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2018 ___ ___ 1100 - MT - y - MATHEMATICS (71) Geometry - SET - C (E)
MT - y
Time : 2 Hours Preliminary Model Answer Paper Max. Marks : 40
Q.P. SET CODE
C A.1.(A) Solve ANY FOUR of the following :
(i) If diagonals of a parallelogram are congruent, then it is a rectangle. 1
(ii) If a transversal intersects two parallel lines then the sum of interior angles on the same side of the transversal is 180o . 1
(iii) In ABC, ABC = 90o B
A CD
Seg BD is the median
on hypotenuse AC.
BD =12
AC
(Median drawn to the hypotenuse is half of it)
7 =12
AC
AC = 14 cm 1
(iv) A quadrilateral is a parallelogram if a pair of opposite sides is parallel and congruent. 1
(v) Equation of the Y – axis is X = 0 1
(vi)o
o
tan40cot50
=o
o
tan40tan(90 50)-
=o
o
tan40tan40
= 1 1
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A.1.(B) Solve ANY TWO of the following :
(i) For a cone Area of the base = 1386 sq. cm Height (h) = 28 cm Volume of a cone = r2 h 1
=13
× 1386 × 28 ½
Volume of a cone = 12936 cm3 ½
(ii) A circle with centre O
O
A BM
chord AB = 24 cm
seg OM chord AB
OM = 5 cm
AM = 12
AB ½
[Perpendicular drawn from the centre to the chord, bisects the chord]
= 12
× 24
AM = 12 cm ½ In OMA, OMA = 90o
OA2 = 52 + 122 [Pythagoras theorem] ½ OA2 = 25 + 144 OA2 = 169 OA = 13 cm Radius of the circle is 13 cm ½
(iii) In FAN,
AF80o 40o
N
F = 80o
A = 40o
N = 60o [Remaining angle] ½ F > N > A ½ AN > AF > FN [In a triangle, ½
side opposite to greater angle is greater] Greatest side is AN and the smallest side is FN ½
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A.2.(A) Select the correct alternative answer and write it :
(i) (c) 22.4 1
(ii) (d) 0 1
(iii) (b) 550 cm2 1
(iv) (a) (1, 3) 1
A.2.(B) Solve ANY TWO of the following :
(i) Line PQ Line RS ...(Given) Slope of line PQ = Slope of line RS ½
½
2 2 = k – 1 ½ 4 + 1 = k k = 5 ½
(ii) A
B P C
D
line AD line BC ...(Given)
ABC and ∆BCD lie between the same two parallel lines AD and BC. ½
Their heights are equal.
Also, they have a common base BC ½
A(ABC) = (BCD) ...(Triangles having equal base and equal height) ½
½
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(iii) PS2 = PQ PR .... tangent secant segments theorem ½
= PQ (PQ + QR)P
Q
R
S
½ = 3.6 [3.6 + 6.4] = 3.6 10 ½ PS2 = 36
PS = 6 ½
A.3.(A) Carry out ANY TWO of the following activites :
(i) Given : In ABC, B = 90o A
B C
D
2 seg BD hypotenuse AC
To prove : ADB ~ BDC Proof :
In ADB and ABC
A A common angle
ADB ABC (each 90o)
ADB ~ ABC ...(i) AA test of similarity
In BDC and ABC
C C common angle
BDC ABC (each 90o)
BDC ~ ABC ...(ii) AA test of similarity
ADB ~ BDC from (i) and (ii)
(ii) Given : In trapezium ABCD, side AB || side CD, diagonal AC and BDintersect each other at point P.
To prove :A( ABP)A( CPD)DD
=AB
C D
P Proof : ABCD is a trapezium (Given) side AB || side CD (Given) 2 on transversal AC
BAC ACD ...(i)
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In APB and CPD,
BAC ACD [from (i)]
APB CPD Vertically opposite angles
APB ~ CPD AA test of similarity
A( ABP)A( CPD)DD
= Theorem of areas of smilar triangle
(iii) Given : Line ET is the tangent at T and E AB is the secnt.
To prove : EAEB = ET2
AE
T
B
Construction : Draw seg AT and seg BT
Proof : In EAT and ETB,
E E Common angle 2
ETA EBT Angle between tangent and secant
EAT ~ ETB AA test of similarity
ETEB
= ...(c.s.s.t)
EAEB = ET2
A.3.(B) Solve ANY TWO of the following :
(i) Analytical fi gure:
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½ mark for drawing circle and locating point P
½ mark for drawing perpendicular bisector of OP
½ mark for drawing circle with centreM
½ mark for drawing Tangents.
(ii) A(8, 9) = (x1, y1) B(1, 2) = (x2, y2)
P(k, 7) = (x, y)
Let point P divide seg AB in the ratio m : n.
By Section formula,
½
7
7 (m + n) = 2m + 9n
7m + 7n = 2m + 9n
7m – 2m = 9n – 7n
5m = 2n
= ½
m : n = 2 : 5
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½
k
k
k
k = 6 ½
(iii) L.H.S. = 2 ½
= (1 + cot2 A) (1 + tan2 A) = cosec2 A × sec2 A ( 1 + cot2 A = cosec2 A
and 1 + tan2 A = sec2 A)
= 2 2
1 1sin A cos A ½
= 2 2
1sin A(1 sin A)- (sin2 A + cos2 A = 1
cos2 A = 1 – sin2 A) ½
= 2 4
1sin A sin A ½
= R.H.S.
2 = 2 4
1sin A sin A
A.4. Solve ANY THREE of the following :
(i) A represents the position of the plane above the ground.
‘C’ is the landing point of the plane on the ground AB represents the
height of the plane from the ground. 1
DAC is the angle of depression
DAC = ACB = 20°
Distance (AC) = speed × time
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= 200 km/ hr × 54 sec
= 200 km/ hr × hr
(1 hr= 3600 sec)
= 200 ×
= 3km ½
AC = 3000 mIn ABC, ABC = 90°
sin ACB = ...(By definition) ½
sin 20° = ½
0.342 =
AB = 0.342 × 3000
AB = 1026 km.
Plane was at a height of 1026 km,when it started landing. ½
(ii) For segment PQR, r = AP = 7.5 units
A
P
QR
= PAR = 30°
A (segment PQR) = r2 ½
= ½
= 56.25 ½
= 56.25 ½
= 56.25 × ½
= 0.66
A (segment PQR) is 0.66 sq. units ½
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(iii) Analytical fi gure:
1 mark for PQY 1 mark for constructing YY5Q YY6Z 1 mark for constructing YQP YZX
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(iv) A
BE C
D
T
ABCD is a parallelogram ...(Given) seg AB seg CD ...(Opposite sides of a parallelogram) seg AT seg CD ...(A - B - T) on transversal TD,
ATD CDT ...(Alternate angles theorem) ½ BTE CDE ...(i) (A - B - T, T - E - D) ½
In BTE and CDE,BTE CDE ...[From (i)]BET CED ...(vertically opposite angles)
∆BTE ∆CDE ...(By AA test of similarity) 1
(c.s.s.t.) ½
DE BE = CE TE ½
A.5. Solve ANY ONE of the following :
(i)
A BC D
NPM
Construction : Draw a common tangent MN at point P
APM ADP [Theorem of Angle between tangent and secant] ½
Let, mAPM = mADP = x ...(i)
BPM BCP [Theorem of Angle between tangent and secant] ½
Let, mBPM = mBCP = y ...(ii)
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mBPM = mAPB + mAPM [Angle addition property] ½
y = mAPB + x [From (i) and (ii)]
mAPB = y – x ...(iii) ½
BCP is an exterior angle of DCDP
mvBCP = mCDP + mCPD [Remote-interior angles theorem] ½
y = x + mCPD
mCPD = y – x ...(iv) ½
mAPB = mCPD [From (iii) and (iv)] ½
APB CPD ½
(ii) Proof : (i) c p = a b
D
A
C Ba
b
p
c
A(ABC) =12
base height
A(ABC) =12
AB CD
A(ABC) =12
c p ...(i) ½
A(ABC) =12
BC AC
A(ABC) =12
a b ...(ii) ½
12
c p = 12
a b [From (i) and (ii)]
c p = a b ½
c p = a b
1
cp= 1
ab [By invertendo] ½
1p
=c
ab
2
1p =
2
2 2 [Squaring on both sides] ½
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In ACB, mACB = 90o [Given]
AB2 = AC2 = BC2 [Pythagoras theorem] ½ c2 = b2 + a2
2
1p
=2
2 2 +2
2 2 ½
2
1p = 2
1a
+ 2
1b
½
A.6. Solve ANY ONE of the following :
(i) P
M
a
a
a
aa
Q S R N
MQ = QR = RN = a...(Given)
Point Q is the midpoint of seg MR ...(i)
In ∆PMR, seg PQ is a median ...[From (i), Definition]
PM2 + PR2 = 2PQ2 + 2QM2 ...(Apollonius theorem) ½
PM2 + a2 = 2a2 + 2a2 ½
PM2 = 4a2 – a2 ½
PM2 = 3a2 ½
PM = 3 × a ...(Taking square roots) ½
Similarly we can prove, PN = 3a
PM = PN = 3 a ½
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(ii) For the sphere, r = 9 cm
For the wire, Thickness (diameter) = 4 mm
Radius (r1) = mm = 2 mm = cm ...[1 cm = 10 mm]
Let the length of wire be h1
Wire is made by melting the sphere,
½
r12 h1 ×r3 ½
× × × h1 × 9 × 9 × 9 ½
h1 ½
h1 24,300 cm
h1 243 m ...[ 1 m = 100 cm] ½
Length of the wire formed is 243 m. ½