C MT - yssc.maheshtutorials.com/images/SSC_Testpapers/... · 5 / MT - y SET - C In APB and CPD, BAC ACD [from (i)] APB CPD Vertically opposite angles APB ~ CPD AA test of similarity

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:

SET - C6 / MT - y

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