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OSWAAL BOOKS“Oswaal House” 1/11, Sahitya Kunj, M.G. Road, AGRA-282002Ph.: 0562-2857671, 2527781, Fax : 0562-2854582, 2527784
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PULLOUT
Worksheets PULLOUT
Worksheets Term 1 (April to September)
SOLUTIONSSOLUTIONS
MathematicsMathematics
Class 9
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(iii)
CONTENTS
Unit-I : Number System
1. Real NumbersSummative Assessment Ø Worksheets 1 - 12 1 - 21
Formative Assessment Ø Worksheets 13 21 - 21
Unit-II : Algebra
2. PolynomialsSummative Assessment Ø Worksheets 14 - 27 42 - 42
Formative Assessment Ø Worksheets 28 42 - 42
Unit-III : Geometry
3. Introduction to Euclid's GeometrySummative Assessment Ø Worksheets 29 - 33 43 - 48
Formative Assessment Ø Worksheets 34 48 - 48
4. Line and Angles Summative Assessment Ø Worksheets 35 - 42 49 - 57
Formative Assessment Ø Worksheets 43 57 - 57
5. Triangle Summative Assessment Ø Worksheets 44 - 55 58 - 77
Formative Assessment Ø Worksheets 56 77 - 77
Unit-IV : Co-ordinate Geometry
6. C0-ordinate GeometrySummative Assessment Ø Worksheets 57 - 64 78 - 90
Formative Assessment Ø Worksheets 65 90 - 90
Unit-V : Mensuration
6. Areas Summative Assessment Ø Worksheets 66 - 71 91 - 101
Formative Assessment Ø Worksheets 72 101 - 101
P-1
1 REAL NUMBERS
SUMMATIVE ASSESSMENT WORKSHEET-1
Section A
1. 4 28 3 7÷ = 4 2 7 3 7× ÷ (1)
= 83
2.5x = 7P
⇒5 5
7×
= 7P (1)
P = 25
7 7× = 257 (1)
Section B
3. x = 3 – 2 2
1x
= 1
3 2 2( – ) ×
( )( )3 2 23 2 2
++
= ( )
–3 2 2
9 8+
= 3 + 2 2 (½)
xx
+FHG
IKJ
12
= xx
+1 + 2 × x × 1
x(½)
= 3 – 2 2 + 3 + 2 2 + 2
xx
+FHG
IKJ
12
= 8 (½)
xx
+FHG
IKJ
1= ± 2 2 (½)
4.
13 41 1
3 35 8 27 +
=
13 41 1
3 33 3(2 ) (3 ) +
(½)
= 1 1
3 34 45(2 3) 5(5) + = (½+½)
= 1
4 14(5 ) 5 5.= = (½)
Section C
5. x = 1p
= 1
5 2 6+=
15 2 6( )+
× ( – )( – )5 2 65 2 6
(1)
LAER RBM EUN S
P-2 T I C S IX--M A T M AEH M - 1 SE RT
=5 2 625 24
––
= 5 – 2 6 (½)
p2 + x2 = (5 + 2 6 )2 + (5 – 2 6 )2
= (5)2 + (2 6 )2 + 2 × 5 × 2 6 + (5)2 + 2 62e j – 2 × 5 × 2 6 (½)
= 25 + 24 + 20 6 + 25 + 24 –20 6 (½)= 98. (½)
6.
√2√3
D
C
–2 –1 O 1BA
P Q 2 3
√3
1
1
–½
Let AB = BC = 1 unit length.
(1)
Using Pythagoras theorem we see that
OC = 1 12 2+ = 2 (½)Construct CD = 1 unit length perpendicular to OC, then using Pythagoras theorem, we seethat
OD = 2 12 2e j + = 3 (½)
Using a compass with centre O and radius OD, draw an arc which intersects the number
line at the point Q, then corresponds to 3. (1)
Section D
7. Let x = 1·32 1·32222..............=10x = 13·222...........
100x = 132·222........... (½)100x – 10x = 132·222........... – 13·222........... (½)
90x = 119·00
x =11990 (½)
Again, Let y = 0·35 0·353535............=100y = 35·3535.............. (½)
100y – y = 35·3535................ – 0·3535 (½)99y = 35
y =3599 (½)
∴ 1·32 0·35 x y+ = + =119 3590 99
+ (½)
=119 11 35 10 1309 350 1659·
990 990 990× + × += = (½)
P-3
SUMMATIVE ASSESSMENT WORKSHEET-2
Section A
1. b2 = a ⇒ a b= . (1)
2. 2 2 2( )( ) ( ) ( )a b a b a b a b+ − = − = − (1)
3. 1 / 4 4 1 / 4 4 1 /4 1 14(81) 81 (3 ) (3 ) 3 3− − −× = × = × (1)
= 3º = 1
Section B
4.1
2 1+= 1 2 1
2 1 2 1−×
+ −(½)
= 2 1 2 12 1 1
− −=
−(½)
= 1·414 – 1
= 0·4145. The two irrational number between 0·5 and 0·55 are
0·51010010001000001............... (1)and 0·502002000200002............. (1)
Section C
6. 2x × (22)x = 3 1 / 3 5 1 / 5(2 ) (2 )× (½)2x · 22x = 21 × 21 (½)2x + 2x = 21 + 1 (½)
23x = 22 (½)Comparing powers on both sides, we get
3x = 2 (½)
x = 23 · (½)
7. Since LCM of 5 and 6 is 30,
∴1 1 56 6 5
= × = 5 5 25
30 5 150× = (½)
and1 1 65 5 6
= × = 6 5 30
30 5 150× = (½)
Hence, four rational numbers between 16 and
15 are :
26 27 28 29, , , ·150 150 150 150 (½+½+½+½)
Section D8. Given that xa = y, yb = z, zc = x
∴ xabc = (xa)bc (1)= (y)bc (½)
LAER RBM EUN S
P-4 T I C S IX--M A T M AEH M - 1 SE RT
= (yb)c (1)= (z)c (½)
xabc = x1 (½)Comparing powers on both sides, we get
abc = 1. (½)
9. x2 + y2 + xy =
2 23 1 3 1 3 1 3 13 1 3 1 3 1 3 1
+ − + −+ + × − + − + (1)
= 3 1 2 3 3 1 2 3 13 1 2 3 3 1 2 3
+ + + −+ ++ − + +
(1)
= ( )( )
( )( )
2 2 3 2 2 31
2 2 3 2 2 3
+ −+ +
− + (½)
= ( ) ( )
( )( )
2 22 3 2 3
12 3 2 3
+ + −+
− + (½)
= 4 3 4 3 4 3 4 3
14 3
+ + + + −+
−(½)
= 14 + 1 = 15. (½)
10. ( ) ( ) ( )2 2 22 2 5 3 2 3 2 1− + + − −
= ( ) ( ) ( )2 2 222 2 (5) 2 2 2 5 3 2 3 2 3 2 3+ − × × + + + × ×
( )2 22 (1) 2 2 1− − + × × (2)
= 8 25 20 2 18 3 6 6 2 1 2 2+ − + + + − − + (1)
= 51 18 2 6 6− + . (1)
SUMMATIVE ASSESSMENT WORKSHEET-3
Section A
1. 12 8 2 3 2 2 4 3 2 4 6× = × = × = (1)
2. Sometimes rational and sometimes irrtational. (1)3. x = 0·777..............
⇒ 10x = 7·777...... ⇒ 10x – x = 7·77........ – 0·777............
⇒ 9x = 7 ⇒ x = 79 (1)
Section B
4. (4 3 3 2) (4 3 3 2)+ × − (½)
= 2 2(4 3) (3 2)− (½)= 48 – 18 (½)= 30 (½)
P-5
5.3 / 4 3 / 281 25
16 9
− − ×
=
3 / 4 3 / 24 23 52 3
− − ×
½
= 3 33 5
2 3
− − ×
1
= 3 32 3
3 5 ×
1
= 3
32 8
1255= 1
Section C
6. (i) (xy + yx)–1 = (52 + 25)–1 (½)
= (25 + 32)–1 (½)
= (57)–1 = 157 (½)
(ii) (xx + yy)–1 = (55 + 22)–1 (½)= (3125 + 4)–1 (½)
= (3129)–1 = 13129 . (½)
7. x = ( 2 ) ( 2 ) 2 2 2
( 2 ) ( 2 )p q p q p q p q
p q p q+ + − + × + × −
+ − −1
x = 2 22 2 4
2 2p p qp q p q
+ −+ − +
⇒ x = 2 2( 4 )
24
p p qq
+ −1
⇒ 2qx = 2 24p p q+ − ⇒ 2qx – p = 2 24p q− 1
On squaring both sides,4q2x2 + p2 – 4pqx = p2 – 4q2
⇒ 4q (qx2 – px) = – 4q2
⇒ qx2 – px + q = 0.
Section D
8. a + b = 2 5 2 52 5 2 5
− +++ −
(1)
= ( ) ( )
( ) ( )
2 22 5 2 5
2 5 2 5
− + +
+ − (1)
= 4 5 4 5 4 5 4 54 5
+ − + + +− (1)
= 18 181 = −− (½)
∴ (a + b)3 = (– 18)3 = – 5832. (½)
LAER RBM EUN S
P-6 T I C S IX--M A T M AEH M - 1 SE RT
9.2 2 2
2 2 2
1 1 1
. .a b c c aa b b c
b c ax x xx x x
++ +
= ( ) ( ) ( )2 2 2 2 2 21 1 1
. .a b b c c a c aa b b cx x x− − − ++ + (1)
= 1 1 1( )( ) ( )( ) ( )( )( ) ( ) ( ). .
a b a b b c b c c a c aa b b c c ax x x− + × − + × − + ×+ + + (1)
= . .a b b c c ax x x− − − (½)
= a b b c c ax − + − + − (1)
= x0 = 1. (½)
10.15
10 20 40 5 80+ + − − = 15
10 2 5 2 10 5 4 5+ + − − (1)
= 15
3 10 3 5− (½)
= ( )( )( )
10 5153 10 5 10 5
+×
− + (1)
= ( )5 10 510 5
× +−
(½)
= 5(3·2 2·2) 5·45+ = . (1)
SUMMATIVE ASSESSMENT WORKSHEET-4
Section A
1.3 / 4
1 / 416
16− = 3 /4 1 / 4 1(16) (16) 16+ = = 1
2. x = 2·9 2·99..........= ⇒ 10x = 29·99..... ⇒ 10x – x = 29·99.............. – 2·99.......... ½
⇒ 9x = 27 ⇒ x = 27 39
= ½
3. Sum of 2 5 and 3 7 = 2 5 3 7+ = 12·4 1
Section B
4. ( ) ( )+ +5 2 3 2 5 2 = ( )+ + +5 2 15 3 2 5 2 2 (1)
= ( )+5 2 17 8 2 (½)
= + ×85 2 40 2
= +85 2 80 . (½)
P-7
9.3
9.3 1
5.
⇒ = 2157 3·4512625
=
Section C
6.5 25 2
+−
⇒ 5 2 5 25 2 5 2
+ +×− ×
1
⇒ 5 2 2 5 2
5 2+ +
− ⇒
7 2 103
+1
⇒ 7 2(3.162)
3+
⇒ 7 6·324
3+
⇒ 13·324
3 ⇒ 4·441 1
7. Mark the distance 9·3 units from a fixed point A on a givenline to obtain a point B such that AB = 9·3 units. From B,mark a distance of 1 unit and mark the new point as C. Find
the mid-point of AC and mark that point as O. Draw a semi-circle with centre O and radius OC.Draw a line perpendicular to AC passing through B and intersecting the semi-circle at D.
Then, BD = 9·3 (1)
To represent 9·3 on the number line, let us treat the line BC as the number line, with B as zero,C as 1, and so on.Draw an arc with centre B and radius BD, which intersects the number line at E. (1)
∴ E represents 9·3 .
Section D
8.( )( )
21 /2
39 3 27
3 2
n n n
m
−+ −× −
×= 1
729
2 1 3
3 3(3 ) 3 (3 )
3 2
n n n
m
+ × −×
= 613 (1)
2 2 3
33 3
3 8n n n
m
+ + −× = 6
13 (1)
3 2
33 [3 1]
3 8n
m−
× = 613 (½)
3 3 83 8n m− × = 3– 6 (½)
LAER RBM EUN S
⋅625 ) 2157 (3 4512 1875 2820 2500 3200 3125 750 625 1250 1250 ×
P-8 T I C S IX--M A T M AEH M - 1 SE RT
Comparing powers on both sides, we get
3n – 3m = – 6 (½)
– 3(m – n) = – 6
m – n = 2. (½)
9. (i)0·714285...
7 5·000000491073028
2014605640355
−
−
−
−
−
−
(1)
∴57 = 0·714285 (1)
and0·81
11 9·0088
20119
−
∴ 911 = 0·81 (½)
Thus, three different irrational numbers between 5 9and7 11 are 0·75075007500075000075.....,
0·767076700767000767.... and 0·80800800080000..... (½)(ii) Number system (½)(iii) Those who are irrational in there approach get fail in their efforts. (½)
SUMMATIVE ASSESSMENT WORKSHEET-5
Section A
1. 1 /2 1 /4 2[{(81) } ]− − = [{(81)–1/2 }]–1/2 = (81)1/4 = (34)1/4 = 3 1
2.150 =
1 1 2 2105 5 2 5 2 2
= × =× ×
½
So, rationalising factor is 2. ½
3.1500
2 15 =
1 1500 1 10100 52 15 2 2
= × = = 1
P-9
Section B4. LCM of 5 and 7 is 35,
∴ 35 = ×3 7
5 7 = 2135 (½)
and 57 = ×5 5
7 5 = 2535 (½)
so, < < < <21 22 23 24 2535 35 35 35 35 (½)
The required three rational numbers are 22 23 24, and35 35 35 . (½)
5. Given, a = 2 and b = 3.
∴ ab + ba = 23 + 32 (½)
= 8 + 9 (1)
= 17. (½)
Section C
6. Let x = 0·328 = 0·3282828......... (½)10x = 3·282828......... (½)
1000x = 328·282828......... (½)1000x – 10x = 328·2828........ – 3·2828........ (½)
990x = 325·00 (½)
x = 325990 = 65
198 . (½)
7.+ +−− +
5 3 5 37 4 3 7 4 3
= ( ) ( ) ( ) ( )
( ) ( )+ + − + −
− +
5 3 7 4 3 5 3 7 4 3
7 4 3 7 4 3(1)
= + + + − + − +−
35 20 3 7 3 12 35 20 3 7 3 1249 48
(1)
= +40 3 241 (½)
= ( )+8 3 5 3 . (½)
Section D8. (1 + x + y–1)–1 + (1 + y + z–1)–1 + (1 + z + x–1)–1
= 1 1 11 1 11 1 1x y zy z x
+ ++ + + + + +
(1)
= + + = ⇒ =+ ++ + + +
∵1 1 1 1, ( 1 )1 1 111 1
xyz xyy xy zx
y xy x
(1)
= 1 1 11 1 1y xy y xy xy y
y xy
+ ++ + + + + + (1)
LAER RBM EUN S
P-10 T I C S IX--M A T M AEH M - 1 SE RT
= 1
1 1 1y xy
y xy y xy xy y+ ++ + + + + + (½)
= ( )( )
11
1y xyy xy
+ +=
+ + . (½)
Value Based Question9. (i) Six rational number between 3 and 4 are 3·125, 3·225, 3·335, 3·445, 3·555, 3·625 1
(ii) Number system. 1(iii) Those who are rational in there approach get fails in their efforts. 1
SUMMATIVE ASSESSMENT WORKSHEET-6
Section A
1. We know that 5 2·236= the irrational number is 5. 1
2.1 / 2
4 3 2x
= ( ) ( )1 / 21 / 4 1 / 8
3 32 2x x
=
½
= 2 1 / 3 1 / 8 2 1 / 2 4 1/12[( ) ] ( )x x x= = ½
Section B
3. Let x = 2·218 = 2·218181818.......
10x = 22·18181818......... (½)
1000x = 2218·181818......... (½)
1000x – 10x = 2218·181818...... – 22·181818.......
990x = 2196·00 (½)
x = × × ×= =× × ×2196 2 3 3 122 122990 2 3 3 55 55 . (½)
4. Given, 17 = 0·142857142857......... (½)
27 = 0·285714285714........ (½)
Number can be 0·160160016000....... (1)
Section C
5.− −
−2 33 4
4 1
(216) (256)=
( ) ( )− −
−
2 33 43 4
4 1
6 4(1)
= − −−2 34 1
6 4 (½)
= 4 × 62 – 43 (½)= 144 – 64 (½)= 80. (½)
P-11
6.
1 1a a
a b b aab ab
++ −
= b ba b b a
++ −
= ( ) ( )
( )( )b a b b b a
b a b a− + +
+ − 1
= 2 2
2 2b ab b ab
b a
− + +
−1
= – 2
2 22b
a b−1
Section D
7.[ ] [ ] [ ]
( )
2 2 2( ) ( ) ( )
4. .a b b c c a
a b c
x x x
x x x
+ + +
= 2( ) 2( ) 2( )
4 4 4. .
. .
a b b c c a
a b cx x x
x x x
+ + +
(1)
= 2 2 2 2 2 2
4 4 4. .
. .
a b b c c a
a b cx x x
x x x
+ + +
(1)
= 2 2 2 2 2 2
4 4 4
a b b c c a
a b cx
x+ + + + +
+ + (1)
= 4 4 4
4 4 4 1a b c
a b cxx
+ +
+ + = . (1)
8.1 1
1 1a b b ax x− −+
+ +=
1 1
1 1a b
b ax x
x x
+
+ + 1
=
1 1b a a b
b ax x x x
x x
++ + 1
= b a
b a a bx x
x x x x+
+ +1
= ( )
1( )
b a
a bx x
x x
+=
+1
9. ( ) ( )7 5
1 2 2 32 22 4 3 5
2 3 2 32 3 2 3
−− −
− −× ××
× ×= ( ) ( )
7 56 82 23 5
3 32 2
−× (1)
=
7 56 82 27 53 52 2
3 3
2 2
× ×−
× ×−× (1)
= 21 20
21 252 2
3 3
2 2
−
−× (½)
= 2 1 2 0 1
21 25 42 2 2
3 3
2 2
−
− −= (½ + ½)
= 3 × 22 = 12. (½)qq
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SUMMATIVE ASSESSMENT WORKSHEET-7
Section A
1.177 59 3 3413 59 7 7
×= =× 1
2. 0·3 0·4+ = 0·33... 0·44... 0·77+ = ½
Let, x = 0·7 ⇒ 10x = 7·77... ⇒ 10x – x = 7·77... – 0·77... ½
⇒ 9x = 7·0 ⇒ x = 79
3. 2 is non-terminating, non reccuring. 1
Section B
4. ( )−
− ÷( )
( )
ca b c b
ab a cx x
xx = −
− ÷ab ac bc
acba bcx x
xx (½)
= xab – ac – ab + bc ÷ xbc – ac (½)= xbc – ac ÷ xbc – ac (½)= 1 (½)
5. − +50 98 162 = × × − × × + × × × ×5 5 2 7 7 2 3 3 3 3 2 (½)
= − + ×5 2 7 2 3 3 2 (½)
= − +2 2 9 2 (½)= 7 2 (½)
6. Given, a = 2 and b = 3.(A) (ab + ba)–1 = (23 + 32)–1 (½)
= (8 + 9)–1 = 17–1 = 117 (½)
(B) (aa + bb)–1` = (22 + 33)–1 (½)
= (4 + 27)–1 = 31–1 = 131 . (½)
7.12 = ×1 2
2 2 (rationalising) (½)
= = =2 1·414 0·7072 2
(½)
∴ + π12 = 0·707 + 3·141 = 3·848. (1)
Section C
8. x = +1 2
1x = ( )
( )( )
( )− −× = −+ −
2 1 2 112 12 1 2 1
= −2 1 (1)
1x x− = ( ) ( )+ − −1 2 2 1 (½)
= + − +1 2 2 1 = 2 (½)
∴ ( )−31x x = 23 = 8. (1)
P-13
9. + ++ + +1 1 2
1 2 2 3 3 5
= ( )
( )( ) ( )
( )( ) ( )
( )( )
− − −× + × + ×
+ − + − + −
2 1 3 2 5 31 1 22 1 2 1 3 2 3 2 5 3 5 3
(1)
= ( ) ( ) ( )− − −+ +− − −
2 1 3 2 2 5 32 1 3 2 5 3
(1)
= − + − + −2 1 3 2 5 3 (½)
= −5 1 . (½)
Section D
10. ( ) ( ) ( )11 1
. .a b cbc caab
c abx x x
x xx= ( ) ( ) ( )
1 1 1. .a b b c c aab bc cax x x− − − (1)
= . .a b b c c aab bc cax x x− − −
(½)
= a b b c c acaab bcx
− − −+ + (½)
= ac bc ab ca bc ababcx
− + − + − (1)
= 0
0abcx x= (½)
= 1. (½)qq
SUMMATIVE ASSESSMENT WORKSHEET-8
Section A1. n – 1 12. 1·011 1·011011011.....= ½
So, 1·011 is smallest ½
Section B3. 7x = 1
x = 17
(½)(7 10 0·14285773028
2014
6056
4035
50491
(1)
∴ x = 0·142857 . (½)
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4. (13 + 23 + 33)–3/2 = (1 + 8 + 27)–3/2 (½)
= (36)–3/2 (½)
= [(6)2]–3/2 = 6–3 (½)
= =31 1
2166 . (½)
5. 3 40 = 3 32 2 2 5 2 5× × × = (½)
3 320 = × × × × × × = × =3 3 32 2 2 2 2 2 5 2 2 5 4 5 (½)
∴ 3 3 40 – − = × − × −3 3 3 3 34 320 5 3 2 5 4 4 5 5 (½)
= − − = −3 3 3 36 5 16 5 5 11 5 . (½)
6.30 29 28
31 30 293 3 3
3 3 ( 1)(3 )
+ +
+ + −=
28 2 1
29 2 13 (3 3 1)
3 (3 3 1)
+ +
+ −1
= 9 3 1
3(9 3 1)+ ++ − ½
= 13 13
3 11 33=
× ½
Section C
7. Given, x = +3 8
∴ x2 = ( )+ = + + × ×2
3 8 9 8 2 3 8 (½)
= +17 12 2 (½)
⇒ 21
x = ( )
( )( )
−×
+ −
17 12 2117 12 2 17 12 2
(½)
= −−
17 12 2289 288 = −17 12 2 (½)
Now, +22
1xx = + + −17 12 2 17 12 2 (½)
= 34. (½)
Section D8. 52x – 1 – 25x – 1 = 2500
⇒ 52x – 1 – [(5)2]x – 1 = 2500 ½⇒ 52x – 1 – 52x – 2 = 2500 ½⇒ 52x – 2 (5 – 1) = 2500 1⇒ 52x – 2 = 625 ⇒ 52x – 2 = 54 ½on comparing both sides, we get ½
2x – 2 = 4 ⇒ x = 3 19. Since L.C.M. of 3, 4, 6, 12 is 12,
∴1 4
3 3 42 2 ×= = 4 12 124122 2 16= = (1)
P-15
1 34 4 35 5 ×= =
3 12 123125 5 125= = (½)
1 26 6 27 7 ×= =
2 12 122127 7 49= = (½)
12 3 = 12 3 (½)
In descending order 12 12 12 12125, 49, 16, 3 (1)
i.e., 4 6 3 125, 7, 2, 3 . (½)
SUMMATIVE ASSESSMENT WORKSHEET-9
Section A
1.32 48 2( 8 12) 2( 2 3) 28 12 2( 2 3) ( 2 3)
+ + += = =+ + +
1
Section B
2. 32 3 3 2×⇒ 2(3)1/3 × 3(2)1/2 ½⇒ 21+1/2 × 31+1/3 ½⇒ 23/2 × 34/3 ½⇒ 6(72)1/6 ½
Section C
3. x = +9 4 5= + +5 4 4 5
= ( ) + + × ×2 25 (2) 2 2 5
x = ( )+2
5 2
x = +5 2 (1)
1x = ( )
( )( )
−×
+ −
5 215 2 5 2
= − = −−5 2 5 25 4 (1)
∴ − 1xx = ( ) ( )+ − −5 2 5 2
= + − +5 2 5 2 = 4. (1)
4. ( ) ( )22 3·3 2x x
= ( )8116
( ) ( )−22 2.3 3x x
= ( )432 (1)
( ) −223
x x= ( )−42
3 (1)
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Comparing the power on both sides, we getx – 2x = – 4 (½)
– x = – 4⇒ x = 4. (½)
5.+−
5 35 3
= ( )( )
++×
− +
5 35 35 3 5 3 (1)
= + + × × +=−5 3 2 5 3 8 2 15
5 3 2 (1)
= ( )+= +
2 4 154 152
(½)
∴ + 15a b = +4 15Comparing on both sides, we get
a = 4, b = 1. (½)
6. +− +4 3
3 3 2 2 3 3 2 2=
( ) ( )( ) ( )
+ + −
− +
4 3 3 2 2 3 3 3 2 2
3 3 2 2 3 3 2 2(½)
= + + −−
12 3 8 2 9 3 6 227 8
(½)
= +21 3 2 219 (½)
= × + ×21 1·732 2 1·41419 (½)
= +36·372 2·82819 (½)
= =39·2 2·063.19 (½)
Section D
7. 3 1 127 4 3
4 5 2
(2187) (256) (1331 )− − −− + = [ ]
3 1 127 4 34(2187) 5(256) 2 (1331)− + (½)
= ( ) ( ) ( )3 1 2
7 4 37 4 34 3 5 4 2 11− + (1)
= 4 × 33 – 5 × 4 + 2 × 112 (1)= 4 × 27 – 20 + 2 × 121 (1)= 108 – 20 + 242 = 330. (½)
SUMMATIVE ASSESSMENT WORKSHEET-10
Section A
1.4 3 2x
= 23 2 1 / 4 1 / 4 2 1 / 3 1 / 43( ) ( ) [( ) ]x x x = = ½
= (x2)1/12 = x1/6 ½
P-17
Section B
2. Let x = =0·237 0·237237237.....
1000x = 237·237237........ (½)
1000x – x = 237·237237........ – 0·237237..... (½)
999x = 237 (½)
x = 237999
. (½)
Section C
3. ( ) ( ) ( )+ − +
+93 1 1 12
3 12= ( ) ( )
( )( )
−− + − + ×
+ −
12 393 36 1 1212 3 12 3
(1)
= ( ) ( )−− + − + −
9 12 33 6 1 12 12 3
(1)
= − − + −3 5 12 12 3 (½)
= – 5. (½)4. (xa – b)a + b · (xb – c)b + c · (xc – a)c + a
= 2 2 2 2 2 2· ·a b b c c ax x x− − − 1
= 2 2 2 2 2 2a b c a b cx + + − − − 1= xº = 1 1
5. ( )− −÷2 1
43 2( )x y xy = ( )−
− ÷
2 11 1 13 242 2 2. ( )x y xy (½)
= − −÷1 1
23 4. ( )x y xy (½)
= − ×1 1
23 4. ( )x y xy (½)
= − + +1 1 123 4 4.x y (½)
= −1 912 4.x y (½)
=
94
112
y
x. (½)
6.−30
5 3 3 5= ( )
( )( )
+×
− +
5 3 3 5305 3 3 5 5 3 3 5
(½)
= ( )
( ) ( )+
−2 2
30 5 3 3 5
5 3 3 5(1)
= ( )+−
30 5 3 3 575 45
(½)
= ( )+30 5 3 3 530
(½)
= +5 3 3 5 . (½)
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P-18 T I C S IX--M A T M AEH M - 1 SE RT
7. ( ) ( )−− −
− −× ××
× ×
7 51 2 2 32 2
2 4 3 55 7 5 75 7 5 7 = ( ) ( )−× ××
× ×
7 54 2 5 32 22 1 3 2
7 7 7 75 5 5 5
(½)
= ( ) ( )−×
7 56 82 23 5
7 75 5
(½)
= ( ) ( )− ×
1 17 52 26 8
3 57 75 5 (½)
= ( ) ( )−
−×1 1
42 402 221 25
7 75 5
(½)
= ( )×1
42 25 221 40
7 55 7
= ( )×1
2 4 27 5 (½)
= × 27 5 = 175. (½)
Section D
8. x = 3 2 2−
1x = ( )
( )( )3 2 21
3 2 2 3 2 2
+×
− +(½)
1x =
( )3 2 29 8+
− = 3 2 2+ (½)
21x = ( )2
3 2 2 9 8 12 2+ = + + = 17 12 2+ (½)
x2 = ( )23 2 2 9 8 12 2− = + − = 17 12 2− (½)
Now, 44
1xx
− = ( ) ( )2 22 2
1 1x xx x
− + (½)
= ( ) ( ) ( ) ( )17 12 2 17 12 2 17 12 2 17 12 2 − − + − + + (½)
= ( ) ( )17 12 2 17 12 2 17 12 2 17 12 2− − − − + + (½)
= 24 2 34− × = 816 2− . (½)
SUMMATIVE ASSESSMENT WORKSHEET-11
Section A1. (16)1/4 – 6(343)1/3 + 18(243)1/5 – (196)1/2
⇒ (2)4/4 – 6(7)3/3 + 18 (3)5/5 – (14)2/2 ½⇒ 2 – 42 + 54 – 14 = 0 ½
Section B2. False.
Justification : 147 147 49 7 ,
75 25 575= = = 1
which is a rational number. 1
P-19
Section C
3. + −+ + +6 3 2 4 3
2 3 6 3 6 2
= ( )( )
( )( )
( )( ) ( )
( )( )
− − −× + × − ×
+ − + − + −
3 2 3 2 6 3 6 26 4 33 2 3 2 6 3 6 3 6 2 6 2 (1)
= − − −+ −− − −18 12 3 12 3 6 4 18 4 6
3 2 6 3 6 2 (1)
= ( ) ( )− −− + −
3 12 6 4 18 618 12 3 4
(½)
= − + − − +18 12 12 6 18 6 = 0. (½)
4. ( ) 1xab
−= ( )2 8xb
a−
⇒ ( ) 1xab
−= ( ) (2 8)xa
b− −
(1)
Comparing powers on both sides, we getx – 1 = – (2x – 8) (½)
x – 1 + 2x – 8 = 0 (½)3x – 9 = 0 (½)
x = 3. (½)5. Since L.C.M. of 3, 2, 4 be 12,
∴ 3 44, 3, 6 = 1 1 13 2 44 , 3 , 6 (½)
= × × ×1 4 1 6 1 33 4 2 6 4 34 , 3 , 6 (½)
= × × ×1 1 14 6 312 12 124 , 3 , 6 (½)
= 1 1 112 12 12256 ,729 ,216 (½)
In ascending order = 1 1 112 12 12(216) ,(256) ,(729)
i.e. = 4 36, 4, 3 . (½ + ½)
Section D
6. x = ( )( )( )2 31
2 3 2 3
+×
− + (½)
= 2 34 3+− = 2 3+ (½)
2x3 – 2x2 – 7x + 5 = ( ) ( ) ( )3 22 2 3 2 2 3 7 2 3 5+ − + − + + (½)
= ( ) ( ) ( )332 2 3 3 2 3 2 3 2 4 3 4 3 + + × × + − + + 14 7 3 5− − + .(1)
= 2 8 3 3 12 3 18 14 8 3 9 7 3 + + + − − − − (½)
= ( )2 26 15 3 23 15 3+ − − (½)
= 52 30 3 23 15 3+ − − = 29 15 3+ . (½)
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P-20 T I C S IX--M A T M AEH M - 1 SE RT
SUMMATIVE ASSESSMENT WORKSHEET-12
Section A
1. (4 3 3 2) (4 3 3 2)+ × − ½
2 2(4 3) (3 2) (16 3) (9 2) 48 18 30− = × − × = − = ½
Section B
2. 2(4 3 3 5)−
= (16 3) (9 5) 2 4 3 3 5× + × − × × × 1
= 48 45 24 15 93 24 15+ − = − 1
Section C3. (729)–1/6 ½
= 1 / 61
(729) ½
= 6 / 61
(3) 1
= 13 1
4. + + −5 2 6 8 2 15 = + + × + + − ×3 2 2 3 2 5 3 2 5 3 (½)
= ( ) ( ) ( ) ( )+ + × × + + − ×2 2 2 2
3 2 2 3 2 5 3 2 5 3 (1)
= ( ) ( )+ + −2 2
3 2 5 3 (½)
= + + −3 2 5 3 (½)
= +2 5 . (½)
5.
3 / 4 3 / 2 381 2516 9
− − − 5 × ÷ 2 ½
=
3 / 4 3 / 2 316 981 25
2 × ÷ 5 ½
=
3 3 32 3 53 5 2
× ×
½
= 8 27 125
27 125 8 × ×
½
= 88 ½
= 1 ½
P-21
Section D
6.1
4 5+ = ( )( )( )
5 41 5 4 5 45 45 4 5 4
− −× = = −−+ − (½)
15 6+ = ( )
( )( )
6 51 6 5 6 56 56 5 6 5
− −× = = −−+ −(½)
16 7+ = ( )
( )( )
7 61 7 6 7 67 67 6 7 6
− −× = = −−+ − (½)
17 8+
= ( )( )( )
( )8 7 8 71 8 78 78 7 8 7
− −× = = −−+ −
(½)
18 9+ = ( )
( )( )
9 81 9 8 9 89 89 8 9 8
− −× = = −−+ −(½)
L.H.S. = 5 4 6 5 7 6 8 7 9 8− + − + − + − + − (1)
= 4 9− + = – 2 + 3 = 1 = R.H.S. Proved (½)qq
FORMATIVE ASSESSMENT WORKSHEET-13
This worksheet contains Activities which are to be perform in the class itself.qq
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P-22 T I C S IX--M A T M AEH M - 1 SE RT
2 POLYNOMIALS
SUMMATIVE ASSESSMENT WORKSHEET-14
Section A1. p(x) = x2 + 11x + k
p(–4) = 0 ⇒ (– 4)2 + 11 × – 4 + k = 0 ½16 – 44 + k = 0 ⇒ k = 28 ½
2. No. of zeros of a cubic polynomial = 3 1
Section B
3. + +2 2 2a b c
bc ac ab=
+ +3 3 3a b cabc
(½)
= 3abcabc
+ + = ∴ + + =
∵3 3 3
03
a b ca b c abc (1)
= 3. (½)4. p(x) = kx2 – x – 4
(x + 1) is a factor of p(x), then p(– 1) = 0 (½)
∴ k (– 1)2 – (–1) – 4 = 0 (½)
k + 1 – 4 = 0 (½)
k = 3 (½)
Section C
5. + − −
3 3
3 5 3 5y yx x
= + − + + + − + + −
2 2
3 5 3 5 3 5 3 5 3 5 3 5y y y y y yx x x x x x
(1)
= + + × × + + − × × + −
2 2 22 2 222 2
5 9 25 3 5 9 25 3 5 9 25y y y y y yx x x x x
(1)
=
× +
2223
5 9 25y yx
(½)
=
+
2225 3 25y yx
(½)
6. x2 – y2 + xy = + − + −− + ×
− + − +
2 23 2 3 2 3 2 3 23 2 3 2 3 2 3 2 (½)
= + −− +− +
5 2 6 5 2 6 15 2 6 5 2 6 (½)
= ( ) ( )
( ) ( )+ − −
+− +
2 25 2 6 5 2 6
15 2 6 5 2 6 (½)
= + + − − +
+−
25 24 20 6 25 24 20 61
25 24(½)
= +40 6 1= 40 × 2·4 + 1 = 96 + 1 = 97. (1)
P-23
Section D7. p(x) = x4 – 2x3 + 3x2 – px + 3p – 7
Put x + 1 = 0 or x = – 1 in p(x), we getp(– 1) = (– 1)4 – 2(– 1)3 + 3(– 1)2 – p(– 1) + 3p – 7 = 19 (½)
1 + 2 + 3 + p + 3p – 7 = 19 (½)4p – 1 = 19 (½)
4p = 20p = 5 (½)
∴ The polynomial p(x) = x4 – 2x3 + 3x2 – 5x + 15 – 7= x4 – 2x3 + 3x2 – 5x + 8 (½)
Put x + 2 = 0 or x = – 2 in p(x)p(– 2) = (– 2)4 – 2(– 2)3 + 3(– 2)2 – 5(– 2) + 8 (½)
= 16 + 16 + 12 + 10 + 8 (½)
= 62. (½)
Value Based Questions8. (i) (998)3 = (1000 – 2)3
We know that(a – b)3 = a3 – b3 – 3ab(a – b) (½)
Hence, (998)3 = (1000)3 – (2)3 – (3) (1000) (2) (1000 – 2) (½)= 1000000000 – 8 – 6000 × 998= 1000000000 – 5988008= 994011992. (1)
(ii) Algebraic Identities (½)(iii) Satisfaction solves the identity crisis among people. (½)
SUMMATIVE ASSESSMENT WORKSHEET-15
Section A1. x – 2 = 0 ⇒ x = 2
p(2) = 2 × (2)2 + 3 × 2 – k = 0 ½⇒ 8 + 6 – k = 0 ⇒ k = 14 ½
2. For zeros put p(x) = 0 ½x (x – 2) (x – 3) = 0,
then x = 0, 2, 3 ½
Section B
3. + − + +
2 42 4
1 1 1 1x x x x
x x x x = − + +
2 2 42 2 4
1 1 1x x x
x x x ,
as (a + b) (a – b) = a2 – b2 (1)
= 4 4
4 41 1
x xx x
− + (½)
= − 8
8
1x
x. (½)
YLOP LIM AON S
P-24 T I C S IX--M A T M AEH M - 1 SE RT
4. − +
2
14 2a b
= + − +
2
14 2a b
(½)
= − + − + + × × − + × × + × ×
222(1) 2 2 (1) 2 (1)
4 2 4 2 2 4a b a b b a
(1)
= + + − − +2 2
1 .16 4 4 2a b ab a
b (½)
5. Let 14 = a, 13 = b, – 27 = ca + b + c = 14 + 13 – 27 = 0 (½)
∴ a3 + b3 + c3 = 3abc (1)(14)3 + (13)3 + (– 27)3 = 3 × 14 × 13 × (– 27)
= – 14742. (½)
Section C6. (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) (1)
= 280 + 2 × 92
(½)
(a + b + c)2 = 280 + 9 = 289
a + b + c = =289 17 (½)
(a + b + c)3 = 173 = 4913. (1)7. p(x) = 3x2 – mx – nx
If (x – a) is a factor of p(x), thenp(a) = 0 (1)
3(a)2 – m × a – n × a = 0 (½)
0,a ≠ a[3a – m – n] = 0 (½)Since 3a – m – n = 0 (½)
∴ a = +3
m n. (½)
8. a7 + ab6 = a(a6 + b6) (½)= a[(a2)3 + (b2)3] (1)= a(a2 + b2) [(a2)2 + (b2)2 – a2 × b2] (1)= a(a2 + b2) (a4 + b4 – a2b2). (½)
Section D
9. +
21xx
= + + × ×22
1 12x x
x x(1)
+
21xx
= 7 + 2 = 9 (½)
+1
xx
= =9 3 (½)
+
33 1( )x
x=
+ + × 2
2
1 1 1–x x x
x x x (1)
+33
1x
x= 3 × (7 – 1) (½)
= 3 × 6 = 18. (½)qq
P-25
SUMMATIVE ASSESSMENT WORKSHEET-16
Section A
1. Degree of a polynomial 3 is 0. 12. It is not defined. 1
Section B3. 1033 = (100 + 3)3 (½)
= 1003 + 33 + 3 × 100 × 3 (100 + 3) (½)
= 1000000 + 27 + 900 × 103
= 1000000 + 27 + 92700 (½)
= 1092727. (½)
4. f(x) = 2x3 – 3x2 + 7x – 6
Put x – 1 = 0 or x = 1 in f(x) (½)
f(1) = 2 × 13 – 3 × 12 + 7 × 1 – 6 (½)
= 2 – 3 + 7 – 6 (½)
= 0 (½)
Hence, (x – 1) is a factor of f(x).
Section C5. 125x3 – 27y3 + z3 + 45xyz= (5x)3 + (– 3y)3 + (z)3 – 3 × (5x) (– 3y) (z) (1)
= (5x – 3y + z) [(5x)2 + (– 3y)2 + (z)2 – (5x)(– 3y) – (– 3y) (z) – (5x) (z)] (1½)= (5x – 3x + z) [25x2 + 9y2 + z2 + 15xy + 3yz – 5xz] (½)
6. 3 – 12(a – b)2 = 3[1 – 4(a – b)2] (1)
= 3[(1)2 – {2(a – b)}2] (1)
= 3(1 + 2a – 2b) (1 – 2a + 2b). (1)7. 9x2 + y2 + z2 – 6xy + 2yz – 6xz= (– 3x)2 + (y)2 + (z)2 + 2 × (– 3x) (y) + 2 × (y) (z) + 2 (– 3x) (z) (1)
= (– 3x + y + z)2 (1)If x = 1, y = 2, z = – 1, then
(– 3x + y + z)2 = (– 3 × 1 + 2 – 1)2 (½)= (– 3 + 2 – 1)2
= 4. (½)
Section D8. p(x) = x4 + 1
Put x – 1 = 0 or x = 1 in p(x)p(1) = (1)4 + 1 (2)
= 1 + 1 = 2Hence, we must be subtracted 2 from x4 + 1 so that it is exactly divisible by (x – 1). (1)Resultant polynomial for divisible by (x – 1)
= x4 + 1 – 2= x4 – 1. (1)
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P-26 T I C S IX--M A T M AEH M - 1 SE RT
Value Based Questions9. (i) (I) Possible length and breadth of the rectangle are the factors of its given area,
Area = 25a2 – 35a + 12= 25a2 – 15a – 20a + 12= 5a(5a – 3) – 4(5a – 3)= (5a – 4)(5a – 3)
So, possible length and breadth are (5a – 3) and (5a – 4) units, respectively. (1)(II) Area = 35y2 + 13y – 12
= 35y2 + 28y – 15y – 12= 7y(5y + 4) – 3(5y + 4)= (7y – 3)(5y + 4)
So, possible length and breadth are (7y – 3) and (5y + 4) units (1)(ii) Factorisation of Polynomials (½)(iii) Expression of one’s desires and news is very necessary. (½)
SUMMATIVE ASSESSMENT WORKSHEET-17
Section A1. Let p(y) = 5y3 – 2y2 – 7y + 1 ½
then, remainder = p (0) = 0 – 0 – 0 + 1 = 1 ½
2.2 2x y
xy+
= – 1 ⇒ x2 + y2 = – xy ½
⇒ x2 + y2 + xy = 0 ½x3 – y3 = (x – y) (x2 + y2 + xy)
= (x – y) (0)= 0
Section B3. x2 – 9 = (97)2 – (3)2 ½
= (97 + 3) (97 – 3) ½= 100 × 94 ½= 9400 ½
4. Area of rectangle = 25a2 – 35a + 12= 25a2 – 20a – 15a + 12 (1)= 5a(5a – 4) – 3(5a – 4) (½)= (5a – 3) (5a – 4) = length × breadth (½)
∴ length and breadth are (5a – 3) and (5a – 4) respectively.
Section C5. p(x) = x6 – ax5 + x4 – ax3 + 3x – a + 2
(x – a) is a factor of polynomial p(x), thenp(a) = 0 (1)
a6 – a × a5 + a4 – a × a3 + 3 × a – a + 2 = 0 (1)a6 – a6 + a4 – a4 + 3a – a + 2 = 0 (½)
2a = – 2 ⇒ a = – 1. (½)
P-27
6. p(x) = 2x3 – 5x2 + px + bx = 2 and x = 0 are zeroes of polynomial, then
p(2) = 2 × 23 – 5 × 22 + p × 2 + b = 0 (½)
⇒ 16 – 20 + 2p + b = 0
⇒ 2p + b = 4 ...(i) (½)
and p(0) = 2 × 0 – 5 × 0 + p × 0 + b = 0 (½)
⇒ b = 0 ... (ii) (½)
From (i) and (ii), we get
2p + 0 = 4 (½)
p = 4/2 = 2.
then, p = 2 and b = 0. (½)
Section D7. Let q(x) = px2 + 5x + r
(x – 2) and ( )12x − are factors of q(x), then
q(2) = p(2)2 + 5 × 2 + r = 04p + 10 + r = 0 (1)
4p + r = – 10 ...(1) .
( )12q + = ( )21 15 02 2p r+ × + =
5 04 2p r+ + = (1)
54 2p r+ = − ...(2) .
Subtracting eqn. (1) from eqn. (2), we get
44p r p r+ − − = 5 102
− +
164
p p−=
5 202
− +
154
p−= 15
2 (1)
p = – 2From eqn. (1), 4 × (– 2) + r = – 10
r = – 10 + 8r = – 2 = p. (1)
SUMMATIVE ASSESSMENT WORKSHEET-18
Section A1. p(0) = 0 – 3 × 0 + 2 = 2 ½
p(2) = 22 – 3 × 2 + 2 = 4 – 6 + 2 = 0 ½p(0) + p(2) = 2 + 0 = 2
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P-28 T I C S IX--M A T M AEH M - 1 SE RT
2.1 32 2
x x + +
= 2 3 1 32 2 4
x x x+ + + ½
= 2 32
4x x+ + ½
Section B3. x4 – y4 = (x2)2 – (y2)2 (1)
= (x2 – y2) (x2 + y2) (½)
= (x – y) (x + y) (x2 + y2). (½)4. (x + 2y)2 = x2 + 4y2 + 2 × x × 2y (1)
= 17 + 4 × 2 (½)
= 17 + 8 = 25
(x + 2y) = =25 5 . (½)5. 249 × 251 = (250 – 1) (250 + 1) (½)
= (250)2 – (1)2 (½)= 62500 – 1 (½)= 62499. (½)
Section C
6. + − −
2 22 15 23 3
a a = + + − + − +
2 1 2 15 2 5 2
3 3 3 3a a a a (1)
= − + + +
2 1 2 17 3
3 3a a (1)
= + +
1 37 3
3 3a a (½)
= + +
17 (3 1)
3a a . (½)
7. a + b + c = 9 ½(a + b + c)2 = 92 ½
a2 + b2 + c2 + 2(ab + bc + ca) = 81 ½35 + 2(ab + bc + ca) = 81 ½
2(ab + bc + ca) = 81 – 35 = 46 ½ab + bc + ca = 46/2 ½ab + bc + ca = 23.
Section D8. (a + b + c)2 – (a – b – c)2 + 4b2 – 4c2
= (a + b + c + a – b – c)(a + b + c – a + b + c) + (2b)2 – (2c)2 (1)= 2a × (2b + 2c) + (2b – 2c)(2b + 2c) (1)= (2b + 2c)(2a + 2b – 2c) (1)= 2(b + c) × 2(a + b – c) (½)= 4(b + c)(a + b – c). (½)
9. Let p(x) = x3 + 2x2 – 5ax – 8
and g(x) = x3 + ax2 – 12x – 6
P-29
When p(x) and q(x) divided by (x – 2) and (x – 3) leave remainders p and q, then
p(2) = p and g(3) = q
p(2) = 23 + 2 × 22 – 5a × 2 – 8 (1½)
= 8 + 8 – 10a – 8
p(2) = 8 – 10a = p ...(1)
g(3) = 33 + a × 32 – 12 × 3 – 6
g(3) = 27 + 9a – 36 – 6
g(3) = – 15 + 9a = q (1½)
If q – p = 10
∴ – 15 + 9a – 8 + 10a = 10
19a – 23 = 10
19a = 33
a = 3319 . (1)
SUMMATIVE ASSESSMENT WORKSHEET-19
Section A
1. Coefficient of x in expression 2 72
x xπ
+ − is 2π
1
2. Put x + 1 = 0 ⇒ x = – 1 in given expression, so ½f(– 1) = (– 1)11 + 101 = – 1 + 101 = 100 ½
Section B3. m (m – 1) – n (n – 1) = m2 – m – n2 + n (½)
= m2 – n2 – m + n= (m – n) (m + n) – (m – n) (1)= (m – n) (m + n – 1). (½)
4. −1
xx
= 3
Squaring both sides, we get
−
21xx
= ( )23 (½)
+ − × ×22
1 12x x
x x= 3 (1)
+22
1x
x= 3 + 2 = 5. (½)
5. p(x) = x3 + x2 + x + 1
Put −12
x = 0 ⇒ =12
x in p(x) (½)
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P-30 T I C S IX--M A T M AEH M - 1 SE RT
Remainder =
12
p (½)
= + + +
3 21 1 1 12 2 2
(½)
= + + +1 1 1
18 4 2
= + + +1 2 4 8
8 =
158
. (½)
Section C6. p(x) = 3x3 – 5x2 + kx – 2
Remainder on dividing p(x) by (x + 2) is :p(– 2) = 3(– 2)3 – 5(– 2)2 + k(– 2) – 2 (½)
= – 24 – 20 – 2k – 2= – 46 – 2k ...(1) (½)
q(x) = – x3 – x2 + 7x + kRemainder on dividing q(x) by (x + 2) is :
q(– 2) = – (– 2)3 – (– 2)2 + 7(– 2) + k (½)= 8 – 4 – 14 + k= – 10 + k ...(2) (½)
Since remainders are same, then from eqs. (1) and (2), we have– 46 – 2k = – 10 + k (½)
3k = – 36k = – 12. (½)
7. x3 + y3 – 12xy + 64= (x)3 + (y)3 + (4)3 – 3 × (x) (y) (4) (1)= (x + y + 4) (x2 + y2 + 16 – xy – 4y – 4x) (1)
= 0 × (x2 + y2 + 16 – xy – 4y – 4x) = 0 (1)
8. a6 – b6 1
(a3)2 – (b3)2 1
⇒ (a3 + b3) (a3 – b3) 1
Section D9. x3 – 8y3 – 36xy – 216
= (x)3 + (– 2y)3 + (– 6)3 – 3(x)(– 2y)(– 6) (1)
= [x + (– 2y) + (– 6)][x2 + (– 2y)2 + (– 6)2 – (x)(– 2y) – (– 2y)(– 6) – (x)(– 6)] (1)
= (x – 2y – 6) (x2 + 4y2 + 36 + 2xy – 12y + 6x) (½)
= 0 × [x2 + 4y2 + 36 + 2xy – 12y + 6x], (∵ x = 2y + 6 or x – 2y – 6 = 0) (1)
∴ x3 – 8y3 – 36xy – 216 = 0. (½)
10. (a) 4a2 – 9b2 – 2a – 3b= (2a)2 – (3b)2 – (2a + 3b) (½)
= (2a – 3b) (2a + 3b) – (2a + 3b) (1)
= (2a + 3b) (2a – 3b – 1) (½)
P-31
(b) a2 + b2 – 2(ab – ac + bc)= a2 + b2 – 2ab + 2ac – 2bc (½)= (a – b)2 + 2c(a – b) (½)= (a – b) [(a – b) + 2c]
= (a – b) (a – b + 2c). (1)qq
SUMMATIVE ASSESSMENT WORKSHEET-20
Section A1. a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
a3 + b3 + c3 – 3abc = 0, as a + b + c = 0 ½a3 + b3 + c3 = 3abc ½
2. (x – 2)3 = (x)3 – (2)3 – 3 × x × 2 (x – 2)= x3 – 8 – 6x2 + 12x ½
coefficient of x2 in the expension of (x – 2)3 = – 6. ½
Section B3. Let x2 + 7 = p and 2x – 1 = q, then given expression
= 12p2 – 8pq – 15q2
= 12p2 – 18pq + 10pq – 15q2 (½)= 6p(2p – 3q) + 5q(2p – 3q) (½)= (2p – 3q) (6p + 5q)= [2(x2 + 7) – 3(2x – 1)] [6(x2 + 7) + 5(2x – 1)] (½)= (2x2 + 14 – 6x + 3) (6x2 + 42 + 10x – 5)= (2x2 – 6x + 17) (6x2 + 10x + 37). (½)
Section C4. 2x3 + 2y3 + 2z3 – 6xyz= 2(x3 + y3 + z3 – 3xyz) (½)
= 2(x + y + z)(x2 + y2 + z2 – xy – yz – zx) (1)= (x + y + z)(2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx) (½)= (x + y + z)[x2 + y2 – 2xy + y2 + z2 – 2yz + z2 + x2 – 2zx] (½)= (x + y + z)[(x – y)2 + (y – z)2 + (z – x)2]. Hence Proved. (½)
5.( ) ( ) ( )
( ) ( ) ( )− + − + −
− + − + −
3 3 32 2 2 2 2 2
3 3 3
a b b c c a
a b b c c a
Both Numerator and Denominator are of the form a3 + b3 + c3 (½)We know that when a + b + c = 0Then a3 + b3 + c3 = 3abc (½)For Numerator a2 – b2 + b2 – c2 + c2 – a2 = 0For Denominator a – b + b – c + c – a = 0 (½)
∴( ) ( ) ( )
( ) ( ) ( )− + − + −
− + − + −
3 3 32 2 2 2 2 2
3 3 3
a b b c c a
a b b c c a=
( ) ( ) ( )( )
× − − −
− − −
2 2 2 2 2 233 ( ) ( )a b b c c a
a b b c b c (½)
= − + − + − +
− − −( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( )a b a b b c b c c a c a
a b b c c a (½)
= (a + b) (b + c) (c + a). (½)
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P-32 T I C S IX--M A T M AEH M - 1 SE RT
6. p(x) = x3 + 3x2 – 2x + 4
p(2) = (2)3 + 3(2)2 – 2 × 2 + 4 (½)= 8 + 12 – 4 + 4 = 20 (½)
p(– 2) = (– 2)3 + 3(– 2)2 – 2(– 2) + 4 (½)= – 8 + 12 + 4 + 4 = 12 (½)
p(0) = 0 + 0 – 0 + 4 = 4 (½)p(2) + p(– 2) – p(0) = 20 + 12 – 4 = 28. (½)
Section D7. 2x3 + 2y3 + 2z3 – 6xyz= 2(x3 + y3 + z3 – 3xyz) (½)
= 2(x + y + z)(x2 + y2 + z2 – xy – yz – zx) (½)= (x + y + z)(2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx) (½)= (x + y + z)[x2 + y2 – 2xy + y2 + z2 – 2yz + z2 + x2 – 2zx] (½)= (x + y + z)[(x – y)2 + (y – z)2 + (z – x)2]. Hence Proved. (½)
Now, 2(13)3 + 2(14)3 + 2(15)3 – 6 × 13 × 14 × 15= (13 + 14 + 15)[(13 – 14)2 + (14 – 15)2 + (15 – 13)2] (½)= 42 × [1 + 1 + 4] (½)= 42 × 6 = 252. (½)
8.
+
− + − + +
− +− + −
− +±∓
2 3 2
3 2
2
2
3 1(Quotient)3 2 3 8 3 2
3 9 6
3 2 – 3 2
0 (Remainder)
xx x x x x
x x x
x xx x (2)
Divisor × Quotient= (x2 – 3x + 2) (3x + 1) (1)= 3x3 – 8x2 + 3x + 2
∴ Dividend= Divisor × Quotient + Remainder (Division algorithm) (1)qq
SUMMATIVE ASSESSMENT WORKSHEET-21
Section A
1.2
21xx
+ = 21
2xx
+ −
= (4)2 – 2 = 16 – 2 = 14 1
2.3 3
2 283 17
83 83 17 17
+
− × +=
2 2
2 2(83 17)(83 83 17 17 )
(83 83 17 17 )
+ − × +
− × + = (83 + 17) = 100 1
Section B3. 2y3 + y2 – 2y – 1 = y2 (2y + 1) – 1 (2y + 1) 1
= (2y + 1) (y2 – 1) ½= (2y + 1) (y – 1) (y + 1) ½
P-33
4. x4 – 125xy3 = x(x3 – 125y3) (½)
= x[(x)3 – (5y)3]= x(x – 5y) [x2 + (5y)2 + x × 5y] (1)
= x(x – 5y) (x2 + 25y2 + 5xy). (½)
Section C
5. +1
xx
= 3
+
31xx
= (3)3 (1)
x3 + + × × +
31 1 13 x xx x x
= 27 ⇒ x3 + 31
x = 27 – 9 = 18 (2)
6. Given a + b + c = 3x
or 3x – a – b – c = 0 (½)
∴ (x – a)3 + (x – b)3 + (x – c)3 – 3(x – a) (x – b) (x – c)
= [x – a + x – b + x – c][(x – a)2 + (x – b)2 + (x – c)2 – (x – a)(x – b) – (x – b)(x – c) – (x – a)(x – c)] (1½)
= [3x – a – b – c][(x – a)2 + (x – b)2 + (x – c)2 – (x – a)(x – b) – (x – b)(x – c) – (x – a)(x – c)] (½)
= 0 × [(x – a)2 + (x – b)2 + (x – c)2 – (x – a)(x – b) – (x – b)(x – c) – (x – c)(x – a)] = 0. (½)
7. (a2 – 2a)2 – 23(a2 – 2a) + 120= (a2 – 2a)2 – 15(a2 – 2a) – 8(a2 – 2a) + 120 (½)
= (a2 – 2a)[a2 – 2a – 15] – 8[a2 – 2a – 15] (½)
= (a2 – 2a – 15)(a2 – 2a – 8) (½)
= (a2 – 5a + 3a – 15)(a2 – 4a + 2a – 8) (½)
= [a(a – 5) + 3(a – 5)][a(a – 4) + 2(a – 4)] (½)
= (a – 5)(a + 3)(a – 4)(a + 2). (½)
Section D
8. (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) (1)(6)2 = a2 + b2 + c2 + 2 × 11 (½)
a2 + b2 + c2 = 36 – 22 = 14 (½)a3 + b3 + c3 – 3abc = (a + b + c) [a2 + b2 + c2 – (ab + bc + ca)] (1)
= 6 × (14 – 11) = 6 × 3 = 18. (1)Alternative method.
a3 + b3 + c3 – 3abc = (a + b + c) [a2 + b2 + c2 – (ab + bc + ca)] (1)= (a + b + c) [(a + b + c)2 – 3(ab + bc + ca)] (1)= 6[62 – 3(11)] = 6[36 – 33) = 6 × 3 = 18. (2)
9. (2x + 3y)3 = (2x)3 + (3y)3 + 3 × 2x × 3y(2x + 3y) (1)= 8x3 + 27y3 + 18(2x2y + 3xy2) (1)= 730 + 18 × 15 (½)= 730 + 270 (½)
(2x + 3y)3 = 1000 (½)
2x + 3y = 3 1000 = 10. (½)qq
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P-34 T I C S IX--M A T M AEH M - 1 SE RT
SUMMATIVE ASSESSMENT WORKSHEET-22
Section A1. Degree of x3 + 5 = 3, Degree of 4 – x5 = 5 ½
∴ Degree of (x3 + 5) (4 – x5) = 3 + 5 = 8 ½2. a = 3 + b ⇒ a – b = 3
(a – b)3 = 33 ⇒ a3 – b3 – 3ab (a – b) = 27 ½⇒ a3 – b3 – 9ab = 27 ½
Section B
3. +3 364 125x y = ( )+3
3(4 ) 5x y (½)
= ( ) ( ) + + − × 224 5 (4 ) 5 4 5x y x y x y (1)
= ( ) + + − 2 24 5 16 5 4 5x y x y xy . (½)
4. x4y4 – 256z4 = (x2y2)2 – (16z2)2 (½)
= ( ) ( )− +2 2 2 2 2 216 16x y z x y z (½)
= ( ) − + 2 2 2 2 2(4 ) 16xy z x y z (½)
= ( ) ( ) ( )− + +2 2 24 4 16xy z xy z x y z . (½)
Section C5. Let p(x) = 3x3 + x2 – 20x + 12
Given (3x – 2) is a factor of p(x).
3x – 2 ) 3x3 + x2 – 20x + 12 ( + −2 6x x 3x3 – 2x2
3x2 – 20x + 12 3x2 – 2x (1½) – 18x + 12 – 18x + 12 ×
x2 + x – 6 = x2 + 3x – 2x – 6 (½)= x(x + 3) – 2(x + 3) (½)= (x + 3) (x – 2) (½)
The other factors are (x – 2) and (x + 3).
Section D6.
3x + 4) 6x4 + 11x3 + 13x2 – 3x + 27 ( 2x3 + x2 + 3x – 5 quotient (½) 6x4 + 8x3
3x3 + 13x2 (½) 3x3 + 4x2 (½) 9x2 – 3x 9x2 + 12x (½)
– 15x + 27 – 15x – 20 (½)
47Thus, quotient = 2x3 + x2 + 3x – 5 and remainder = 47.
– +
– +
– –
– –
– –
+ –
P-35
7. R.H.S. = (x – y) (x2 + y2 + xy)= x3 + xy2 + x2y – x2y – y3 – xy2 (1)
= x3 – y3 = L.H.S. (1)
Now, 216x3 – 125y3 = (6x)3 – (5y)3 (½)
= (6x – 5y) [(6x)2 + (5y)2 + 6x × 5y] (1)
= (6x – 5y) (36x2 + 25y2 + 30xy]. (½)
SUMMATIVE ASSESSMENT WORKSHEET-23
Section A1. On expanding we get
⇒ 3 3 2 21 8 12 6
27 27 27 27x y xy x y− + − ½
= 3
3 2 28 4 227 27 9 9x y xy x y− + − ½
Section B2. (2x – 3y + z)2 = [2x + (– 3y) + z]2 (½)
= (2x)2 + (– 3y)2 + z2 + 2 × 2x × (– 3y) + 2 × (– 3y) × z + 2 × 2x × z (1)= 4x2 + 9y2 + z2 – 12xy – 6yz + 4xz. (½)
Section C
3. 3 21 9 127 216 2 4p p p− − + = ( ) ( )3 23 21 1 1(3 ) 3 (3 ) 3 36 6 6p p p− − × × + × (1½)
= − − × × −
33 1 1 1(3 ) 3 3 3
6 6 6p p p (½)
= −
3136
p . (1)
Section D4. Let p(x) = x2 + px + q and q(x) = x2 + mx + n
Since x + a is a factor of p(x) and q(x), then p(– a) = 0 and q( – a) = 0 (1)p(– a) = 0 ⇒ (– a)2 + p(– a) + q = 0
a2 – pa + q = 0 ...(1) (½)q(– a) = 0 ⇒ (– a)2 + m(– a) + n = 0
a2 – ma + n = 0 ...(2) (½)Subtracting eqn. (2) from eqn. (1), we get
a2 – pa + q – a2 + ma – n = 0 (1)
a(m – p) = n – q
a = n qm p
−−
. (1)
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5. b = ( )
( )( )
−= ×
+ −
5 2 61 15 2 6 5 2 6a
(½)
= ( )−
= −−
5 2 65 2 6
25 24(½)
∴ a + b = + + − =5 2 6 5 2 5 10 (½)
Also b = 1 1aba
⇒ = (½)
a2 + b2 = (a + b)2 – 2ab (½)= (10)2 – 2 × 1= 100 – 2 = 98 (½)
(a3 + b3) = (a + b)3 – 3ab(a + b) (½)= (10)3 – 3 × 1 × 10= 1000 – 30 = 970. (½)
6. Volume of cuboid = 8x3 + 12x2 – 2x – 3= 4x2(2x + 3) – 1 (2x + 3) (½)= (2x + 3) (4x2 – 1) (½)= (2x + 3) (2x – 1) (2x + 1) (½)
The possible dimensions are (2x – 1), (2x + 1) and (2x + 3). (½)Verification : For x = 5Dimensions are (2 × 5 – 1), (2 × 5 + 1) and (2 × 5 + 3) i.e., 9, 11, 13 units. (½)
Volume = 9 × 11 × 13 = 1287 cubic units (½)Given volume = 8(5)3 + 12(5)2 – 2(5) – 3 (½)
= 1000 + 300 – 10 – 3= 1287 cubic units. (½)
7. x4 + 2x3y – 2xy3 – y4 = (x2)2 – (y2)2 + 2x3y – 2xy3 (½)= (x2 – y2) (x2 + y2) + 2xy(x2 – y2) (1)= (x2 – y2) (x2 + y2 + 2xy) (1)= (x – y) (x + y) (x + y)2 (1)= (x – y) (x + y)3. (½)
SUMMATIVE ASSESSMENT WORKSHEET-24
Section A1. 4x2 + 3x – 6
x + 1 ) 4x3 + 7x2 – 3x – 6 4x3 + 4x2
3x2 – 3x So, quotient is 4x2 + 3x – 6 1 3x2 + 3x – 6x – 6 – 6x – 6 ×
P-37
2.3 3 343
729p q + =
33 7
( )9
pq +
½
= + + −
2 2 77 499 81 9
pqpq p q ½
Section B
3.3 3 38 1 1
15 3 5 − −
= 75x
⇒512 1 1
3375 27 125− − = 75
x½+½
⇒ on solving we get, x = 8 ½+½
Section C4. Factors of 12 = (+ 1, + 2, + 3, + 4, + 6, + 12) (½)
p(x) = 6x3 – 25x2 + 32x – 12p(2) = 6(2)3 – 25(2)2 + 32 × 2 – 12
= 48 – 100 + 64 – 12 = 112 – 112 = 0 (½)∴ x = 2 is a zero of p(x) or (x – 2) is a factor of p(x) (½)
6x3 – 25x2 + 32x – 12 = 6x2(x – 2) – 13x(x – 2) + 6(x – 2) (½)= (x – 2)(6x2 – 13x + 6)= (x – 2)(6x2 – 9x – 4x + 6) (½)= (x – 2)[3x(2x – 3) – 2(2x – 3)]= (x – 2)(2x – 3)(3x – 2). (½)
5. (i) + + − −22
1 22 2x x
x x=
+ + × × − −
22 1 1 2( ) 2 2x x x
x x x(1)
= + − +
21 12x xx x
(1)
= + + −
1 12x x
x x . (1)
(ii) x4 – x4 = (x2)2 – (y2)2
= (x2 + y2) (x2 – y2)= (x2 + y2) (x – y) (x + y)
Section D6. Let p(x) = ax4 + 2x3 – 3x2 + bx – 4
x2 – 4 or (x – 2) (x + 2) is a factor of p(x), then put x = 2 in p(x)p(2) = a(2)4 + 2(2)3 – 3(2)2 + b × 2 – 4 = 0
16a + 16 – 12 + 2b – 4 = 0 16a + 2b = 0 8a + b = 0 ...(1) (1)
Again put x = – 2 in p(x)p(– 2) = a(– 2)4 + 2(– 2)3 – 3(– 2)2 + b × – 2 – 4 = 0
16a – 16 – 12 – 2b – 4 = 016a – 2b = 32 (1)8a – b = 16 ...(2) .
Adding eqs. (1) and (2)16a = 16 ⇒ a = 1 (½)
+ +
− + − − −
−− +
+ − −−
− +
−−
− +
2
2 4 3 2
4 2
3 2
3
2
2
2 14 2 3 8 4
4
2 8 4 2 8
44
0
x xx x x x x
x x
x x xx x
xx
YLOP LIM AON S
P-38 T I C S IX--M A T M AEH M - 1 SE RT
By equation (1)8 × 1 + b = 0 ⇒ b = – 8 (½)
p(x) = x4 + 2x3 – 3x2 – 8x – 4= (x2 – 4) (x2 + 2x + 1) (1)
= (x – 2) (x + 2) (x + 1)2.7. Let f(x) = ax3 – 3x2 + 4
and g(x) = 2x3 – 5x + aWhen divided by (x – 2)
f(2) = p and g(2) = q (1½)f(2) = a × 23 – 3 × 22 + 4
p = 8a – 12 + 4p = 8a – 8 ...(1)
g(2) = 2 × 23 – 5 × 2 + aq = 16 – 10 + a (1½)q = 6 + a ...(2)
p – 2q = 4, (given)8a – 8 – 12 – 2a = 4
6a – 20 = 4a = 4. (1)
SUMMATIVE ASSESSMENT WORKSHEET-25
Section A1. The zero of a polynomial is
p(0) = 3(0)2 + 2(0) – 1 = – 1 12. 4x3 – x8 + 20 – 5x5
∴ degree = 8 1
Section B3. 8 – 27a3 – 36a + 54a2 1
(2)3 – (3a)3 – 3(2)2 (3a) + 3 (3a)2 (2) ½= (2 – 3a)3 ½
4. a2 + 4b2 + 9c2 + 4ab + 12bc + 6ac – a2 – 4b2 – 9c2 + 4ab – 12bc + 6ac – 6b2 – 9b 1= 8ab + 12ac – 6b2 – 9b 1
Section D5.
− −
− −
+ +
+ −
+ + − −
+− −+
− −− −
−
2
3 2
3 2
2
2
3 61 4 3 10
3 3 10
3 36 106 6
4
x xx x x x
x xx xx x
xx
= Remainder
(2)
P-39
Verification :Let f(x) = x3 + 4x2 – 3x – 10 (1)
By remainder theorem, remainder
f(– 1) = (– 1)3 + 4(– 1)2 – 3(– 1) – 10
= – 1 + 4 + 3 – 10 = – 4. (1)
6. Let p(x) = bx3 + 3x2 – 3Put x – 4 = 0 or x = 4 in p(x), we get
p(4) = b(4)3 + 3(4)2 – 3 = R1 (½)64b + 48 – 3 = R1
64b + 45 = R1 (½)Let q(x) = 2x3 – 5x + bAgain put x – 4 = 0 or x = 4 in q(x), we get
q(4) = 2(4)3 – 5(4) + b = R2 (½)= 128 – 20 + b = R2
108 + b = R2Given that 2R1 – R2 = 0 (½)⇒ 2(64b + 45) – (108 + b) = 0 (1)
128b + 90 – 108 – b = 0 (½)127b – 18 = 0
b = 18127
. (½)
7. (x + y)3 + (y + z)3 + (z + x)3 – 3(x + y)(y + z)(z + x)= (x + y + y + z + z + x) [(x + y)2 + (y + z)2 + (z + x)2
– (x + y)(y + z) – (y + z)(z + x) – (x + y)(z + x)]= 2(x + y + z)[x2 + y2 + 2xy + y2 + z2 + 2yz + z2 + x2 + 2zx – xy – xz – y2 – yz – yz – xy – z2 – zx – xz – x2 – yz – xy]= 2(x + y + z) [x2 + y2 + z2 – xy – yz – zx]= 2(x3 + y3 + z3 – 3xyz). (1+1+1+1)
SUMMATIVE ASSESSMENT WORKSHEET-26
Section A1. (x + y + z) (x2 + y2 + z2 + 2xy + 2yz + 2xz)
⇒ It is the expandable form of ½(x + y + z)3 ½
Section B2. (2x)3 + (3y)3 = (2x + 3y) (4x2 + 9y2 – 6xy)
= (8) (4x2 + 9y2 – 12) 1= 32x2 + 72y2 – 96 1
Section C3. (x2 – 4x)(x2 – 4x – 1) – 20 = (x2 – 4x)2 – (x2 – 4x) – 20 (½)
= (x2 – 4x)2 – 5(x2 – 4x) + 4(x2 – 4x) – 20 (½)
= (x2 – 4x)[x2 – 4x – 5] + 4[x2 – 4x – 5] (½)
YLOP LIM AON S
P-40 T I C S IX--M A T M AEH M - 1 SE RT
= (x2 – 4x – 5) (x2 – 4x + 4)
= (x2 – 5x + x – 5)[(x)2 – 2 × 2 × x + (2)2] (½)
= [x(x – 5) + 1(x – 5)][x – 2]2 (½)
= (x – 5)(x + 1) (x – 2)(x – 2). (½)
Section D4. g(x) = x2 – 3x + 2
= x2 – 2x – x + 2= x(x – 2) – 1(x – 2)= (x – 2)(x – 1) (1)
Zero of x – 2 is 2Zero of x – 1 is 1
f(x) = 2x4 – 6x3 + 3x2 + 3x – 2f(2) = 2(24) – 6(23) + 3(22) + 3(2) – 2 (1)
= 32 – 48 + 12 + 6 – 2 = 0f(1) = 2(1)4 – 6(1)3 + 3(1)2 + 3(1) – 2 (1)
= 2 – 6 + 3 + 3 – 2 = 0⇒ (x – 1) and (x – 2) are factors of f(x).∴ f(x) is exactly divisible by g(x). (1)
5. (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx) (1)
(10)2 = 40 + 2(xy + yz + zx)
100 – 40 = 2(xy + yz + zx) (1)
xy + yz + zx = 60 302 = (½)
and x3 + y3 + z3 – 3xyz = (x + y + z) [x2 + y2 + z2 – (xy + yz + zx)] (1)
= 10 [40 – 30]
= 10 × 10 = 100. (½)
6. x – 3 = 0 ⇒ x = 3 ½= a (3)3 + 4(3)2 + 3 (3) – 4 ½= 27a + 41 ...(1) 1= (3)3 – 4(3) + a= 15 + a ...(2) ½
As (1) = (2)∴ 27a + 41 = 15 + a ½
a ⇒ – 1 1qq
SUMMATIVE ASSESSMENT WORKSHEET-27
Section A1. ∴ Given x + 2 = 0 ⇒ x = – 2
3(– 2)3 – 5 (– 2)2 + k (– 2) – 2 = – (– 2)3 – (– 2)2 + 7 (– 2) + k– 24 – 20 – 2k – 2 = + 8 – 4 – 14 + k ½
k = – 12 ½
P-41
Section B2. (x + 2)2 + p2 + 2p(x + 2) = (x + 2)2 + (p)2 + 2 × (x + 2) × p (1)
= (x + 2 + p)2, [∵ a2 + b2 + 2ab = (a + b)2] (1)
Section C3. 250x3 – 432y3 = 2[125x3 – 216y3] (½)
= 2[(5x)3 – (6y)3] (½)
= 2(5x – 6y) [(5x)2 + (6y)2 + 5x × 6y] (1½)
= 2(5x – 6y) (25x2 + 36y2 + 30xy). (½)
4. p(x) = x3 – 3x2 – 9x – 5Put x = – 1
p(– 1) = (– 1)3 – 3(– 1)2 – 9(– 1) – 5 (½)= – 1 – 3 + 9 – 5= 0 (½)
Hence, (x + 1) is a factor of p(x), thenx3 – 3x2 – 9x – 5 = x2(x + 1) – 4x(x + 1) – 5(x + 1) (½)
= (x + 1) (x2 – 4x – 5) (½)= (x + 1) (x2 – 5x + x – 5)= (x + 1) [x(x – 5) + 1(x – 5)] (½)= (x + 1) (x – 5) (x + 1). (½)
Section D5. (a + b)3 + (a – b)3 + 6a(a2 – b2) = (a + b)3 + (a – b)3 + 3 × 2a (a – b)(a + b) (1)
= (a + b)3 + (a – b)3 + 3(a + b)(a – b)[(a + b) + (a – b)] (1)= [(a + b) + (a – b)]3 (1)= (2a)3
= 8a3 (1)
6. p3 – q3 = (p – q)(p2 + q2 + pq) (1)
= (p – q)[(p – q)2 + 3pq] (1)
= + ×
210 10 539 9 3 (1)
= +
10 1005
9 81
= +
10 100 4059 81
= 5050729
. (1)
7. Let p(x) = 2x3 – 5x2 + x + 2Put x = 1 in p(x)
p(1) = 2(1)3 – 5(1)2 + 1 + 2= 2 – 5 + 1 + 2 = 0 (1)
Hence, (x – 1) is a factor of p(x).
YLOP LIM AON S
P-42 T I C S IX--M A T M AEH M - 1 SE RT
Put x = 2 in p(x)
p(2) = 2(2)3 – 5(2)2 + 2 + 2
= 16 – 20 + 2 + 2 = 0 (1)Hence, (x – 2) is a factor of p(x).
Put x = 12
− in p(x) (1)
12
p − = 2 × − − − + − +
3 21 1 15 22 2 2
(1)
= − × − × − +1 1 1
2 5 28 4 2
= − − − +1 5 1
24 4 2
= − − − +=
1 5 2 80
4
Hence, (2x + 1) is a factor of p (x).
8. p(x) = x3 – 2x2 – 5x + 6Put x = 1 in p(x)
p(1) = (1)3 – 2(1)2 – 5 × 1 + 6 (1)= 1 – 2 – 5 + 6= 0
Hence, (x – 1) is a factor of p(x), thenx3 – 2x2 – 5x + 6 = x2(x – 1) – x(x – 1) – 6(x – 1) (1)
= (x – 1)(x2 – x – 6) (½)= (x – 1)(x2 – 3x + 2x – 6) (½)= (x – 1)[x(x – 3) + 2(x – 3)] (½)
= (x – 1)(x – 3)(x + 2). (½)
FORMATIVE ASSESSMENT WORKSHEET-28
Section AThis worksheet contains Activities which are to be perform in the class itself.
P-43
3 INTRODUCTION TO EUCLID’S GEOMETRY
SUMMATIVE ASSESSMENT WORKSHEET-29
Section A1. AB + BC = AC
x + 3 + 2x = 4x – 5⇒ 3x + 3 = 4x – 5 ½⇒ 4x – 3x = 3 + 5 ⇒ x = 8 ½
Section B2. AC = BD (given)
AB + BC = BC + CD 1AB = CD 1
3. (a) Infinite if they are collinear. 1(b) Only one if they are non-collinear. 1
Section C4. Given,
AC = DC and CB = CE 1AC + CB = CD + CE 1
⇒ AB = DE 15. Since, x + y = 10 and x = z,
∵ x + y = z + y 1⇒ 10 = z + y 1Hence, z + y = 10. 1
Section D6. To prove : ∠A > ∠C 1
Cons. : Join ACProof : In ∆ DAC, CD > AD
∠1 > ∠3 ...(i) 1In ∆ ABC : BC > AB
∠2 > ∠4 ...(ii) 1Adding equations (1) and (2), we get
∠1 + ∠2 > ∠3 + ∠4∠A > ∠C 1
SUMMATIVE ASSESSMENT WORKSHEET-30
Section A1. Dimensions of surface = Length and Breadth which is 2. 1
A
BD
E
C
A
D
B C
12
34
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P-44 T I C S IX--M A T M AEH M - 1 SE RT
Section B2. AC = DC
CB = CE (given)Adding AC + CB = DC + CE ½
AB = DE ½If equals are added to equals, the wholes also equal. 1
3.A
P BLet AB be a ⊥ to a line L and AP is any other line segment.In right ∆ ABP, ∠B > ∠P (∵ ∠B = 90º) 1⇒ AP > AB or AB < AP 1
Section C4. Since, ∠1 = ∠3 and ∠2 = ∠4, therefore
⇒ ∠1 + ∠2 = ∠3 + ∠4 1⇒ ∠BAD = ∠BCD 1⇒ ∠A = ∠C 1
Section D5. (i) The terms need to be defined are :
Polygon : A simple closed figure made up of three or more line segments. (1)Line segment : Part of a line with two end points.Line : Undefined term.Point : Undefined term.Angle : A figure formed by two rays with a common initial point.Ray : Part of a line with one end point.Right angle : Angle whose measure is 90°.Undefined terms used are : Line, Point. (1)Euclid’s fourth postulate says that ‘‘all right angles are equal to one another.’’In a square, all angles are right angles, therefore, all angles are equal (From Euclid’s fourthpostulate).Three line segments are equal to fourth line segment. (Given)Therefore, all the four sides of a square are equal (by Euclid’s first axiom ‘‘things which are equalto the same thing are equal to one another.’’) (2)
SUMMATIVE ASSESSMENT WORKSHEET-31
Section A1. Only one line can pass through two distinct point : 1
A
B
P-45
Section B2. There are two undefined terms, line and point. They are consistent, because they deal with two
different situations.
(i) Says that given two points be A and B, there is a point C, lying on the line which is betweenthem. (1)
(ii) Says that given A and B, we can take C not lying on the line passing through A and B.These ‘Postulates’ do not follow from Euclid’s postulates.However, (ii) follow from given postulate (i). (1)
3. AB = BC (given)∵ AX + BX = BY + CYSince, BX = BY ½∴ AX + BY = BY + CY ½
AX = CYAxiom : If equals are subtracted from the equals, the remainders are equals. 1
Section C
4. Here, OX = 12 XY, PX = 1
2 XZ (1)
⇒ XY = 2(OX), XZ = 2(PX)
X
Z Y
O P
Also, OX = PX, (Given) (1)XY = XZ,
(Because things which are double of the same things are equal to one another.) (1)
Section D5. (i) False : Because infinitely many lines can pass through a single point. (½)
(ii) False : Because only one line can pass through twodistinct points.
A B (1)(iii) True : Because a terminated line i.e., line segment AB can
be produced indefinitely on both the sides (Euclid’spostulate 2).
A B (1)(iv) True : Two circles are equal if :
(a) their circumferences are equal, or (1)(b) their radii are equal.
(v) True : Things which are equal to the same thing are equal to one another (Euclid’s axioms).(½)
P
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P-46 T I C S IX--M A T M AEH M - 1 SE RT
Value based question6. (i) The terms need to be defined are :
Polygon : A simple closed figure made up of three or more line segments. (1)Line segment : Part of a line with two end points.Line : Undefined term.Point : Undefined term.Angle : A figure formed by two rays with a common initial point.Ray : Part of a line with one end point.Right angle : Angle whose measure is 90°.Undefined terms used are : Line, Point. (1)Euclid’s fourth postulate says that ‘‘all right angles are equal to one another.’’In a square, all angles are right angles, therefore, all angles are equal (From Euclid’s fourthpostulate).Three line segments are equal to fourth line segment. (Given)Therefore, all the four sides of a square are equal (by Euclid’s first axiom ‘‘things which areequal to the same thing are equal to one another.’’) (1)
(ii) Introduction to Euclid’s Geometry. (½)
(iii) Equality leads to democracy. (½)qq
SUMMATIVE ASSESSMENT WORKSHEET-32
Section A1. To form a straight line. 1
Section B
2. Axion : If C be mid-point of line AB, then AC = 12
AB
AC = 12
AB ⇒ AD = 12
AC 1
⇒ AD = 1 12 2
AB
½
⇒ AD = 14
AB ½
3. x – 15 = 25On adding 15 to both sides, we have
x – 15 + 15 = 25 + 15 (½)x = 40 (½)
Euclid’s Axiom : If equals are added to equals, the wholes are equal. (1)4.
Let AB has 2 mid points, say X and Y, then (½)
2AB = AX and 2
AB = AYY (½)
∴ AX= AY (Things which are equal to the same thing are equal to one another) (½)
Hence, X and Y coincide. Thus, every line segment has one and only one mid point. (½)
P-47
Section C5. Given, ∠ABC = ∠ACB
⇒ ∠1 + ∠4 = ∠2 + ∠3 (1)⇒ ∠1 + ∠4 – ∠4 = ∠2 + ∠3 – ∠3 (As ∠3 = ∠4) (1)⇒ ∠1 = ∠2. (1)
6. AB = 2AE (E is the mid point of AB) (1)CD = 2DF (F is the mid point of CD) (1)
Also AE = DF (Given)Therefore AB = CD (Things which are double of the same thing are equal to one another) (2)
Value based question7. (i) If a straight line l falls on two straight lines m and n such that the sum of the interior angles
on one side of l is two right angles then by Euclid’s fifth postulate the lines will not meet on thisside of l. Next, we know that the sum of the interior angles on the other side of line l will also betwo right angles, Therefore, they will not meet on the other side also. So, the lines m and n nevermeet and are, therefore, parallel. (2)
(ii) Introduction to Euclid’s Geometry. (½)
(iii) Universal truth. (½)qq
SUMMATIVE ASSESSMENT WORKSHEET-33
Section A1. Only one line can pass through two distinct points. 1
Section B2. AB = AD
AC = AD ½AB = AC ½
Axiom : Things which are equal to the same thing are equal. 13. Given : ∠1 = ∠4 and ∠3 = ∠2
Again ∠2 = ∠4 ½Axiom : Things which are equal to the same thing are equal to one another, 1∴ ∠1 = ∠3 ½
4. In a circle, having centre at P, we have
PR = PQ = radius (½)In a circle having centre at Q, we have
QR = QP = radius (½)Euclid’s first axiom : Things which are equal to the same thing are equal to one another. (½)
∴ PR = PQ = QR. (½)
Section C5. Given, C is the mid-point of AB.
∴ AC = CB …(1) (1)If possible, let D be another mid-point of AB.∴ AD = DB …(2) (1)A B CD
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Now, subtracting (1) from (2), we getAD – AC = DB – CB
⇒ – CD = CD⇒ 2CD = 0⇒ CD = 0∴ C and D coincide.Hence, every line segment has one and only one mid-point. (1)
6. Given,: AC = BC (1)
A B C (1)
So, AC + AC = AC + BC (Equals are added to equals)⇒ 2AC = AB, (... AC + CB coincides with AB)
∴ AC = 12
AB. (1)
Section D7. Given : Three lines l, m and n in a plane such that l || m and m || n. (1)
To prove : l || n.Proof : If possible, let l be not parallel to n, then l and n should intersect in a unique point, say A.
(1)
m
l
n
Thus, through a point A, outside m, there are two lines l and n, both parallel to m. (1)This contradicts the parallel line axiom.So, our assumption is wrong.Hence, l || n.
8. ∠1 = ∠2 ...(1) (½)∠3 = ∠4 ...(2) (½)
Adding (1) + (2), we get∠1 + ∠3 = ∠2 + ∠4 (½)
∠ABC = ∠DBC (½)Euclid’s axiom used : If equals are added to equals, wholes are equal. (1)
FORMATIVE ASSESSMENT WORKSHEET-34
(B) Match1. D2. E3. A4. F5. C6. H7. G8. B
P-49
4 LINES AND ANGLES
SUMMATIVE ASSESSMENT WORKSHEET-35
Section A1. Let ∠a = 2x, ∠b = 3x
Then 2x + 3x = 5x = 180° (½)x = 36°
∠a = 2x = 72°∠h + ∠a = 180° (Corresponding exterior angles) (½)
∠h = 180° – 72° = 108°.
Section B
2.43 of a right angle =
4 90º 120º3
× = 1
Supplement of 120º = 180º – 120º = 60º. 1
Section C3. ∠BMN + ∠DNM = 180°, (Interior angles) (1)
⇒ 2 2BMN DNM∠ ∠+ = 180
2° ⇒ ∠1 + ∠2 = 90° (1)
∠1 + ∠2 + ∠3 = 180° ⇒ ∠3 = 90°.
PM and PN intersect at right angle. (1)
Section D4. AB || DC
y + 35° + 80° = 180° (Corresponding interior angles) (1)y = 180° – 115°y = 65° (½)
∠ABD = ∠CDB ⇒ x = 35° (Alternate angles) (1)In ∆BCD,
35° + y – 30° + z = 180° (1)z = 180° – 5° – yz = 175 – 65°
z = 110°. (½)
ENIL NDN AAS G SEL
P-50 T I C S IX--M A T M AEH M - 1 SE RT
Value based question
5. (i) ∠TOP = 12 ∠TOB (½)
∠ORS = 12 ∠ORD (½)
But ∠TOB = ∠ORD (l || m and corresponding angles) (½)∴ ∠TOP = ∠ORS (½)
But they are corresponding angles w.r.t. transversal TR and lines OP and RS.
Hence, OP || RS. (½)
(½)
(ii) Lines and angles. (½)
(iii) Bisectoring among human beings gives rise to deterioration in the society. (½)qq
SUMMATIVE ASSESSMENT WORKSHEET-36
Section A1. In ∆ABC, ∠A + ∠B + ∠C = 180º
Also, in ∆ DEF, ∠D + ∠E + ∠F = 180º ½∴ ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360º
= 4 × 90º ⇒ k = 4 ½
Section B2. In ∆ ABC, ∠A + ∠B + ∠C = 180º
2x – 5º + 5x + 5º + 3x + 50º = 180º10x + 50º = 180º ⇒ x = 13º 1
∠A = 2x – 5º = 26º – 5º = 21∠B = 5x + 5º = 65º + 5º = 70º 1
∠C = 3x + 50º = 39º + 50º = 89º
Section C3. Given, ∠POR : ∠ROQ = 5 : 7,
Let ∠POR = 5x∠ROQ = 7x ⇒ 5x + 7x = 180º ⇒ 12x = 180º 1
x = 15º∠POR = d = b = 5x = 5 × 15º = 75º 1∠ROQ = c = a = 7x = 7 × 25º = 105º 1
P-51
Section D4. 4b + 75º + b = 180º
5b = 180º – 75º = 105º (Linear pair) 1
b = 105º 21º
5= 1
4b = 4 × 21º = 84º ⇒ a = 4b = 84º 12c + a = 180º
⇒ 2c = 180º – 84º = 96º ⇒ c = 48º 1
Value based question5. (i) Two plane mirrors PQ and RS are placed parallel to each other i.e. PQ || RS. An incident ray
AB after reflection takes the path BC and CD.BN and CM are the normals to the plane mirrors PQ and RS respectively. (½)
Since BN ⊥ PQ, CM ⊥ RS and PQ || RS (½)∴ BN ⊥ RS ⇒ BN || CM (½)Thus BN and CM are two parallel lines and transversal BC cuts them at B and C respectively.∴ ∠2 = ∠3But ∠1 = ∠2 and ∠3 = ∠4∴ ∠1 + ∠2 = ∠2 + ∠2 and ∠3 + ∠4 = ∠3 + ∠3 (½)⇒ ∠1 + ∠2 = 2 (∠2) and ∠3 + ∠4 = 2(∠3)⇒ ∠1 + ∠2 = ∠3 + ∠4 (½)⇒ ∠ABC = ∠BCD.Thus, lines AB and CD are intersected by transversal BC, such that
∠ABC = ∠BCDi.e. alternate interior angles are equal.Therefore, AB || CD (½)(ii) Lines and angles (½)(iii) Similarity leads to unanimity. (½)
SUMMATIVE ASSESSMENT WORKSHEET-37
Section A1. No, The following statement is not correct because ∠2 and ∠8 are not supplementary. 1
Section B2. ∠1 = ∠x = 75º (Alternate angles) ½
∠2 = 180º – x = 180º – 75º = 105º 1
∠2 = 105º = 75º + 30º = 175º 90º3
+ × ½
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Section C3. Since AC || DE, ∠ACB = ∠DEC ½
∴ y = 55º (Corresponding angles) 1
∠ABC = 180º – (70º + 55º) = 55º ½
∴ x = ∠ABC = 55º (Corresponding angles) 1
Section D4. y + 20º = 58º (Cross angle)
y = 58º – 20º = 38º 1½
∠PRQ = 180º – (58º + 22º) (Linear pair)
= 180º – 80º = 100º (Angle sum property) 1½
x = 180º – (100º + 38º)
= 180º – 138º = 42º 1qq
SUMMATIVE ASSESSMENT WORKSHEET-38
Section A1. x + 10º + x + x + 20º = 180º
3x = 180º – 30º = 150º ½
x = 150º 50º
3= ½
Section B2. ∠FEC + ∠DCE = 155º + 25º = 180º ½
(Angle on the same side of transversal)⇒ EF || CD ½
∠BCD = ∠BCE + ∠ECD
= 35º + 25º = 60º = ∠ABC (Alternate angles) ½∴ AB || CD
So, AB || CD || EF
Thus, AB || EF ½
Section C3. 5x – 20º + 2x – 10º = 180º
7x = 180º + 30º = 210º
x = 30º 1
y = 180º – (5x – 20º) = 180º – (150º – 20º)
y = 180º – 130º = 50º 1
z = 2x – 10º
= 60º – 10º = 50º 1
P-53
Section D4. ∠ACD = ∠ABC = x (say) ½
∠BCD = ∠CPD = y (say) ½
∠APC = x + y (Exterior angle) 1½
∠ACP = x + y 1
∠ACP = ∠APC ½
SUMMATIVE ASSESSMENT WORKSHEET-39
Section A
1. x + 50º + 90º = 180º ⇒ x = 40º ½
q = x = 40º ½
Section B
2. ∠a = 2x and ∠b = 3x
∠a + ∠b = 90º
⇒ 2x + 3x = 90º
x = 18º
a = 2 × 18º = 36º ½
b = 3 × 18º = 54º ½
c = 180º – b = 180º – 54º ½
= 126º ½
Section C3. Draw AS ⊥ to PQ
Draw ⊥ from A to M
In ∆ PSA
∠ PSA + ∠SAP + ∠ APS = 180°
90° + ∠SAP + 15° = 180° (1)
∠ SAP = 75°
∠ BAS = 75° + 35°
= 110°
∠ NMA = 90°
∠ BMA = 90° – 10°
= 80° (1)
(Exterior angle) ∠ BAS = ∠ ABM + ∠ AMB
110° = x + 80°
x = 30° (1)
Q
S
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Section D
4. x = 42° (Alternate angles) (1)∠1 = 24° + 42° = 66° (Exterior Angle) (½)
y = ∠1 = 66° (∵ CE = AC) (1)z = 180° – (x + y) = 180° – 108° (1)
= 72°. (½)qq
SUMMATIVE ASSESSMENT WORKSHEET-40
Section A1. AB = AC
∴ ∠C = ∠BThen, ∠A + ∠B + ∠C = 180º ½
50º + ∠B + ∠B = 180º ½2∠B = 130º ⇒ ∠B = 65
Section B2. Let x = 3k, y = 2k
Then, x + y = 3k + 2k = 180º ½(Angle on the same side of transversed)
5k = 180ºk = 36º
∴ x = 3k = 108º (½)y = 2k = 72º (½)
Thus, ∠z = ∠x = 108°. (½)
Section C3.
∠ACD = ∠A + ∠B (Exterior angle is the sum of opp. interior angle) (½)
12 ext. ∠ACD = 2 2
∠ ∠+A B
∠2 = ∠1 + 12 ∠A ...(1) (½)
Also, ∠2 = ∠1 + ∠E(Exterior angle is the sum of opp. interior angle) ...(2) (½)From eqs. (1) and (2), we get (½)
∴ ∠1 + 2A∠ = ∠1 + ∠E (1)
∠E = 2A∠ . Proved.
z
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Section D4. Since, AE || DC
∠D + ∠1 = 180° (Angles on the same side of transversal)∠1 = 180° – 110° = 70° (1)
∠4 = ∠1 = 70° (Alternate angles) (½)Again, 97° + ∠2 = 180° (Angle on the same side of transversal) (½)
∠2 = 180° – 97° = 83°∠3 = ∠2 = 83° (Vertically opp. angles) (½)
In ∆BEF,∠3 + ∠4 + ∠EBF = 180° (Angle sum property) (1)
83° + 70° + ∠EBF = 180°∠EBF = 180° – 153°∠EBF = 27°. (½)
SUMMATIVE ASSESSMENT WORKSHEET-41
Section A1. AB = AC ⇒ ∠ACB = ∠B ⇒ 80º + ∠ACB + ∠B = 180º ½
2 ∠ACB = 100 ⇒ ∠ACB = 50º, Again, ∠ACB + x = 180º ½50º + x = 180º ⇒ x = 130º
Section B2. x + y + z + w = 360° (1)
(x + y) + (z + w) = 360°x + y + x + y = 360° (Since z + w = x + y) (½)
∴ 2(x + y) = 360°x + y = 180°
⇒ AOB is a straight line. (½)
Section C3. EF ⊥ CD ⇒ ∠CFE = 90° (1)
90° + z = ∠CFGz = 130° – 90°
= 40° (1)x = ∠CFG (Alt. int. angles)
= 130°x + y = 180° (Linear pair) (1)
130° + y = 180°y = 50°.
4. In ∆ABC,∠A + ∠B + ∠C = 180° (½)
∠A + ∠C = 180° – ∠B = 180° – 90° = 90° (½)
12 (∠A + ∠C) = 45° (½)
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In ∆AOC,12 ∠A + 1
2 ∠C + ∠AOC = 180° (½)
45° + ∠AOC = 180° (½)∠AOC = 180° – 45° = 135°. (½)
Section D5. To prove : Sum of all the angles of ∆ABC is 180°. (1)
Construction : Draw a line l parallel to BC.Proof : Since l||BC, we have ∠2 = ∠y (Alternate angles are equal) ... (i) (1)Similarly l||BC
∠1 = ∠z (Alternate angles are equal) ...(ii)Also, sum of angles at a point A on line l is 180°. (½)∴ ∠2 + ∠x + ∠1 = 180°i.e. ∠y + ∠x + ∠z = 180° (from (i) and (ii))∴ ∠x + ∠y + ∠z = 180° (½)
∠A + ∠B + ∠C = 180°Sum of all angles of a ∆ is 180° (1)∴ 5x + 6x + 7x = 180° ⇒ x = 10°
Angles are 50°, 60° and 70°. Hence proved.qq
SUMMATIVE ASSESSMENT WORKSHEET-42
Section A1. Since, 110º + x + 120º = 360º ½
∴ x = 360º – 230º = 130º ½
Section B2. 3z – 42° = 2z + 13° (Alternate interior angles)
z = 42° + 13° (½)z = 55° (½)
∠DNM = 2z + 13° = 110° + 13° = 123° (½)∠CNM = 180° – ∠DNM
= 180° – 123° = 57°. (½)
Section C3. y = 88° (Corresponding angles) (½)
a = 180° – 88° = 92° (Linear Pair) (½)
b = 180° – 110° = 70° (Linear Pair) (½)
x = 180° – (92 + 70)°, (Angle sum property) (½)
= 180° – 162° (½)
= 18°. (½)
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4. Given AB || CD, CD || EF, EA⊥AB. (½)∠BEF = 55°
y = 180° – 55° = 125° (Co-interior angles) (½)x = y = 125° (Corresponding angles) (½)
EA ⊥ AB∴ z = 90° – 55° = 35°. (½)
Section D5. Ext. angle ∠PBC = x + z
2∠OBC = x + z (½)Ext. angle ∠QCB = x + y
2∠OCB = x + y (½)∠BOC + ∠OCB + ∠OBC = 180° (Angle sum property) (1)
2∠BOC + 2∠OCB + 2∠OBC = 360°2∠BOC + x + y + x + z = 360°
2∠BOC + 180° + x = 360°2∠BOC = 180° – x° (1)
∠BOC = 90° – 12 x. (1)
FORMATIVE ASSESSMENT WORKSHEET-43
This worksheet contains activities which due to be perform in the class itself.qq
A
R C
P O Q
x
y z
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5 TRIANGLES
SUMMATIVE ASSESSMENT WORKSHEET-44
Section A1. ∠ABC = 180º – 125º = 55º ½
AB = AC ½So, ∠ABC = ∠ACB = 55°In ∆ABC ∠A + ∠B + ∠C = 180°
x + 55° + 55° = 180°x = 70°
Section B2. In ∆PQS and ∆PRS
PQ = PR (given) ½
PS = PS (Common)
∠PSQ = ∠PSR = 90º ½
By R.H.S. rule, ∆PQS ≅ ∆PRS ½
⇒ ∆QPS = ∆RPS
Hence, PS bisects ∠P (by c.p.c.t) ½
3. In ∆ABP and ∆ACP, AB = AC (Given) (½)AP = AP (Comman)
∠APB = ∠APC = 90º (AP ⊥ BC) (½)By RHS rule, ∆ABP ≅ ∆ACP (½)⇒ ∠B = ∠C (½)
A
B CP
Section C4. AB = AC (½)
∴ ∠B = ∠C = xBA = BC
∴ ∠A = ∠C = xAlso, AC = BC (½)
∠A = ∠B = xBut ∠A + ∠B + ∠C = 180° (1)
3x = 180°x = 60°
∴ ∠A = ∠B = ∠C = 60°. (1)
P
Q RS
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5. Construction : Produce QS to meet PR in T (½)In ∆PQT, PQ + PT > QT
PQ + PT > QS + ST ...(1) (½)In ∆SRT TR + ST > SR ...(2) (½)Adding (1) and (2), we get
PQ + PT + TR + ST > QS + ST + SR (1)PQ + PR > QS + SRQS + SR < PQ + PR. (½)
Section D6. ∠XAK + ∠KAH = 180° (Linear pair) .
∠KAH = 180° – 137° = 43°AB = AC (1)
∴ ∠ABC = ∠ACB∠ABC + ∠ACB = 137°, (ext. angle)
∴ ∠ABC = ∠ACB = 1372
° = 68·5° (½)
CH = CB (Given) .⇒ ∠CBA = ∠CHB = 68·5° (½)∴ ∠HCB = 180° – 137° = 43° (1)
∠CHK = ∠HCB = 43°. (Alternate angles) (1)
Value based questions7. (i) AB and CD intersect at O
∴ ∠AOD = ∠BOC (Vertically opp. angles) ...(i)In ∆AOD and ∆BOC, we have (½)
∠AOD = ∠BOC From (ii) .∠DAO = ∠CBO = 90° (Given) (½)
and AD = BC (Given) .∴ ∆AOD ≅ ∆BOC (By AAS congruence criterion) .⇒ OA = OB (By c.p.c.t.) .i.e., O is the mid-point of ABHence, CD bisects AB. (1)(ii) Congruency of triangles. (½)(iii) Equality is the sign of democracy. (½)
SUMMATIVE ASSESSMENT WORKSHEET-45
Section A1. By SAS or side Angle side congruency. 1
Section B2. AX || BY
⇒ ∠BAX = ∠ABY (Alternate angles) ...(1) (½)∠AXY = ∠BYX (Alternate angles) ...(2) (½)
In ∆PAX and ∆PBY, AX = BY (Given) ...(3) (½)From (1), (2) and (3), we get
∆APX ≅ ∆BPY, (By ASA) Proved. (½)
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3. OA = OB (O is the mid-point of AB)
∠AOC = ∠BOD (Vertically opposite angles) (½)
OC = OD (O is the mid-point of CD) (½)
∆AOC ≅ ∆BOC (By SAS) (½)
⇒ AC = BD. (By c.p.c.t.) Proved. (½)
Section C4. In ∆ABC, AB > BC ⇒ ∠C > ∠A ...(1) .
AB > AC ⇒ ∠C > ∠B ...(2) (1)On adding (1) and (2), we get
2∠C > ∠A + ∠B2∠C + ∠C > ∠A + ∠B + ∠C (1)
3∠C > 180°∠C > 60°. (1)
5. Since AB||DC ∠ABP = ∠CDP (Alternate interior angles) (1)∠APB = ∠CPD (Vertical opposite angles) .
PD = PB (Given) (1)∆APB ≅ ∆CPD (By ASA) .
AP = PC (By c.p.c.t) (1)
⇒ P is also the mid-point of AC.
Section D6. Construction : Take BD = AB. Join AD (1)
Proof : In ∆ABD,AB = BD ⇒ ∠1 = ∠3 and ∠4 > ∠1 ⇒ ∠4 > ∠3 (1)In ∆ADC, ∠3 > ∠2∴ ∠4 > ∠3 > ∠2⇒ ∠4 > ∠2. (1)⇒ AC > DC (1)⇒ AC > BC – BD
AC > BC – AB∴ BC – AB < AC
Similarly, AC – AB < BC
and BC – AC < AB.
Value based question7. (i) l and m are two parallel lines intersected by another pair of parallel lines p and q.
AD || BC
and AB || CD. (½)
⇒ ABCD is a parallelogram.i.e. AB = CDand BC = AD (½)
P-61
Now in ∆ABC and ∆CDA, we haveAB = CDBC = AD
and AC = AC∴ We have ∆ABC ≅ ∆CDA (By SSS criterion of congruence) (1)(ii) Congruency of triangles. (½)
(iii) Equality is the sign of democracy. (½)qq
SUMMATIVE ASSESSMENT WORKSHEET-46
Section A1. AD = BC (given)
∠BAD = ∠ABC (given)AB = AB (common) ½
∴ ∆DAB ≅ ∆CBA (by SAS)∴ ∠BDA = ∠ACB (by c.p.c.t) ½
Section B2. In ∆PAB and ∆PDC, PA = PD
AB = CD 1½by RHS ∆PAB ≅ ∆PCD ½
PB = PC ⇒ ∠PCB = ∠PBC Proved3. In ∆ABY and ∆ACX
AB = AC (P is mid point of AD)AY = AX (Given) 1∠A = ∠A (Common)
∴ By SAS, ∆ABY ∆ ≅ ACX Proved 1
Section C4. PQ = PR ⇒ ∠PQR = ∠PRQ (Angles opp. to equal sides) (½)
ST||QR ⇒ ∠PST = ∠PQR (Corresponding angles) (1)∠PTS = ∠PRQ (Corresponding angles) (½)
∴ ∠PST = ∠PTS (½)⇒ PS = PT. (Sides opp. to equal angles) (½)
5. ∠DBC = ∠DCB (Given)
∴ DC = DB(Sides opp. to equal angles are equal) ...(i) (½)
In ∆ABD and ∆ACD,AB = AC (Given)
BD = CD [from (i)] (½)
AD = AD (Common) .∆ABD ≅ ∆ACD (SSS) (1)
∠BAD = ∠CAD (By c.p.c.t.) (½)
Hence, AD is bisector of ∠BAC. Proved. (½)
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Section D6. AD = BD
⇒ ∠ABD = ∠DAB = 59°(Angles opp. to equal sides are equal) (1)
In ∆ABD,
59° + 59° + ∠ADB = 180°
∠ADB = 180° – 118° = 62° (1)
∠ACD = 62° – 32° = 30° (Exterior angle is equal to sum .
of interior opposite angles) .
In ∆ABD,
AB > BD (Side opp. to greatest angle is the longest) (1)
Also in ∆ABC, AB < AC
⇒ BD < AC. (1)7. Given, ∠BAB = ∠EAC
Adding ∠CAD on both sides, we get
∠BAD + ∠DAC = ∠EAC + ∠DAC (1)
∠BAC = ∠EAD
In ∆ABC and ∆ADE,
AB = AD (Given)
∠BAC = ∠EAD (Proved)
AC = AE (given)
∴ ∆ABC ≅ ∆ADE (By SAS) (2)
⇒ BC = DE (By PCS) (1)qq
SUMMATIVE ASSESSMENT WORKSHEET-47
Section A1. OA = OB
OP = OP ½∠AOP = ∠BOP ½∆OAP ≅ ∆OBP
Section B2. In ∆BED and ∆CFD,
∠DEB = ∠DFC = 90° (½)
BD = DC (D is the mid point)
ED = FD (Given)
∴ ∆BED ≅ ∆CEF (By RHS) (1)
∠B = ∠C. (By c.p.c.t.) Proved. (½)
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3. In ∆ADC and ∆ABC, AC is common
∠DAC = ∠BAC, (½)
∠DCA = ∠BCA (Given) (½)
Hence, ∆ADC ≅ ∆ABC (By AAS rule) (½)
⇒ CD = BC. Proved. (By c.p.c.t.) (½)
Section C4. PQRS is a square.
(i) SRT is an equilateral triangle.
∴ ∠PSR = 90°, ∠TSR = 60° (1)
⇒ ∠PSR + ∠TSR = 150°. Similarly ∠QRT = 150°
∆PST and ∆QRT, we have PS = QR (S)
∠PST = ∠QRT = 150° (A) and ST = RT (S) (1)
By SAS, ∆PST ≅ ∆QRT
⇒ PT = QT (By c.p.c.t.) Proved.(ii) In ∆TQR, QR = RT (Square and equilateral are ∆ on same base) (½)
⇒ ∠TQR = ∠QTR = x (½)
∴ x + x + ∠QRT = 180°2x + 150°= 180° ⇒ 2x = 30° ∴ x = 15°. Proved.
5. ∠BDC + ∠CDA = 180° (Linear pair)∠BDC + x = 180°
∠BDC = 180° – x (½)Similarly, ∠BEA = 180° – y (½)Since, x = y (Given)∴ ∠BDC = ∠BEA (½)
∠B = ∠B (Common)AB = BC (½)
∴ ∆BAE ≅ ∆BCD (By AAS) (½)∴ AE = CD. (By c.p.c.t.) Proved. (½)
Section D6. Join OD and OB.
In ∆AOB and ∆COB,
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AO = CO (Given) .OB = OB (Common) .AB = BC (∵ ABCD is a rhombus) .
∆AOB ≅ ∆COB (SSS Congruence.) .∠AOB = ∠COB (By c.p.c.t.) ...(1) (1)
Similarly, ∆AOD ≅ ∆COD (½)∠AOD = ∠COD (By c.p.c.t.) ...(2) (½)
But, ∠AOD + ∠COD + ∠COB + ∠AOB = 360° (½)2(∠AOD + ∠AOB) = 360° (½)
∠AOD + ∠AOB = 180° (½)⇒ D, O, B are collinear. (½)
SUMMATIVE ASSESSMENT WORKSHEET-48
Section A1. ∠ADB = ∠ACB = 90º (given)
AD = BC (given) (½)AB = AB (common)
∴ ∆ABD ≅ ∆BAC (by R.H.S.) (½)
Section B2. Let ∠DAC be ∠3,
∠1 = ∠2 (Given) (½)
∠1 + ∠3 = ∠2 + ∠3
∠BAC = ∠EAD ...(i) (½)Given that, BD = CE
BD + DC = CE + DC (½)BC = DE ...(ii) .
∠B = ∠E (Given) ...(iii) .
From (i), (ii) and (iii), we get∆ABC ≅ ∆AED. (By AAS rule) (½)
Section C3. Given, ∠B > ∠A and ∠C > ∠D,
∴ OA > OB ...(1)
OD> OC (Side opposite to greater angle is longer) ...(2) (1½) Adding (1) and (2), we get
OA + OD > OB + OCAD > BC. (1½)
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4. AB = AC 2 2AB AC⇒ = (½)
⇒ AE = AF,Since E and F are mid-points of AB and AC.In ∆ABF and ∆ACE, AB = AC (Given) (½)
∠A = ∠A (Common) (½)AF = AE (Proved) (½)
∴ ∆ABF ≅ ∆ACE (By SSA cong.) (½)∴ BF = CE. (By c.p.c.t) (½)
Section D5. Proof : In ∆ADE, we have
AD = AE (½)∠ADE = ∠AED
180° – ∠ADE = 180° – ∠AED (½)∠ADB = ∠AEC ...(i) (½)
Consider ∆ABD and ∆ACEAD = AE
∠ADB = ∠AEC From (i) .BD = EC (1½)
By SAS congruence, ∆ABD ≅ ∆ACEBy c.p.c.t., AB = AC (1)∴ ∆ABC is an isosceles triangle.
6. ∠FEC = ∠ECD = 10° (Alternate angles are equal) (½)∠BCA = 90°
∠BCD + ∠DCA = 90°10° + ∠DCA = 90° (1)
∠DCA = 80°AC = DC
∠DAC = ∠ADC = 180 802
° − ° = 50° (½)
In ∆BAC,∠A + ∠B + ∠BCA = 180°
50° + ∠B + 90° = 180°∠B = 40° (1)
In ∆BDE,∠B + ∠BED + ∠BDE = 180°
40° + 90° + ∠BDE = 180°∠BDE = 50°. (1)
SUMMATIVE ASSESSMENT WORKSHEET-49
Section A1. AB = AC
⇒ ∠C = ∠B ½∠A + ∠B + ∠C = 180°
⇒ 90° + ∠B + ∠B = 180° ⇒ 2∠B = 90° ∴ ∠B = 45°. ½
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Section B2. (i) In ∆AOD and ∆BOC,
OA = OB (Given)OD = OC (Given)
∠AOD = ∠BOC (Vertically opposite angles)So, by SAS criteria,
∆AOD ≅ ∆BOC (1)(ii) ∠CBA = ∠DAB (By c.p.c.t.)AD and BC are two lines intersected by AB such that ∠CBA = ∠DAB and they form pair ofalternate angles.Hence, AD || BC. (1)
Section C
3. ∠1 + ∠5 = 180° = ∠2 + ∠6 ⇒ ∠1 = ∠2 (∵ ∠5 = ∠6) (1)
In ∆CAP and ∆BAP,
∠1 = ∠2 (Proved)
∠3 = ∠4 (AD is bisector of ∠BAC)
AP = AP
∴ ∆CAP ≅ ∆BAP (By AAS) (1)
⇒ CP = BP. (By c.p.c.t.) Proved. (1)
4. BC = DE (Given)Add CD both side
BC + CD = CD + DEBD = CD (1)
In ∆ABD and ∆FECAB = EF (Given)BD = CD (1)
∠ABD = ∠FEC = 90°So, By RHS esiteria
∆ABD ≅ ∠FEC (Proved) (1)
Section D
5. In ∆ALB and ∆NCMAL = CN (given)BL = CN (given)
∴ ∆ALB ≅ ∆NCM (By SAS)AB = NM∠A = ∠N (c.p.c.t.) (½)
∴ In ∆ABC and ∆NML,Now, AL = CN⇒ AL + LC = LC + CN
AC = LN (½)
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AB = NM (Proved)AC = LN (Proved)
LA = LN (Proved) (1)∴ ∆ABC ≅ ∆NML. (SSS) (½)
6. (i) AB = AC, BD = CD, AD = DA (1)∆ABD ≅ ∆ACD (By SSS) ...(i)
∴ ∠BAD = ∠CAD (By c.p.c.t.) .(ii) AB = AC, ∠BAP = ∠CAP, AP = AP
∆ABP ≅ ∆ACP (By SAS) (1)∴ BP = CP (By c.p.c.t.) .(iii) From (i), ∠BAD = ∠CAD⇒ AP is the bisector of ∠A (1)
BD = CD, BP = CP, DP = DP∆BDP ≅ ∆CDP ⇒ ∠BDP = ∠CDP
DP is the bisector of ∠D. (1)qq
SUMMATIVE ASSESSMENT WORKSHEET-50
Section A
1. Let the angles of triangle are 5x, 3x and 7x, then5x + 3x + 7x = 180°
15x = 180°x = 12° (½)
∴ Angles are 60°, 36°, 84° ∵ Each angle is less than 90°
∴ The triangle is an acute angled triangle. (½)
Section B
2. AB = AD ⇒ ∠ABD = ∠ADB ...(i) (½)BC = CD ⇒ ∠CBD = ∠CDB ...(ii) (½)
Adding eq. (i) and (ii), we get∠ABD + ∠CBD = ∠ADB + ∠CDB (½)
⇒ ∠ABC = ∠ADC. Proved. (½)
Section C3. In ∆PAM and ∆QBM, (½)
AM = BM (Given)
∠1 = ∠2 (V.O.A.) (½)
∠3 = ∠4 = 90°
∴ ∆PAM ≅ ∆QBM (By ASA Cong.) (1½)
∴ PA = BQ. (By c.p.c.t.) Proved. (½)
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4. (i) In ∆ABC and ∆DCB,AB = DC (Given)
∠ABC = ∠DCB (Given) (1)BC = CB (Common)
∴ ∆ABC ≅ ∆DCB (By SAS) (1)(ii) ⇒ AC = DB. (By c.p.c.t.) (1)
Section D5. (i) In ∆ABD and ∆BAC
∠DAB = ∠CBA (Given) .AB = BA (Comman) (1)AD = BC (Given) .
∴ ∆ABD ≅ ∆BAC (By SAS) Proved. (1)(ii) ∴ BD = AC (By c.p.c.t.) Proved. (1)(iii) ∠ABD = ∠BAC. (By c.p.c.t.) Proved. (1)
6. Proof : We are given two triangles ABC and PQR in which :∠B = ∠Q, ∠C = ∠R
and BC = QRWe need to prove that ∆ABC ≅ ∠PQR (1)There are three cases.
Case I : Let AB = PQIn ∆ABC and ∆PQR,
∠B = ∠Q (Given) .BC = QR (Given) .AB = PQ (Assumed) .
∴ ∆ABC ≅ ∆PQR (By SAS rule) (1)Case II : Suppose AB ≠ PQ and AB < PQTake a point S on PQ such that QS = ABJoin RS.
In ∆ABC and ∆SQR,AB = SQ (By construction) .BC = QR (Given) .∠B = ∠Q (Given) (1)
∴ ∆ABC ≅ ∆SQR (By SAS rule) .∠ACB = ∠QRS (By c.p.c.t.) .
P-69
But ∠QRP = ∠ACB⇒ ∠QRP = ∠QRSWhich is impossible unless ray RS coincides with RP.∴ AB must be equal to PQ. (½)So, ∆ABC ≅ ∆PQRCase III : If AB > PQ.We can choose a point T on AB such that TB = PQ and repeating the arrangements as given inCase II, we can conclude that AB = PQ and so,
∆ABC ≅ ∆PQR (½)qq
SUMMATIVE ASSESSMENT WORKSHEET-51
Section A1. Let the angle of triangle are 3x, 4x and 3x
then 3x + 4x + 3x = 180º10x = 180º
x = 180º 18º10
= ½
So, Angles are 3 × 18 = 54º4 × 18 = 72º and 54º ½
Section B2. RP = RQ (Given) .
⇒ ∠RQP = ∠RPQ (Proved) (½)⇒ ∠MQP = ∠NPQ
In ∆PQN and ∆QPM,
∠NPQ = ∠MQP (½)
PN = QM (Given) .
PQ = PQ (Common) .∴ ∆PQN ≅ ∆QPM (By SAS) (½)
⇒ ∠PQN = ∠QPM (By c.p.c.t.) .
⇒ ∠PQO = ∠QPO (½)
⇒ OP = OQ. (Sides opposite to equal angles are equal) Proved.
Section C3. Proof : In ∆ABC and ∆BAD,
BC = AD (Given)∠CBA = ∠BAD (Given)
AB = AB (2)∴ ∆ABC ≅ ∆BAD (By SAS cong.)
∴ BD = AC (By c.p.c.t.) (1)4. In ∆ABC,
AB + BC > AC ...(1) (1)
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In ∆ACD,CD + DA > AC ...(2) (1)
Adding (1) and (2), we getCD + DA + AB + BC > 2AC. (1)
Section D5. In ∆RTS, RT = ST
∠TSR = ∠TRS (Angles opposite to equal sides are equal) ...(1) (½)∠1 = ∠4 (V.O.A) .
⇒ 2∠2 = 2∠3⇒ ∠2 = ∠3 ...(2) (½)Subtract eqn. (2) from eqn. (1), we get
∠TRS – ∠2 = ∠TSR – ∠3∠TRB = ∠TSA (1)
In ∆RBT and ∆SAT,∠RTB = ∠STA (Common angles) .
RT = ST (Given) .∠TRB = ∠TSA (Proved) (1)
∴ ∆RBT ≅ ∆SAT (By ASA congruence) .⇒ RB = AS (By c.p.c.t.) (1)
6. ∠PQT = ∠RQU and ∠TQS = ∠UQS∴ ∠PQT + ∠TQS = ∠RQU + ∠UQS (1)⇒ ∠PQS = ∠RQS
PQ = RQ, QS = QS (1)∴ ∆PQS ≅ ∆RQS (By SAS) (½)∴ ∠P = ∠R (By c.p.c.t.) .Again, in ∆PQT and ∆RQU,
∠P = ∠R∠PQT = ∠RQU
PQ = QR∴ ∆PQT ≅ ∆RQU (By ASA) (1)
∴ QT = QU. (By c.p.c.t.) (½)qq
SUMMATIVE ASSESSMENT WORKSHEET-52
Section A1. Then, if AC = BC (given an isosceles triangle)
So, Acc. to SAS propertythe angles B and C are also equal.i.e., ∠B = ∠CSo, Let x be the angle,
x + x + 90º = 180º ½x = 45º
So, ∠B = 45º ½A
C
B90°
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Section B2. ∠AOB = ∠COD (given)
∠AOB – ∠COB = ∠COD – ∠COB 1In ∆ AOC and ∆BOD
AO = OBOC = OD
∠AOC = ∠BOD∴ ∆AOC ≅ ∆BOD (SAS) (½)
AC = BD (By CPCS) (½)
Section C3. Proof : In ∆PQR,
PQ = PR∠PQR = ∠PRQ (Angles opp. to equal sides are equal) ...(i) (1)
In ∆PQS, ∠PQR > ∠PSQ (Ext. angle of a ∆ is greater than .each of interior opp. angles) .(½)
∴ ∠PRQ > ∠PSQ, using (i) (½)⇒ ∠PRS > ∠PSR ⇒ PS > PR (½)
PS > PQ (∵ PR = PQ) (Side opp. to greater angle is larger) (½)4. Proof :
In ∆ABD and ∆BAC,
AD = BC (Given)
BD = AC (Given)AB = AB (Common side) (2)
By SSS congruence axiom,∆ABD ≅ ∆BAC
⇒ ∠ADB = ∠BCA (By c.p.c.t.) (1)∠DAB = ∠CBA. (By c.p.c.t.).
Section D5. AL = CN (Given) .
AL + LC = CN + LCi.e. AC = LNIn ∆ALB and ∆NCM, (1)
AL = CN (Given) .∠ALB = ∠NCM = 90°
BL = CM (Given) .∴ ∆ALB ≅ ∆NCM (SAS) .⇒ AB = NM (c.p.c.t.) (1)
In ∆BLC and ∆MCL,BL = MC (Given) .
∠BLC = ∠MCL = 90°
LC = CL∴ ∆BLC ≅ ∆MCL (By SAS) .
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⇒ BC = ML (c.p.c.t.) (1)∴ In ∆ABC and ∆NML,
AB = NM (Proved) .AC = NL (Proved) .BC = ML (Proved) .
∴ ∆ABC ≅ ∆NML. (SSS) (1)6. ∵ ∠1 + ∠3 = 180° (Linear pair)
∵ ∠2 + ∠4 = 180° (Linear pair)∵ ∠1 + ∠3 = ∠2 + ∠4
But ∠1 = ∠2⇒ ∠3 = ∠4
In ∆OAC and ∆ODB, (1)∠3 = ∠4
∠AOC = ∠DOB (V.O.A.) (½)
OA = OD (Given) (½)
∴ ∆OAC ≅ ∆ODB (ASA) (1)
OC = OB (c.p.c.t.) (½)
⇒ ∆OCB is an isosceles triangle (½)qq
SUMMATIVE ASSESSMENT WORKSHEET-53
Section A1. ∠ABC = 180° – 125° = 55° (½)
AB = AC
⇒ ∠ACB = ∠ABC = 55°
∴ x = 180° – 55° – 55° = 70°. (½)
Section B2. Given, ∠DCA = ∠ECB
Adding ∠DCE to both sides, we get∠DCA + ∠DCE = ∠ECB + ∠DCE
∠ECA = ∠DCB (1)In ∆ACE and ∆BCD,
AC = BC, (Given)∠ECA = ∠DCB, (Proved) .∠EAC = ∠DBC, (Given) .
∴ ∆ACE ≅ ∆BCD, (By AAS cong.) (½)∴ BD = AE. Proved. (½)
Section C3. In ∆ABD and ∆CDB,
AB = CD (Given) (1)∠ABD = ∠CDB (Given)
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BD = DB (Common) (1)∴ ∆ABD ≅ ∆CDB (By SAS)⇒ AD = CB. (By c.p.c.t.) (1)
4. In ∆COD and ∆BOA,OD = OA (Given)OC = OB (Given) (1)
∠DOC = ∠AOB = 90°∴ ∆COD ≅ ∆BOA (By SAS) (1)⇒ CD = AB. (By c.p.c.t.) (1)
Section D5. In ∆ABC,
AB = BC (Given) (1)∴ ∠1 = ∠2 (Angles opposite to equal sides)In ∆ABD and ∆CBD,
AB = BC∠1 = ∠2AD = CD (1)
∴ ∆ABD ≅ ∆CBD (By SAS cong.) .∠3 = ∠4 (By c.p.c.t.) (½)
∠3 + ∠4 = 180°∴ ∠4 = 90°∴ ∠ADE = ∠4 (V.O.A) .∴ ∠ADE = 90° Proved. (½)
In ∆EAD and ∆ECD, ∠A = ∠CAD = CD (Given)
∠ADE = ∠CDE = 90° (½)
and DE = DE (Common)So, ∆EAD = ∆ECD . (By AAS)
⇒ AE = CE. (By cpct) (½)
6. In ∆AMB and ∆PNQ,
AB = PQ (Given) .
AM = PN (Given) .
∠1 = ∠2 = 90°
⇒ ∆AMB ≅ ∆PNQ (By RHS) .
⇒ ∠3 = ∠4 (By c.p.c.t.) (2)
Similarly, ∆AMC ≅ ∆PNR
⇒ ∠5 = ∠6
∴ in ∆ABC and ∆PQR, AB = PQ (Given) .
AC = PR (Given) .
∠A = ∠P (∠3 + ∠5 = ∠4 + ∠6) (Proved) .
⇒ ∆ABC ≅ ∆PQR. (By SAS) (2)qq
A
B CM
5 3
1
P
Q RN
6 4
2
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SUMMATIVE ASSESSMENT WORKSHEET-54
Section A1. AD = BC
∠BAD = ∠ABCAB = AB
∴ ∆DAB ≅ ∆CBA (By cpct) ½∴ ∠BDA = ∠ACB ½
Section B
2. Given, ∠BAD = ∠EAC
Adding ∠CAD on both sides, we get
∠BAD + ∠CAD = ∠EAC + ∠CAD (½)
∠BAC = ∠DAE
AC = AE (Given) (½)
AB = AD (Given)
∆BAC ≅ ∆DAE (By SAS) (½)
∴ BC = ED (By c.p.c.t.) (½)
Section C
3. In ∆ABC, AB = AC ⇒ ∠1 = ∠2 ...(1)
Angles opp. to equal sides are equal.In ∆ADC, AB = AD∴ AC = AD (½)
In ∆BCD, ∠3 = ∠4 ...(2) (½)∠1 + (∠2 + ∠3) + ∠4 = 180° (½)
∠2 + ∠2 + ∠3 + ∠3 = 180° (½)
2(∠2 + ∠3) = 180°∠2 + ∠3 = 90° (½)
⇒ ∠BCD is a right angle. (½)
4. In ∆PQS,
PQ + QS > PS (Sum of a any two sides is greater than the third side) ...(1) (2)
In ∆PSR,
PR + SR > PS ...(2) (1)
∴ On adding (1) & (2)
PQ + PR + QS + SR > PS + PS
PQ + QR + RP > 2 PS
Section D5. In ∆BCE and ∆CBF, we have
∠BEC = ∠CFB (each 90°)BC = BC (common)
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BE = CF (given)By RHS congluence.,
∆BCE ≅ ∆CBF∠BCE = ∠CBF (2)
In ∆ABC∠C = ∠B ⇒ AB = AC
Similarly, we prove AB = BC (1)Then, AB = BC = ACSo, ∆ABC is equilateral triangle. (1)
6. By triangle inequality property,
In ∆ABC, AB + BC > AC ...(1) (½)
In ∆BCD, BC + CD > BD ...(2) (½)
In ∆CDA, CD + DA > AC ...(3) (½)
In ∆DAB, DA + AB > BD ...(4) (½)Adding (1), (2), (3) and (4), we getAB + BC + BC + CD + CD + DA + DA + AB > AC + BD + AC + BD (1)
2(AB + BC + CD + DA) > 2(AC + BD)
Hence, Perimeter > Sum of its diagonals. (1)qq
SUMMATIVE ASSESSMENT WORKSHEET-55
Section A
1. ∠ADB = ∠ACB = 90º
AD = BC (½)
AB = AB
∴ ∆ABD ≅ ∆BAC (½)
Section B2. In ∆ABC,
∠A + ∠B + ∠C = 180°∠A = 180° – 20 – 70 = 90° (½)
So, ∠BAN = ∠NAC = 45°In ∆ABM
∠BAM = 180 – 90 – 70 = 20 (½)∠BAN = ∠BAM + ∠MAN
45 = 20 + ∠MAN (½)∠MAN = 25°
Section C3. Given, BR = CQ
BR + BQ = CQ + BQQR = BC (½)
A
F E
B C
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In ∆ABC and ∆PQR,QR = BC (Proved above) .AB = PQ (Given) (1)
∠BAC = ∠QPR = 90° (Given) .∴ ∆ABC ≅ ∆PQR (By RHS) (1)
⇒ AC = PR. (By c.p.c.t.) (½)4. In ∆ABE and ∆ACD,
AB = AC (Given) (½)∠C = ∠B (Angle opp. to equal sides) (½)BE = CD (Given) (½)
∴ ∆ABE ≅ ∆ACD (By SAS) (1)⇒ AD = AE. (By c.p.c.t.) (½)
Section D
5. In ∆ABM and ∆PQN, AB = PQ (Given)
BC = QR (Given)
⇒ 12 BC = 1
2 QR
i.e. BM = QN (½)
AM = PN (Given) .
∴ ∆ABM ≅ ∆PQN (SSS) (½)
⇒ ∠ABM = ∠PQN (c.p.c.t.) .
i.e. ∠ABC = ∠PQR (1)
Now, in ∆ABC and ∆PQR,AB = PQ (Given) .
∠ABC = ∠PQR (Proved) .(½)
BC = QR (Given) .∴ ∆ABC ≅ ∆PQR. (SAS) (1½)
6. Produce AD to E such thatAD = DE. Join EC. (½)
In triangles ADB and EDC,AD = DE (Const) (½)
BD = DC (Given) .∠ADB = ∠EDC (V.O.A) (½)
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∴ ∆ADB ≅ ∆EDC (SAS Congruence axiom) (½)⇒ AB = EC (By c.p.c.t.) .In ∆AEC, AC + EC > AE [Triangle Inequality Property] (½)∴ AC + AB > AE (∵EC = AB) .
AC + AB > AD + DEAC + AB > AD + AD (∵DE = AD) (½)
AC + AB > 2AD. Proved. (1)qq
FORMATIVE ASSESSMENT WORKSHEET-56
This worksheet contain activities that are to be perform in the class itself.qq
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P-78 T I C S IX--M A T M AEH M - 1 SE RT
6 COORDINATE GEOMETRY
SUMMATIVE ASSESSMENT WORKSHEET-57
Section A1. Clearly, (– 2, 5) i.e., x coordinate is – ve of y coordinate is + ve, so it lies the IInd quadrant.
12. So, clearly, from y-axis, the ⊥ distance is 7. 1
Section B
3. On x-axis : C (– 3, 0), F (6, 0), G (3, 0) 1
On y-axis : A (0, 2), D (0, – 3), E (0, 4) 1
4. (i) Since the point lies on x-axis at a distance of 9 units from y-axis. Hence its co-ordinates are(9, 0) (1)
(ii) According to question, the required co-odinates are (0, –9). (1)
5. The point on y-axis has x-cordinate 0.
Since it lies 4 units distance in negative direction of y-axis. (1)
∴ The point is (0, – 4). (1)
Section C
6. According to question, plot the points A(0, 3), B(5, 3), C(4, 0) and D(–1, 0) on graph paper.
(2)
From figure, ABCD is a parallelogram. The point (2, 2) will be inside the figure. (1)
Section D
7. (i) E (–1, 2) (1)
(ii) D (–4, – 1) (1)
(iii) 2 – 4 = – 2 (1)
(iv) 3 + (– 4) = – 1 (1)
P-79
Value based questions8. (i) Draw X´ OX and Y´ OY
as the co-ordinate axes andmark their point of intersection Oas the origin (0, 0).
In order to plot the points (– 2, 8), we take 2 units on OX´ and then 8 units parallel to OY to obtainthe point A(– 2, 8).
Similarly, we plot the point B(– 1, 7). (½)
In order to plot (0, – 1·25) we take 1·25 units below x-axis on the y-axis to obtain C(0, – 1·25).(½)
In order to plot (1, 3) we take 1 unit on OX and then 3 units parallel to OY to obtain the point D(1, 3)(½)
In order to plot (3, – 1), we take 3 units on OX and then 1 unit below x-axis parallel to OY ́toobtain the point E(3, – 1) (½)
(ii) Co-ordinate geometry. (½)
(iii) Co-ordination among people is good for progress. (½)
SUMMATIVE ASSESSMENT WORKSHEET-58
Section A1. Mirror image is (1, 2) 12. The co-ordinater of a point whose ordinate is 6 of lies on y-axis is (0, 6). 1
Section B3. Plotting of P(2, – 6)
Coordinates of M(2, 0) (½)
Coordinates of N(0, – 6). (½)
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C
P-80 T I C S IX--M A T M AEH M - 1 SE RT
1
-1-2-3-4-5-6
2 3 4 5 6 7 8
-1
1234567
-2-3-4-5-6-7
xx`
y`
y
(2,-6)N P
M
(1)
Section C4.
1
-1-2-3-4-5-6
2 3 4 5 6 7 8
-1
12345
-2-3
xx`
y`
y
A (4, 0)
B (0, 4)
O
(2)
From figure, OA = 4 unitsOB = 4 units
Area of ∆OAB = 12 × OA × OB
= 12 × 4 × 4 = 8 square unit. (1)
Section D5. According to question (– 2, 9) = (1 + x, y2)
⇒ – 2 = 1 + x or x = – 3 (1)and 9 = y2 or y = 3, as y > 0 (1)So, P(y, x) = P(3, – 3),which will be in IV quadrant.
Q(2, x) = Q(2, – 3),which will be in IV quadrant.
R(x2, y – 1) = R(9, 2),which will be in I quadrant
S(2x, – 3y) = S(– 6, – 9),which will be in III quadrant. (½+½+½+½)
P-81
X X´
Y
Y´
A
B
C
D
O
6. Since ∆ABC is equilateral triangle with side 2a units, thereforeAB = BC = CA = 2a units
O is mid point of AC, then (1)
OA = 12 AC
= 12 (2a) = a units (1)
Now, in right ∆AOB
OB = 2 2–AB OA
= 2 2(2 ) –a a
= 2 24 –a a
= 23 3 unitsa a= (1)
Thus, co-ordinates of B are (0, 3).a (1)
Value based question7. (i) Let OB = a
⇒ AB = 2a⇒ AC = 2 × a = 2aUsing Pythagoras theorem in ∆COA (1)
A B
D
C
O xx'
y'
y
(– , 0)a ( , 0)a
(2a)2 = a2 + (OC)2
⇒ (OC)2 = 3a2
⇒ OC = 3a
So co-ordinates of C are (0, 3)a
Similarly, co-ordinates of D are −(0, 3)a (1)
(ii) Co-ordinate geometry. (½)(iii) Co-ordinate among people is the first condition of success in any society. (½)
SUMMATIVE ASSESSMENT WORKSHEET-59
Section A1. It is (0, – 7). 12. It lies on IV and II quadrant. 1
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P-82 T I C S IX--M A T M AEH M - 1 SE RT
Section B
3. (A) P(0, 5) (½)
(–2, 0)
(B) Q(0, – 3) (½)
(C) R(5, 0) (½)
(D) S(–2, 0) (½)
Section C4. Drawing the points, A(3, 2), B(2, 3), C(–4, 5) and D(5, – 3). Joining AB, BC, CA, AD, we get a
quadrilateral ABCD.
1
-1-2-3-4-5-6
2 3 4 5 6 7 8
-1
1234567
-2-3-4-5-6-7
xx`
y`
y
(2,3)
(3,2)
(5,-3)
(-4,5)
AB
C
D
Section D5. From figure the co-ordinate of vertices of a rectangle are :
O(0, 0) (1)
A(– p, 0) (1)
B(– p, – q) (1)
C(0, – q) (1)
A (–, , 0)p (0, )q
C (0, – )qB p q(– , – )
(O, O)
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–5 –4 –3 –2 –1–1–2
–3 –4
–5
O1 2 3 4 5
54
32
1
Y
Y´
X´ X
(–3, 5)
(–3, –5) (2, –5)
6. (i) (– 3, – 5) lies in III quadrant, as x < 0 and y < 0.
(ii) (2, – 5) lies in IV quadrant, as x > 0 and y < 0.(iii) (– 3, 5) lies in II quadrant, as x < 0 and y > 0. (½+½+½)Verification : The points (– 3, – 5), (2, – 5) and
(– 3, 5) are plotted as shown in figure :
Result is verified : (i) (– 3, – 5) lies in III quadrant.(ii) (2, – 5) lies in IV quadrant.(iii) (– 3, 5) lies in II quadrant. (1)
SUMMATIVE ASSESSMENT WORKSHEET-60
Section A1. It lies on the IV quadrant. 12. It lies on the x-axis. 1
Section B3. Co-ordinates of A = (5, 0) (½)
Co-ordinates of B = (5, 3) (½)
Co-ordinates of C = (–2, 4) (½)
Co-ordinates of D = (0, –3). (½)
Section C4.
1
-1-2-3-4-5-6-7-8
2 3 4 5 6 7 8
-1
1234567
-2-3-4-5-6-7
xx`
y`
y
(-3,-3) (3,-4)
(5,-2)
(0,7)(-1,7)
(-8,0)3
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P-84 T I C S IX--M A T M AEH M - 1 SE RT
Section D5. (i) Required points are A(0, 3) and L(0, – 4). (2)
(ii) Required points are G(5, 0) and I(– 2, 0). (1)(iii) Required points are D(– 5, 1) and H(– 5, – 3). (1)
6. The points A(– 3, – 4), B(– 2, 0),C(– 1, 4) and D(1, 0)
are plotted as shown below :(i) A(– 3, – 4) lies in III quadrant.(ii) B(– 2, 0) lies on x-axis.(iii) C(– 1, 4) lies in II quadrant.
(iv) D(1, 0) lies on x-axis. (½+½+½+½)qq
SUMMATIVE ASSESSMENT WORKSHEET-61
Section A1. In II and IV quadrant. 1
2. AB = 2 2(0 0) (4 0) 0 16 4− + − = + =
BC = 2 2(0 4) (0 0) 16 0 4− + − = + =
Area of triangle = 12 × base × height (½)
= 12 × 4 × 4
= 8 sq. units (½)
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Section B3. (a) IV quadrant (b) II quadrant (½+½)
(c) III quadrant (d) I quadrant (½+½)
Section C4. From the graph,
QR = 2 + 2 = 4
So, PQ = QR = RP = 4 (1)
Now OR = 2
So, from right triangle OPR,
OP2 = PR2 – OR2 (1)
OP2 = 42 – 22 = 12
So. OP = 2 3 (½)
Therefore, co-ordinates of P are (0, 2 3 ). (½)
Section D5.
6.
–5 –4 –3 –2 –5 1 2 3 4 5
4
3
2
1
–1
–2
–3
–4
(I quadrant)
(IV quadrant)(III quadrant)
(II quadrant)
(T)(B)
(E)(S)
(4)
(1)
(1)
(1)
(1)
ROOC AN TID OEGE RT YEM
P-86 T I C S IX--M A T M AEH M - 1 SE RT
SUMMATIVE ASSESSMENT WORKSHEET-62
Section A1. (x + 2, 4) = (5, y – 2) (½)
⇒ x + 2 = 5 and 4 = y – 2⇒ x = 5 – 2 = 3 and y = 4 + 2 = 6 (½)The co-ordinates (x, y) = (3, 6)
2. The co-ordinates of the origin is (0, 0). 1
Section B3. (A) (–3, – 2) (½) (B) 0 (½)
(C) 0 (½) (D) Q and T. (½)
Section C4. (i)
(1)
(ii) Rhombus ABCD. (1)(iii) AB lies in II quadrant and AD lies in I quadrant. (1)
Section D5. The points A(– 3, 0), B(5, 0) and C(0, 4) can be plotted as shown below :
Figure formed is a triangle ABC.
1 2 3 4 5 –1
–2
–3 –2 –1
4
3
2
1
Y
X
Y´
X´ O
C(0, 4)
(–3, 0)A
B
(5, 0)
Area of ∆ABC = 12 × AB × OC
= 12 × 8 × 4;
(as AB = 8 and OC = 4)= 16 sq. units. (2) + (2)
1
-1-2-3-4-5
2 3 4 5
-1
12345
-2-3-4-5
xx`
y`
y
A (0,4)
D (3,0)
C (0,-4)
B (-3,0)
P-87
6. (i) Plot the points M(4, 3), N(4, 0), O(0, 0) and P(0, 3) on graph paper.
1
-1-2-3-4
2 3 4 5 6 7 8
-1
1234567
-2-3-4
xx`
y`
y
O (0,0)
M (4,3)
N (4,0)
P (0,3)
(2)
(ii) By joining MNOP, the figure will be rectangle. (1)(iii) Perimeter of rectangle = 2(l + b) = 2(4 + 3) = 14 units. (1)
SUMMATIVE ASSESSMENT WORKSHEET-63
Section A1. At the origin. 12. Point C (3, a – 3) lie on x-axis
∴ a – 3 = 0 ½a = 3 ½
Section B
3. OA = OB = 6 32 = units and BC = AD = 3 units.
Co-ordinates of A be (–3, 0) (½)Co-ordinates of B be (3, 0) (½)Co-ordinates of C be (3, 3) (½)Co-ordinates of D be (–3, 3). (½)
Section C4. In ∆POQ,
∠POQ = 90º
QR = 8
OQ = 4 (1)
OP2 + OQ2 = PQ2 (½)
OP2 + 42 = 82, (PQ = QR = RP) (½)
OP2 = 48
ROOC AN TID OEGE RT YEM
P-88 T I C S IX--M A T M AEH M - 1 SE RT
OP = 48 or 4 3 (½)
The co-ordinates of P are ( 48,0) or (4 3,0). (½)
Section D5. (i) Plot the point M(5, – 3) and N(–3, –3) on graph paper.
(ii) Length of MN = 3 + 5 = 8 units(iii) From figure, (½)
A(3, – 3)B(1, – 3)C(–1, – 3) (½+½+½)
6.
Plot the three vertices of the rectangle as A(3, 2),B(– 4, 2), C(– 4, 5). (2)
We have to find the co-ordinates of the fourth vertex Dso that ABCD is a rectangle. Since the opposite sides of arectangle are equal, so the abscissa of D should be equal toabscissa of A, i.e., 3 and the ordinate of D should be equal tothe ordinate of C, i.e., 5.
So, the co-ordinates of D are (3, 5). (2)
SUMMATIVE ASSESSMENT WORKSHEET-64
Section A1. On comparing we get, b = 2 ½
So, from 2 – a + b = 6, we get ½a = – 2 (1)
2. We get square.
Section B3. In, II quadrant, III quadrant, ½+½+½+½
IV quadrant and I quadrant.
–5 –4 –3 –2 –1–1
–2
–3 –4 –5
O1 2 3 4 5
5
432
1
Y
Y´
X´ X
(–4, 5)C D
A(3, 2)(–4, 2)B
(2)
P-89
Section C4. (i)
(2)
(ii) Graph is a straight line. (1)5. OB = a, OA = a
A B
D
C
O xx'
y'
y
(–a, 0) (a, 0)
⇒ AB = 2a⇒ AC = 2aUsing Pythagoras Theorem in ∆COA, (1)
(2a)2 = a2 + (OC)2
⇒ (OC)2 = 3a2
⇒ OC = 3a (1)So, co-ordinates of C are (0, 3)a (½)
Similarly, co-ordinates of D are (0, 3)a− . (½)
Section D
6. Points Quadrants
(–2, 4) II
(1, 2) I
(–3, –5) III
(3, –1) IV
(–1, 0) X-axis
54321
– 1– 2– 3– 4– 5
1 2 3 4 5–4 –3 –2 –1
(–2, 4)
(–1, 0)(1, 2)
(3, –1)
(–3, –5)
y'
xx'
y
ROOC AN TID OEGE RT YEM
P-90 T I C S IX--M A T M AEH M - 1 SE RT
–4 –3 –2 –1–1–2 –3 –4
O1 2 3 4 5 6
6
543
2
1
Y
Y´
X´ X
D(2, 6)
C(2, 0)
(–2, 3)A
B(–2, 0)
7. The given points A(– 2, 3), B(– 2, 0), C( 2, 0) and D(2, 6) can be plotted as under :
On joining AB, BC, CD and DA, we get a trapezium ABCD.
From figure, BC = 2 – (– 2)
= 4 units (2)
and AB = 3 – 0
= 3 units (1)
Now, area of ∆ABC = 12 × BC × AB
= 12 × 4 × 3
= 6 sq. units. (1)qq
FORMATIVE ASSESSMENT WORKSHEET-65
Section AThis worksheet contains activities that are to be performed in the class itself.
P-91
7 AREAS
SUMMATIVE ASSESSMENT WORKSHEET-66
Section A
1. x = 2 225 15 40 10− = × (½)
x = 20 cm.
Area of right triangle = 12 × base × height
= 12 × 15 × 20 = 150 cm2. (½)
2. Area of equilateral triangle= 23 16 34 a× =
a2 = 4 × 16 ⇒ a = 8 m (½)Perimeter = 3a = 3 × 8 = 24 m. (½)
Section B3. Perimeter of an equilateral triangle = 3a = 60 cm
a = 20 cm (1)
Area = 23 3 20 20 100 34 4a = × × = cm2. (1)
4. Sides of an isosceles triangle are a, a and b, then
a + a + b = 32 (½)2a + 12 = 32
a = 10 cm
s = 32 162 = cm (½)
Area = 16 (16 12)(16 10)(16 10)× − − − (½)
= 16 4 6 6× × × = 48 cm2. (½)
Section C5. Draw DE || BC and DF ⊥ AB
AE = 25 – 10 = 15 m
For ∆AED, s = 13 14 15 212+ + = m
Area of ∆AED = 21 8 7 6 84× × × = m2
Also, area of ∆AED = 1 152 × × DF = 84 (1)
⇒ DF = 2 84 5615 5× = (½)
Area of parallelogram EBCD = 5610 1125× = m2 (1)
Area of trapezium = 84 + 112 = 196 m2. (½)
AERA S
P-92 T I C S IX--M A T M AEH M - 1 SE RT
Section D6. Perimeter of the triangle = 84 cm
Ratio of its sides = 13 : 14 : 15Sum of the ratio = 13 + 14 + 15 = 42 cm (1)
∴ a = 1342 × 84 = 13 × 2 = 26 cm
b = 1442 × 84 = 14 × 2 = 28 cm
c = 1542 × 84 = 15 × 2 = 30 cm
s = 26 28 302
+ +
s = 842 = 42 cm (1)
Using Heron’s formula,
Area of ∆ = ( – )( – )( – )s s a s b s c
= 42(42–26)(42–28)(42–30) cm2 (1)
= 42 16 14 12× × × cm2
= 2 3 7 16 2 7 3 4× × × × × × × cm2
= 2 × 3 × 7 × 4 × 2 cm2 (1)= 336 cm2.
SUMMATIVE ASSESSMENT WORKSHEET-67
Section A
1. ( )( )( )S S a S b S c− − − 1
2. Perimeter = 4x + 5x + 6x = 150⇒ 15x = 150⇒ x = 10 ½Sides are 4 × 10 = 40 cm, 5 × 10 = 50 cm and 6 × 10 = 60 cm ½
Section B3. Let the sides are 5x, 12x and 13x then,
Perimeter = 5x + 12x + 13x = 12030x = 120
x = 4 cm (½)Then sides are 20 cm, 48 cm, 52 cm
s = 120 602 = cm (½)
Area = ( )( )( )s s a s b s c− − −
= 60(60 20)(60 48)(60 52)− − − (½)
= 60 40 12 8× × × = 10 6 10 4 6 2 4 2× × × × × × ×= 10 × 6 × 4 × 2 = 480 cm2. (½)
P-93
4. Let third side = x
Then 24 + 10 + x = 60
x = 60 – 34 = 26 cm (½)
s = 60 302 = cm (½)
Area = ( )( )( )s s a s b s c− − −
= 30(30-24)(30-10)(30-26) (½)
= 30 6 20 4× × × = 10 3 3 2 10 2 4× × × × × ×
= 10 × 3 × 4 = 120 cm2. (½)
Section C
5. Sides of a triangle are a = 6 cm, b = 6 cm, c = 4 cm
s = 6 6 4 16 82 2 2a b c+ + + += = = cm (½)
A = ( )( )( )s s a s b s c− − − (1)
= 8 2 2 4× × ×
= 8 2 cm2 (½)
Area of two black triangles = 8 2 2 16 2× = cm2 (½)
Area of two white triangles = 16 2 cm2. (½)
Section D6. For one identical triangular leaf, let
a = 28 cm, b = 15 cm and c = 41 cm
Also, s = 28 15 412 2
a b c+ + + +=
842= = 42 cm (1)
Using Heron’s formula,
Area of one triangular leaf = ( – )( – )( – )s s a s b s c
= 42(42–28)(42–15)(42–41) (1)
= 42 14 27 1× × ×
= 3 14 14 3 9× × × × = 9 × 14= 126 cm2 (1)
There are 6 leaves in a circle.So, total number of leaves in 2 circles = 2 × 6 = 12∴ area of 12 leaves = (12 × 126) cm2 = 1512 cm2
Hence, total area to be painted red = 1512 cm2. (1)qq
AERA S
P-94 T I C S IX--M A T M AEH M - 1 SE RT
SUMMATIVE ASSESSMENT WORKSHEET-68
Section A
1. Clearly, the length of the altitude of an equilateral triangle of side ‘a’ is 3
.2
a 1
2. x = 2 210 6 8− = cm
Area of right triangle = 12 × base × height (½)
= 12 × 6 × 8 = 24 cm2. (½)
Section B3. Let the longest side be 90 cm and other side BC be 54 cm
AB = 2 290 54 72− = cm (½)
∴ Area of the triangle = 12 b h× ×
= 1 54 722 × × (1)
= 1944 cm2. (½)4. Perimeter of rhombus = 4 × side = 200
side = 50 m (½)
For ∆ABC, s = 50 50 80 902+ + = m (½)
Area of ∆ABC = ( )( )( )s s a s b s c− − −
= 90 (90 50)(90 50)(90 80)× − − −
= 90 40 40 10× × ×= 1200 m2. (½)
∴ Area of rhombus = 2 × Area of ∆ABC= 2 × 1200 = 2400 m2. (½)
Section C
5. For ∆ABC, s = 96 52 52 200 1002 2+ + = = m (½)
Area = ( )( )( )s s a s b s c− − −
= 100 4 48 48× × × (½)= 10 × 2 × 48 = 960 m2 (½)
P-95
Area of rhombus = 960 × 2 = 1920 m2 (½)
Area to be grazed by 20 cows = 1920 m2
Area to be grazed by 1 cow = 192020 (½)
= 96 m2. (½)
Section D6. a = 120 m, b = 170 m, c = 250 m (½)
s = 120 170 250 540 2702 2 2a b c+ + + += = =
Area = ( ) ( ) ( )s s a s b s c− − − (½)
= 270 (270 120)(270 170)(270 250)× − − −
= 270 150 100 20× × × (½)
= 81000000 (½)= 9000 m2 (½)
Area = 12 × base × height (½)
9000 = 12 × 250 × h (½)
h = 72 m. (½)
Value based question7. (i) Let the sides of the triangle be a, b and c.
a : b : c = 12 : 17 : 25
12 17a b= = 25
c k= (say)
⇒ a = 12k, b = 17k, c = 25k (½)Perimeter = 540 cma + b + c = 540
12k + 17k + 25k = 54054k = 540
k = 540 1054 = (½)
Hence a = 12 × 10 = 120 cm, b = 17 × 10 = 170 cm, c = 20 × 10 = 250 cm
Now s = 1 5402 × = 270 cm
(s – a) = (270 – 120) cm = 150 cm(s – b) = (270 – 170) cm = 100 cm(s – c) = (270 – 250) cm = 20 cm (½)
AERA S
P-96 T I C S IX--M A T M AEH M - 1 SE RT
∴ Area of triangle = ( – )( – )( – )s s a s b s c
= 270 150 100 20× × × cm2
= 100 27 15 10 2× × × cm2
= 100 3 3 3 3 5 2 5 2× × × × × × × cm2
= 100 × 3 × 3 × 5 × 2= 9000 cm2 (½)
(ii) Heron’s formula. (½)(iii) A balanced ratio leads to good result. (½)
SUMMATIVE ASSESSMENT WORKSHEET-69
Section A
1. Perimeter of an equilateral triangle = 3a = 60 m
a = 20 m
Area = 23 3 20 20 100 34 4a = × × = m2. 1
2. S = + + + +
= =8 7 15
15 cm2 2
a b c1
Section B
3. a = 15 cm, b = 15 cm, c = 12 cm
s = 15 15 12 212 2a b c+ + + += = cm (½)
Area = ( )( )( )s s a s b s c− − − (½)
= 21 (21 15)(21 15)(21 12)× − − −
= 21 6 6 9× × × (½)
= 18 21 cm2. (½)
Section C
4. s = 20 50 50 1202 2
+ + = = 60 cm (½)
Area of one triangular piece = ( )( )( )s s a s b s c− − − (½)
= 60(60 20)(60 50)(60 50)− − −
= 60 40 10 10× × × (½)
= 200 6 cm2 (½)
Required cloth = 10 × 200 6 = 2000 × 2·45 (½)= 4900 cm2. (½)
P-97
Section D5. Rhombus,
Perimeter = 52 cm (1)
Side = 52 134 = cm
Diagonal = 24 cm (1)OB = OD = 12 cm
OA = 2 213 12 169 144 25 5− = − = = cm (1)
Area of rhombus = 14 5 122× × × (1)
= 120 cm2
Value based question
6. (i) If a, b, c be the lengths of sides BC, CA and AB of ∆ABC and if s = 1( ),2 a b c+ + then
area of ∆ABC = ( – )( – )( – )s s a s b s c
∴ Here a = b = c (= a) and s = 12 (a + a + a) = 3
2a (½)
s – a = 32 2a aa− =
s – b = 32 2a aa− =
s – c = 32 2a aa− = (½)
area = ( ) ( ) ( ) ( )32 2 2 2a a a a = 32 2
a a× ×
= 23
4a . ...(1) (1)
(ii) To find the area, when its perimeter = 180 cm2
Here, a + a + a = 180⇒ 3a = 180
⇒ a = 180 603 =
Hence, required area = 23 (60)4 ×
= 900 3 cm2. (1)(iii) Heron’s formula. (½)(iv) We should follow the traffic rules to save our life. (½)
SUMMATIVE ASSESSMENT WORKSHEET-70
Section A
1. Let the side of an isosceles right angle triangle be x, x and 5 2.
So, x2 + x2 = 2(5 2) ½2x2 = 25 × 2
x = 5 cm ½
AERA S
P-98 T I C S IX--M A T M AEH M - 1 SE RT
2. x2 + x2 = ( )212 2
2x2 = 144 × 2
x = 12 cm (½) Given 3a = 4x (a, be a side of triangle)
3a = 4 × 12a = 16 cm
Area of equilateral triangle = 23 3 16 164 4a = × ×
= 64 3 cm2. (½)
Section B
3. Diagonals bisect each other at O.
Draw LOM ⊥ AB.OL = OM = 3 cm
Area of shaded region OAB = 1 8 32 × × (2)
= 12 cm2.
Section C
4. s = 51 37 20 1082 2
+ + = = 54 cm (½)
Area = ( )( )( )s s a s b s c− − − (½)
= 54(54 51)(54 37)(54 20)− − −
= 54 3 17 34× × × (½)
= 306 cm2 (½)
No. of rose beds = Total areaoftriangularfield
Areaoccupiedbyeachrosebed (½)
= 306 516 = . (½)
5. s = 26 28 30 842 2
+ + = = 42 cm (½)
Area of ∆ = ( )( )( )s s a s b s c− − − (½)
= 42(42 26)(42 28)(42 30)− − − (½)
= 42 16 14 12× × × (½)
= 336 cm2
Again, area of parallelogram = Area of triangle (1)b × h = 336 (½)
28 × h = 336 (½)
h = 12 cm (½)
P-99
Section D
6. ar (∆BCD) = × × =1
12 5 302
m2 (½)
BD = 2 212 5 13+ = m2 (½)
For ∆ABD, s = 13 8 9 152+ + = cm
Area of ∆ABD = 15 2 7 6× × ×
= 6 35 m2 (1½)
ar (quad. ABCD) = ( )+30 6 35 m2. (½)
7. Let the equal sides be x, then
32x x x+ + = 42
72x = 42 (1)
x = 12 cm
Then sides are 12 cm, 12 cm and 18 cm
s = 12+12+18
2 = 42 212 = cm (1)
Area = ( )( )( )s s a s b s c− − −
= 21(21 12)(21 12)(21 18)− − − (1)
= 21 9 9 3× × ×
= 27 7 cm2. (1)
SUMMATIVE ASSESSMENT WORKSHEET-71
Section A
1. Sides are 3x, 4x, 5x, then,
3x + 4x + 5x = 36
⇒ 12x = 36 ⇒ x = 3 ½
∴ Sides are a, b, c are 9 cm, 12 cm, 15 cm. ½
2. Area of ∆ABC = 214 4 8 cm
2× × = 1
Section B
3. Let a = 20 cm, b = 30 cm, c = 40 cm
Then s = 20 30 402 2
a b c+ + + += = 45 cm
AERA S
P-100 T I C S IX--M A T M AEH M - 1 SE RT
Now, using Heron’s formula,
Area of ∆ = ( – )( – )( – )s s a s b s c (1)
= 245(45–20)(45–30)(45–40) cm
= 245 25 15 5 cm× × ×
= 23 3 5 5 5 3 5 5 cm× × × × × × ×
= 23 5 5 15 cm× × ×
= 75 × 3·873 cm2
= 290·48 cm2. (1)
Section C
4. Area of the equilateral triangle = 23 3 10 104 4a = × × (½)
= 25 3 sq. cm = 25 × 1·732
= 43·3 sq. cm (1)
In ∆BDC, DC = 2 210 8 6− = cm (½)
∴ Area of ∆BDC = 1 6 8 242 × × = sq. cm (½)
∴ Area of unshaded portion = 43·3 – 24
= 19·3 sq. cm. (½)
Section D
5. Let the equal sides be ‘a’ units, then
s = 24 2 242 2
a a a+ + += (½)
= a + 12
Area = ( )( )( )s s a s b s c− − − (½)
60 = ( 12)( 12 )( 12 )( 12 24)a a a a a a+ + − + − + −
= ( 12)(12)(12)( 12)a a+ − (½)
60 = 2( 144) 12a − ×
5 = 2 144a − (½)
On squaring both sides, we get25 = a2 – 144a2 = 25 + 144 = 169a = 13 cm (½)
Perimeter = 13 + 13 + 24 = 50 cm (½)
P-101
6. In ∆ABC, AC2 = AB2 + BC2
AC2 = 92 + 402
= 81 + 1600 = 1681 (½)
AC = 41 m
Area of ∆ABC = 12 × b × h (½)
= 12 × 9 × 40
= 180 m2
For ∆ACD, s = 28 41 15 84 422 2+ + = = m (½)
Area = − − −( )( )( )s s a s b s c
= − − −42(42 28)(42 41)(42 15) (½)
= 42 14 1 27× × × (½)
= 126 m2 (½)Area of quadrilateral = Area of ∆ABC + Area of ∆ACD (½)
= 180 + 126 = 306 m2. (½)qq
FORMATIVE ASSESSMENT WORKSHEET-72
Section AThis worksheet contains activities that are to be performed in the class itself.
AERA S