl,i.e., Volume of required water is 668.67 m3.

30. Draw perpendicular bisectors OL of ABand OM of CD, O being the centre ofthe circle. Since AB || CD. O, M, L arecollinear. Suppose OM = x cm, thenOL = (x + 3) cm. Let r be the radius forthe circle. We have, OD = OB = r.

We know longer chord is nearer to thecircle.

From right triangles ODM and OBLOD2 = OM2 + MD2 andOB2 = OL2 + LB2

(By Pythagoras theorem)

⇒ r2 = x2 +�211

2⎛ ⎞⎜ ⎟⎝ ⎠

...(i)

and r2 = (x + 3)2 +�25

2⎛ ⎞⎜ ⎟⎝ ⎠

...(ii)

� � �


x2 + 121

4= x2 + 9 + 6x +�

254

⇒ 6x =121

4 –�

254

�– 9

⇒ 6x =964

– 9��⇒ 6x = 24 – 9

119��

⇒ 6x = 15�⇒ x =�156

⇒� x =�52

From (i), r2 =�25

2⎛ ⎞⎜ ⎟⎝ ⎠

�+�211

2⎛ ⎞⎜ ⎟⎝ ⎠

= 254

�+� 1214

�=�146

4�= 36.5

∴ r = ± 36.5 = ± 6.04

But r cannot be negative.

∴ r = 6.04 cm.

31. (i) and (ii) try yourself

Practice Paper –3

SECTION-A1. Given: AOB is the diameter of the circle

⇒ ∠ACB = 90° (Angle in a semicircle)and AC = BC (Given)⇒ ∠CAB = ∠CBA = x (say) (... Angles opposite to equal sides are equal)

⇒ x + x + 90° = 180° (ASP)⇒ 2x = 90° ⇒ x = 45°⇒ ∠CAB = 45°.

2. (c) Line 2x + 3y = 6 will meet the x-axisif y = 0.Putting y = 0, we have

2x = 6 ⇒ x = 3Thus, coordinates of the required pointare (3, 0).

3. 9��%�� !��9�� !��<��$�� !��

;��-��5��E� ��%��

!��

π��

π��

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π�� !��

π�� * ��

!��

π��

�� 5��

!� � �

�

� ��

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� �

�

� ��

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� �

� �

�

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SECTION-B

5. ∠��! ∠.�!��2D

��

��D 7 �2D 7�� !��32D ��6�

⇒ � !�+�D�

6. Mean is 9

∴ x x x x x3 5 7 105

� � � � � � � �= 9

⇒ 5x + 25 = 45

⇒ 5x = 20 ⇒ x = 4

⇒ Last three observations are

4 + 5 = 9, 4 + 7 = 11, 4 + 10 = 14

∴ Mean of last three observations

= � �9 11 143

= 343

= 111

3.

��

7. Σfi = 7 + 8 + 10 + 15 + 10 = 50

(i) P(being 30) = Favourable outcomes

if∑

= 10 150 5

= or 0.2

(ii) P(being 25) = Favourable outcomes

if∑

= 15 350 10

= or 0.3

OR

No, since the number of trials in whichthe event can happen cannot be greaterthan the total number of trials.

8. True, as AC – AB < BC,i.e., AC < AB + BC.

9. �� [Q��!�8 �� 2� �

<��%�� ! +

��%�%�5�� !�

∴ � �[�[Q�� !��

�

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SECTION-C

11. � .��!��!��

�

⇒��!��

!��

!��

��!��2 ��

;�� 5��%� ��%��$��! �π��

! � *��

*��

* �2��

! ��2�� /��%��!��2�*��2

�� !��22�12. Let the equation of the line be y = a

which is parallel to x-axis. Where a isarbitrary constant.According to question,

a = – 3 (... Graph of the line is below to x-axis)

Putting a = – 3 in y = a, we havey = – 3

121��

Hence the graph can be shown as inthe figure.

OR

The given equation is 2x + 1 = x – 3.

Transposing, 2x – x = – 3 – 1

⇒ x = – 4.

(i) Since x = – 4 represents a point onthe number line so it is a solution of thegiven equation.

(ii) Consider x = – 4 is of the form x = a.Graph of such equation is a line parallelto y-axis in a cartesian plane. Therefore,many points that lie on the line thesolution of the give equation.

13. Let ∠A = 3x, ∠B = 7x, ∠C = 6x, ∠D = 4x.Also, ∠A + ∠B + ∠C + ∠D = 360°

3x + 7x + 6x + 4x = 360°20x = 360°

∴ x = 18°∠A + ∠B = 3x + 7x = 10x = 10 × 18°

= 180°and ∠C + ∠D = 6x + 4x = 10x

= 10 × 18° = 180°∴ AD || BC.

14. 9��%��'��%��.��%��

!��

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⇒ �� !��

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⇒ �� !��

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15. Let ABCD be the parallelogram andABEF be the rectangle on the samebase AB and have equal areas.⇒ They lie between same parallels AB = CD (... Opposite sides of ||gm)

AB = EF (... Opposite sides ofrectangle)

⇒ CD = EF ... (i)

Also BE < BC ... (ii) (... In righttriangle BEC, BC is hypotenuse)

Similarly, AF < AD ...(iii)Adding (ii) and (iii), we get

BE + AF < AD + BCBE + AF + AB < AD + BC + AB

(Adding AB to both sides)BE + AF + AB + EF < AD + BC + AB + CD

[Using (i)](� AB = CD, AB = EF ⇒ CD = EF)

⇒ Perimeter of rectangle < Perimeterof the parallelogram.

ORL is the mid-point of BC,

⇒ BL = CL ...(i)

DQ || AB and CB is transversal

⇒ ∠QCL = ∠LBP ...(ii)(Alternate interior angles)

Now in ΔQCL and ΔPBL,

∠QLC = ∠PLB

(Vertically opposite

angles)

CL = BL [Using (i)]∠QCL = ∠LBP [Using (ii)]

∴ ΔQCL ≅ ΔPBL

(ASA criterion of congruence)

��

∴ ar(ΔQC L) = ar(ΔPBL) ...(iii)

(Congruence triangles have equal area)Now adding ar(APLCD) to both sides in(iii), we getar(APLCD) + ar(QCL) = ar(APLCD)

+ ar(PBL)⇒ ar(APQD) = ar(ABCD)

Hence proved.16. �� !

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17. :��!� ��

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⇒��!��

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�� 8 � !�+ 8 �!��<��%��5��5��+��3��,��<�� J��%��

∴:��%��5��!��I� ��!��+��!��

18. 9��F1�!��⇒ �F��!�F/�!�F4�!�� 7 � � ��N�� -�� <�� ΔF�1�$��

��7�� !��7��

⇒ �� 7 �� 7 �� !�� 7 ��⇒ �� !��8,!+

⇒ ��!�76

!��16� �

⇒ F��!��7��!� � �� 1

1 36

��!��16� ��

19. ��

<��&�� +��

20. Volume of hemisphere = 323

r�

⇒539

6=

r323� ⇒ πr3 =

5392 2�

⇒ r3 = 539 7

2 2 22�

� �

⇒ r3 = 49 7

2 2 2�

� �

⇒ r = 37 7 72 2 2� �

� �

⇒ r =72

cm

Its curved surface area = 2πr2

= 2 ×222 7

7 2� �� = 77 cm2.

OR

On revolving the given right triangleabout the side 8 cm, we find a rightcircular cone with

Radius = r = 6 cm, height = h = 8 cmAnd slant height = l = 10 cm.

123��

Therefore, volume of the cone

= 13

πr2h = 13

π × 6 × 6 × 8

= 96 π cm3

and curved surface = πrl

= π × 6 × 10

= 60 π cm2.

SECTION-D

21.

(i) From the table given in question,– 3 + 5 = – 1 + 3 = 1 + 1 = 2So the required equation is

x + y = 2(ii) Substituting x = – 2, y = p in

x + y = 2, we get– 2 + p = 2

⇒ p = 4Substituting x = 0, y = q in x + y = 2,we get

0 + q = 2 ⇒ q = 2

Thus, p = 4, q = 2.

22. Given: A parallelogram ABCD.

E is the mid-point of AB

F is the mid-point of CD

To prove: AF and CE trisect the diagonalBD, i.e., DM = MN = BN.

Construction: Join AF and CE

Proof: AB = CD and AB || CD

(Opposite sides of a parallelogram)

⇒12

AB = 12

CD

⇒ AE = CF and AE || CF

Hence quadrilateral AFCE is a parallelo-gram.

∴ AF || EC (Opposite sides of aparallelogram)

In ΔABM, E is the mid-point of AB andEN || AM

⇒ N is the mid-point of BM.

(Converse of mid-point theorem)

∴ BN = NM ... (i)

Also in ΔDNC, F is the mid-point of CDand MF || NC.

∴ M must be the mid-point of DN

or DM = MN ... (ii)

From (i) and (ii) we conclude thatDM = MN = NB.

⇒ M and N are the points of trisection ofBD.

Hence, the segments AF and CE trisectthe diagonal BD.

23. Inner radius of the well = r

=102

m

= 5 mAnd depth = h = 14 m

Volume of earth taken out = πr2h

Outer radius = R = (5 + 5) m = 10 m

��

Area of ring of the embankment

= π (R2 – r2) = 227

(10 – 5) (10 + 5)

=227

× 5 × 15 m2.

Let the height of the embankment = HThen volume of the embankment

= π(R2 – r2)HNow, volume of the embankment= volume of the earth taken out⇒ π(R2 – r2) H = πr2h

⇒ (102 – 52) H = 52 × 14

⇒ H =5 5 14

15 5� �

�

⇒ H = 143

⇒ H = 423

m.

ORLet r be the radius of the base of thecone.Then, area of the base = πr2

= 28.26 sq. cm.⇒ 3.14 r2 = 28.26

⇒ r2 = 28.263.14

= 1413157

⇒ r2 = 9⇒ r = 3, h = 4 m (Given)

∴ l = r h2 2+ = 2 23 4+

= 9 16+ = 25 = 5 cm.

Volume = 13�r2h =

13

× 3.14 × 9 × 4

= 3.14 × 12 cu. cm= 37.68 cu. cm.

Curved surface area = �rl= 3.14 × 3 × 5 = 3.14 × 15= 47.10 sq. cmNo, the two are not numerically equal.

24. Radius (r) of base = 3 cm.Height (h) of cylinder = 10.5 cm.Surface area of the pen holders= area of the base + curved surface area

= πr2 + 2πrh = πr(r + 2h)

= π(3) [3 + 2(10.5)] = 3π[3 + 21]= 3π[24] cm2 = 72π cm2.

Area of the cardboard required to make35 pen holders

= 35 × 72π cm2 = 35 × 72 ×227

cm2

= 7920 cm2.25. ��%��-��-��!��

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28. We are given an arc PQ of a circle whichis subtending angles POQ at the centreO and PAQ at a point A on the remainingpart of the circle. We need to prove that∠POQ = 2∠PAQ.Consider the three different cases asgiven in figures. In figure (i), arc PQ isminor; in figure (ii), arc PQ is asemicircle and in figure (iii), arc PQ ismajor.Let us begin by joining AO and extendingit to a point B.

(i) (ii) (iii)

In all the cases,

∠BOQ = ∠OAQ + ∠AQO

(Exterior angle theorem)Also in ΔOAQ, OA = OQ

(Radii of a circle)

Therefore, ∠OAQ = ∠OQAThis gives ∠BOQ = 2 ∠OAQ ... (i)Similarly, ∠BOP = 2 ∠OAP ... (ii)Adding equations (i) and (ii), we get

∠BOP + ∠BOQ = 2 (∠OAP + ∠OAQ)⇒ ∠POQ = 2 ∠PAQ.

Hence proved.In given figure, using above theorem

∠BAD =12

∠BOD

��

⇒ ∠BAD =12

× 136° = 68°

Now ABCD is a cyclic quadrilateral⇒ ∠A + ∠C = 180°

(Opposite angles of a cyclicquadrilateral are supplementary)

⇒ 68° + ∠C = 180°⇒ ∠C = 180° – 68°⇒ ∠BCD = 112°.

ORGiven: P is any point on circumcircle ofΔABC, PL, PM and PN are perpendi-culars from P on BC, AC and ABrespectively.

To prove: Points L, M

and N are collinear.

Construction: Join

PA, PC, MN and ML.

Proof: Consider

quadrilateral AMPN.

Since PM ⊥ AC and PN ⊥ AB

∴ ∠PMA + ∠ANP = 90° + 90° = 180°

So AMPN is a cyclic quadrilateral.(If opposite angles of a quadrilateral are

supplementary, then it is cyclic)∴ ∠PMN = ∠PAN ...(i)

(Angles in the same segment)Now, consider quadrilateral PCLM.

∠PMC = ∠PLC = 90° (Construction)

Since line segment joining P and Csubtends equal angles at points M and Llying on its same side.∴ Points P, M, L and C are concyclic.� PMLC is a cyclic quadrilateral� ∠PML + ∠PCL = 180° ...(ii)Clearly, PABC is a cyclic quadrilateral

and NAB is a straight line.� ∠PCB = ∠PAN

(Exterior angle of a cyclicquadrilateral is equal

to interior opposite angle)� ∠PCL = ∠PAN ...(iii)From (i) and (iii), we get� ∠PCL = ∠PMN ...(iv)From (ii) and (iv), we get∠PML + ∠PMN = 180°� LMN is a straight line.∴ L, M and N are collinear.

29. Given: In an isosceles triangle ABC, D,E and F as the mid-points of sides BC,CA and AB respectively, AB = AC. ADintersects FE at O.To prove: AD ⊥ EF and AD is bisectedby FE.Construction: Join DE and DF.Proof: D, E, F are the mid-points ofsides BC, AC and AB respectively.Since the segment joining the mid-points of two sides of a triangle isparallel to third side and is half of it,

∴ DE || AB and DE = 12

AB

Also, DF || AC and DF = 12

AC

But AB = AC [Given]

⇒12

AB = 12

AC

⇒ DE = DF

Also, DE || AF and DF || AE

⇒ AFDE is a parallelogram.

In parallelogram AFDE, adjacent sidesDE and DF are equal.

⇒ DEAF is a rhombus.

127��

⇒ Diagonals AD and FE bisect eachother at right angle.

⇒ AD ⊥ FE and AD is bisected by FE.

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2 10×� !

2110

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31. (i) The required frequency distributionTable is given below:

Date Tally Marks Frequency

0-5 ||| 3

5-10 |||| |||| | 11

10-15 | 1

15-20 |||| 4

20-25 |||| | 6

25-30 || 2

(ii) Range = Highest observation – Leastobservation

= 29 – 1 = 28

(iii) Blood donor, Helpful, Thankful,Grateful.

Practice Paper–4

SECTION-A

1. 9��%��!��

E��%��!��,�2��

⇒��

π��!��,�2��⇒ ��

*��

* ��!��,�2�

⇒ �� !�=\(?( ) @

A AA� �

�

⇒ �� !�,��⇒ ��!��

��$��%� ��%��

!��π��!��*��

*��*��!��3� ��

2. As the difference of any two sides of atriangle can’t be more than the third side.

3. ��%��%��%��$��

�. * 4� !��4 * .:

⇒ �.�*�� !��*� ,

⇒ �. !� �� ! ��

��

4. True

SECTION-B

5. OA = OD = 5 cmAB = 8 and OD ⊥ AB

⇒ AC =12

AB = 4 cm

So, ΔOCA is a right triangle

⇒ OC = 2 2OA AC�

= 2 25 4 9 3� � � cm.

CD = OD – OC = 5 – 3 = 2 cm.OR

/�� F� �%� �� R��4��/�� R� �� $�� %�� %�� %��4��4/�

��

4: !��

�4�!��

F: !�4��!��

4/�!�3� �

;��4:F�4F� !�4:��7�F:�

!�� 7 3� ! �� 7 �� ! �22⇒ 4F !��2� �I� ��%�� 2� ��

6. Mean of 5 observations = 30∴ Sum of the 5 observations = 150

Mean of 4 observations = 28∴ Sum of 4 observations = 112∴ Excluded observations = 150 – 112

= 38.7. <��&�� ! +� 7 ��%��-��-

�� %� �� 5�&�� %��%��-��5�-��%��&��

8. Total number of students = 80(i) Number of students getting less than

40 marks = 8 + 16 = 24∴ P(getting less than 40 marks)

= 2480

= 0.3

(ii) Number of students getting 60% ormore marks = 10 + 6 = 16.

∴ P(getting 60% or more marks)

= 1680

= 0.2.

9. ��Δ�4. !12

��Δ�4/ �

�:��%��5��

��$��&��

]]]]]]] ⇒ � !12

�� ⇒��8��! 2�

10. �� $��%��%�� %��%��%��5�

��

�

SECTION-C

11. Since ABCD is a parallelogram BC ||| AD.⇒ 5x = 25°

⇒ x = 5°

Similarly, AB || DC

∴ 6y = 60°

⇒ y = 10°

Hence x = 5°, y = 10°.

We know that the area of a parallelo-gram is twice that of a triangle if boththe figures are on the same base andbetween the same parallels.Here, ΔACD and parallelogram ABCDare on the same base DC and betweenthe same parallels AB and CD.∴ ar(ΔACD) : ar(parallelogram ABCD)

= 1 : 2.

12. ��!�+ �� !�� ⇒ � !��7��!�� 7�+�

!��+� 7 �,�!��⇒ ! �� ⇒ � �!�� 6��&��%��

!��

* + * �� ! ��2� ��

6��&��%��2�� ! �2 * ��2� ��!��22� ��

13. The coordinates of the points lying onthe line parallel to the y-axis, at adistance 4 units from the origin and inthe positive direction of the x-axis areof the form (4, a).Putting x = 4, y = a in the equationx + y = 6, we get4 + a = 6 ⇒ a = 6 – 4 = 2Thus the required point is (4, 2).

ORSince the point (3, 4) lies on the graphof 3y = ax + 7, therefore, x = 3 and y = 4satisfy the equation.⇒ 3 × 4 = a × 3 + 7

129��

⇒ 12 – 7 = 3a ∴ a =53

.

Now, equation will be 3y =53

x + 7

Putting y = – 1, 3(–1) =53

x + 7

⇒ – 3 – 7 =53

x

(Transposing)

⇒ – 10 × 3 = 5x(Cross-multiplication)

∴ x = – 10 × 3

5= –6.

14. Given: ABCD is a �gm in which X andY are the mid-points of opposite sidesAD and BC respectively.

To prove: AP = PQ = QC.

Proof: Since AD and BC are oppositesides of �gm ABCD.

∴ AD = BC and AD�BC

⇒12

AD = 12

BC

and AX or XD ��BY or YC

⇒ XD = BY and XD ��BY

∴ BXDY is a parallelogram.

⇒ XB ��DY

Now, ΔBPC, Y is the mid-point of BC

and YQ ��BP (� XB ��DY proved above)

Using mid-point theorem,Q is the mid-point of CP,i.e., PQ = QC …(i)Similarly, from ΔAQD, P is the mid-point of AQ i.e., AP = PQ …(ii)

From (i) and (ii), AP = PQ = QC.

Hence Proved.

15. K�5��!�� !��

∴�E��%�� -�� !� π��

!�227

25 21× ×

!��

!��2� ��/��5��%� ��%�� -��

!��π�� !��

227

5 21× ×

!�� 2 � � ��

!��2� ��

16. ∠ABC =12

∠AOC

So angle subtended by an arc at thecentre of the circle is double the anglesubtended by it at any other point ofthe circle.

i.e., ∠AOC = 2∠ABC = 2 × 45° = 90°Hence, OA ⊥ OC.

17. "� ��#� %�&�� -� �� %��5��$U

�� <��-�:��#� 0�&�� -

� �� 3 � �, IIII �� IIII �� IIII �� 3�� + �� , ��

<��!��

I��5��J�R��%�&�� -��<��%��%��

��

18. Given: In ΔABC, BC = 7 cm, AB + AC = 13 cm and ∠ABC = 75°.Steps of construction:

1. Draw a ray BX.

2. Cut BC = 7 cm from BX.

3. Construct ∠YBC = 75°.4. With B as centre and radius = 13 cm, draw an arc which cuts BY at D.5. Join DC.6. Draw perpendicular bisector of DC, which intersects BD in A.7. Join AC.

ΔABC is the required triangle.OR

Steps of construction:1. Draw a line segment BC = 3 cm.2. Make ∠RBC = 40° at B.3. Cut BP = 1.5 cm from BR.4. Join C and P.

131��

5. Draw perpendicular bisector ofPC, meeting BR at A.

6. Now, join A and C.Thus, the required triangle ABC isformed.

19. Let the ratio constant be x,then the radius = r = 3xand the height = h = 4x

Volume =13

πr2h

⇒ 301.44 =13

× 3.14 × (3x)2 × 4x

⇒301.44 × 33.14 × 9 × 4

= x3

⇒ x3 = 8⇒ x = 2 cm.∴ Radius = 3x = 6 cm

Height = 4 × 2 = 8 cm

Slant height = r h2 2�

= 2 26 8�

! �36 64

! 100 10 cm.�

OR

C.S.A. = 12320 cm2

Let l be the slant height.

r = 56 cm.

∴ πrl = 12320

⇒ l22

567

� � = 12320 ⇒ l = 1232022 8�

⇒ l = 70 cm

Now, h = l r2 2�

= 2 270 56�

= (70 56)(70 56)� �

= 126 14� = 2 63 2 7� � �

= 2 7 3 3 2 7� � � � �

∴ Height = 2 × 7 × 3 = 42 cm.

20. Let number of girls be x

⇒ Sum of marks obtained by girls = 67x

Let number of boys be y

Sum of marks obtained by boys = 63y

x + y = 80

⇒ y = 80 – x ...(i)

Sum of marks obtained by boys and

girls together = 64.5 × 80

∴ 67x + 63y = 64.5 × 80

⇒ 67x + 63 × 80 – 63x = 64.5 × 80[From (i)]

⇒ 4x = (64.5 – 63) × 80

⇒ 4x = 1.5 × 80

⇒ x = 1.5 × 20

⇒ x = 30

∴ Number of girls = 30

Number of boys = 80 – 30 = 50.

SECTION-D

21. Here the class intervals are in inclusiveform. Therefore, the lower limit and theupper limit both are included in thecorresponding class interval.Total number of students = 30One student is chosen out of 30 studentsat random.

(i) Number of students whose weight isless than or equal to 45 kg

= 9 + 5 + 14 = 28

∴ Probability that his weight is less

than or equal to 45 kg =2830

=1415

(ii) Number of students whose weight isat most 40 kg, that is, 40 kg or less

= 9 + 5 = 14

��

∴ Probability that his weight is atmost 40 kg

=1430

= 7

15(iii) Number of students whose weight is

at most 50 kg, that is, 50 kg or less= 9 + 5 + 14 + 2 = 30

∴ Probability that his weight is atmost 50 kg

=3030

= 1

(iv) Number of students whose weight ismore than 50 kg = 0

∴ Probability that his weight is morethan 50 kg

=030

= 0.

OR

The experimental probability of an eventis given by

P =

Number of trials in which theevent has happened

Total number of trials

Total number of trials = 150Number of times 3 heads appeared = 24Number of times 2 heads appeared = 45Number of times 1 head appeared= 72Number of times no head appeared = 9Let E1, E2, E3 and E4 be the events ofgetting, 3 heads, 2 heads, 1 head and nohead respectively. Then

P(E1) =

Number of times 3 headsappeared


= 24

150 = 0.16

P(E2) =

Number of times 2 headsappeared


= 45150

= 0.30

P(E3) =

Number of times1 headappeared


= 72

150 = 0.48

P(E4) =

Number of times no headsappeared


= 9150

= 0.06

Sum of all these probabilities

= P(E1) + P(E2) + P(E3) + P(E4)

= 0.16 + 0.30 + 0.48 + 0.06

= 1. Hence proved.

22. Two parallelograms PQRS and MNRS, onthe same base SR and between the sameparallels PN and SR are given. We needto prove that ar(PQRS) = ar(MNRS)

Proof: In ΔPMS and ΔQNR,∠1 = ∠4 (Corresponding angles)∠2 = ∠5 (Corresponding angles)

∴ ∠3 = ∠6 (Angle sum property)

Also, PS = QR (Opposite sides of ||gm

PQRS)So, ΔPMS ≅ ΔQNR (By ASA rule)Therefore,ar(ΔPMS) = ar(ΔQNR)

(Congruent figures have equal areas)

⇒ ar(ΔPMS) + ar(MQRS) = ar(ΔQNR)+ ar(MQRS)

[Adding ar(MQRS) to both sides]⇒ ar(PQRS) = ar(MNRS)

Hence proved.

133��

(i) We know that a rectangle is also aparallelogram.Thus, parallelograms ABCD and EFCDstand on the same base DC and liebetween the same parallels DC and EB.∴ ar(ABCD) = ar(EFCD)

(ii) ar(ABCD) = Base × Height= (DC) (AM).

23. Rectangle is rolled along its length⇒ l = 44 cm = circumference of the

base of the cylinder⇒ 2πr = 44 cm ⇒ πr = 22

⇒ r = 22�

=2222

× 7

∴ r = 7 cm and h = 20 cm.C.S.A. = 2πrh

= 2 × 227

× 7 × 20 = 880 cm2

Volume = πr2h

= 227

× 7 × 7 × 20 = 3080 cm3.

24. Given: A circle whose centre is O and�ACB and��ADB are two angles formedin the same segment of the circle.

To prove:��ACB = �ADBConstruction: Join OA and OB.

Proof:��AOB = 2�ACB ...(i)and �AOB = 2�ADB ...(ii)

(Angle subtended by an arc at the centreis double the angle subtended by it atany other point on the remaining partof the circle.)


2�ACB = 2�ADB

⇒ �ACB = �ADB

Thus, angles in the same segment of acircle are equal.

OR

Given: AB = 2AC and radius OA = r.

Let OM ⊥ AB and ON ⊥ AC. Hence, OM= p and ON = q.

To prove: 4q2 = p2 + 3r2

Proof: In right-angled triangle OAM,

OM2 + AM2 = OA2

(Using Pythagoras Theorem)

⇒ p2 +2AB

2� ��

= r2

(... AM = MB =AB2

)

⇒ p2 + 22AC

2� ��

= r2 (... AB = 2AC)

⇒ p2 + AC2 = r2

⇒ p2 + (2AN)2 = r2 (... AN = CN = AC2

)

⇒ p2 + 4AN2 = r2 ...(i)

In right-angled triangle OAN,

AN2 + ON2 = OA2

(Using Pythagoras Theorem)

⇒ AN2 + q2 = r2

∴ AN2 = r2 – q2

Putting this value in (i), we have

p2 + 4(r2 – q2) = r2

⇒ p2 + 4r2 – 4q2 = r2

⇒ p2 + 3r2 = 4q2

i.e., 4q2 = p2 + 3r2.

Hence proved.

��

� �E��%��%��:�� I��-�

27. ;�� 4/.�

⇒ AB = CD

⇒12

AB = 12

CD

⇒ AP = DR ...(i)

(P and R are the mid-points of ABand CD respectively)

As S is the mid-point of AD,

AS = SD ...(ii)

In ΔAPS and ΔDRS,

AP = DR [From (i)]

∠PAS = ∠RDS (Each 90°)

AS = SD [From (ii)]

⇒ ΔAPS ≅ ΔDRS

(SAS criterion of congruence)

⇒ PS = RS ...(iii) (CPCT)

Similarly, we can prove that

PQ = RQ, SP = QP and QR = SR ...(iv)

Using equations (iii) and (iv), weconclude that

PQ = QR = RS = SP

⇒ PQRS is a rhombus.

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29. Steps of construction:

1. Draw a line segment MN = 13 cm.

2. Construct an angle of 60° at M and45° at N such that ∠PMN = 60° and∠QNM = 45°.

3. Draw the bisectors of 60° and 45°which intersect each other at A.

4. Draw the perpendicular bisectors ofMA and NA, which meet MN at Band C respectively.

5. Join AB and AC.

ΔABC is the required triangle such that AB + BC + CA = 13 cm and base angles ∠B= 60° and ∠C = 45°.

��

30. Let the total distance covered be x kmand the total fare charged be � y. Thenfor the first km, fare charged is � 10 andfor remaining (x – 1) km fare charged is� 4 (x – 1).

Therefore, y = 10 + 4(x – 1)

= 4x + 6

The required equation is y = 4x + 6.

x 0 – 112

y 6 2 8

"�� (i)

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(ii)

Practice Paper–5

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�� there are 100 tickets and one ticket isdrawn, so there are 100 possibleoutcomes.

Let A be an event that ‘‘the numberdrawn is a multiple of 3 or 5’’.

Favourable outcomes for a multiple of 3are 3, 6, 9, ....99, i.e., 33.

Favourable outcomes for a multiple of 5are 5, 10,.......100, i.e., 20.

Favourable outcomes for a multiple of15 are 15, 30, ......., 90, i.e., 6.

Therefore, favourable outcomes for anevent A are 33 + 20 – 6 = 47.

Hence, probability of a ticket drawnWhose number is a multiple of 3 or 5 is

P(A) =

Favourable outcomes foran event A

Total possible outcomesfor an experiment

= 47

100.

ORAs three coins are tossed once. Thereforethere are 8 possible outcomes, i.e., HHH,HHT, HTH, THH, THT, HTT, TTH,TTT.Let A be an event of ‘‘at least onehead’’. At least one head implies that inan event there can be one head, twoheads or three heads.Therefore, favourable cases for an eventA are seven, i.e., HHT, HTH, THH,TTH, THT, HTT, HHH.

∴ P(A) =

Favourable outcomes foran event A

Total possible outcomesfor an experiment

=78

.

13. Sum of observations = 50 × 80.4!��2�2

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OR

Let radius, height and slant height of thecone be r metres, h metres and l metresrespectively.

Curved surface area = 188.40 m3

⇒ πrl = 188.40

⇒ 3.14 × r × 2 64r � = 188.4

(... l = 2 2r h� and h = 8 m)

⇒ r 2 64r �= 60

⇒ r2(r2 + 64) = 3600

(Squaring both sides)

⇒ r4 + 64r2 = 3600⇒ x2 + 64x – 3600 = 0,

where x = r2

⇒ x2 + 100x – 36x – 3600 = 0⇒ x(x + 100) – 36(x + 100) = 0⇒ (x + 100) (x – 36) = 0⇒ x = – 100 or 36⇒ r2 = – 100 or 36⇒ r2 = 36

(Square of a real number can'tbe negative)

⇒ r = 6 m

Volume =13

πr2h

=13

× 3.14 × 6 × 6 × 8

= 301.44 m3.�"� F�� $��,2D 8 �2D�! �2D

�� .��$��C�!��2� ��

141��

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ORSteps of construction:

1. Draw a rayBY and cutB C = 8 c mfrom it.

��

2. Make ∠YBX = 90° at B.3. From BX, we cut a line segment

BD = 12 cm.4. Join C and D.5. Draw the perpendicular bisector of

CD, which intersects BX at A.6. Join A and C.

Thus, ΔABC is the required triangle.

�� Curved surface area of the conical tent= Area of cloth used in it, i.e., 165 m2

⇒ πrl = 165

⇒227

× 5 × l = 165

(... Base radius = 5 m)

⇒ l = 165×722×5

= 212

m.

∴ h = 2 2–l r =

2221

– (5)2

� ��

= 441

– 254

= 441 –100

4

= 3412

m.

(i) Base area = πr2 = 227

× (5)2

=22 25

7�

m2

For a student, area required 57

m2

So, the number of students that can sit

in the tent = 2

Base area of tent5

m7

=22 25 7

7 5�

� = 110.

(ii) Volume of the cone = 13

πr2h

=13

× 227

× (5)2 ×3412

=275 341

21=

275 18.4621`

= 241.7 m3.

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PULLOUT WORKSHEETS Material...2015/12/29 · Mathematics New Saraswati House (India) Pvt. Ltd. New...

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Transcript of PULLOUT WORKSHEETS Material...2015/12/29 · Mathematics New Saraswati House (India) Pvt. Ltd. New...