2018 ___ ___ 1100 - MT - z - MATHEMATICS (71) Geometry - SET - D (E)

MT - z

Time : 2 Hours Preliminary Model Answer Paper Max. Marks : 40

Q.P. SET CODE

D 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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Q.2.(A) Select the correct alternative answer and write it : 4

(i) (b) 3.5 1

(ii) (d) sec

(iii) (a) 44 1

(iv) (a) (6, 13) 1

A.2.(B) Solve ANY TWO of the following :

(i) P(–12, –3), Q(4, k) ...(Given)

Slope of PQ = ...(Given)

Slope of PQ = ...(Given) ½

= ½

= k + 3

k + 3 = 8 ½ k = 8 – 3

k = 5 ½

(ii) Proof : InABP andCDP

A

B

C

D

P

....(Given) ½

APB CPD ....(vertically opposite angles) ½

...(SAS test for similarity) 1

(iii) By theorem on interesecting chords, M

N

S

R

P

PN PM = PR PS... (I) ½

let PN = x. PM = 11 – x ½

substituting the values in (I),

x (11 – x) = 6 4

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11x – x2 – 24 = 0

x2 – 11x + 24 = 0

(x – 3) (x – 8) = 0

x – 3 = 0 or x – 8 = 0 ½

x = 3 or x = 8

PN = 3 or PN = 8 ½

Q.3.(A) Carry out ANY TWO of the following activites :

(i) Given : In ABC, ABC = 90o

To prove : AC2 = AB2 + BC2

Construction : Draw seg BD hypotenuse AC. A – D – C

Proof : In ABC, ABC = 90o (Given)

seg BD hypotenuse AC (Construction)

ABC ~ ADB ~ BDC (Similarity of right angle triangle) 2

ABC ~ ADB A

B C

D

ABAB

=

AB2 = AC × AD ...(i)

ABC ~ BDC

BCDC

= ...(ii)

BC2 = AC × DC ...(ii)

Adding (i) and (ii),

AB2 + BC2 = AC × AD + AC × DC

AB2 + BC2 = AC (AD + DC)

AB2 + BC = AC AC(AD + DC)

AB2 + BC2 = AC2

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(ii) Given : In ABC, A

P Q l

B C

Line l || side BC

To prove : =AQQC

Construction : Draw seg PC and seg BQ Proof :

A( APQ)A( PQB)DD = AP

PB ....(i) Triangles with equal heights 2

A( APQ)A( PQC) =

AQ

QC ....(ii) Triangles with equal heights

seg PQ || side BC (Given)

PQB and PQC have equal heights.

Also they have common base PQ

A PQB = A PQC ....(iii)

=AQQC from (i), (ii) & (iii)

(iii) Given : PA and PB are tangent segments drawn from external point P to thecircle

To Prove : seg PA seg PB Construction : Draw seg OA, seg OB and seg OP. 2 Proof : In OAP and OBP

A

B

PO

OAP = OBP = 90o Tangent theorem

hypt. OP hypt. OP Common side

seg OA seg OB radii of same circle

OAP OBP Hypotenuse side theorem

seg PA seg PB csst

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A.3.(B) Solve ANY TWO of the following :

(i) Analytical fi gure:

1 mark for drawing circle and P = 100º

1 mark for drawing tangents at A and B

(ii) Let A(h, –6) = (x1, y1) B(2, 3) = (x2, y2) and C(–6, k) = (x3, y3) G(1, 5) = (x, y) Point G is the centroid of ∆ABC. By centroid formula,

x = and y = ½

and ½

1 3 = h – 4 and 5 3 = –3 + k ½ 3 + 4 = h and 15 + 3 = k h = 7 and k = 18 h = 7 and k = 18 ½

(iii) sin + sin = 1 (Given) sin = 1 – sin ½ sin = cos2 sin2 = cos4 (Squaring both sides) ½ 1 – cos2 = cos4 ½ cos + cos4 = 1 ½

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A.4. Solve ANY THREE of the following :

(i) AB represents the height of the lighthouse. AB = 100 m

C represents the position of ship. 1DAC is the angle of depression.

DAC = ACB = 30° ...(BA alternate angle theorem

In ABC, ABC = 90°

tan ACB = ...(By definition) ½

tan 30° = ½

=

BC = 100

BC = 1001.73 ½ BC = 173 m

The distance of the ship from thelighthouse is 173 m. ½

(ii) (i) For arc RXQ, = ROQ = 60°

Q

R

M

N

YXO

½ OR (r) = 7 cm Length of arc RXQ =

× 2r

= × 2 × × 7 ½

= 7.33 cm

Length of arc RXQ is 7.33 cm ½

(ii) For arc MYN, OM (r) = 21 cm, = MON = 60° ½

Length of arc MYN =

× 2r

= × 2 × × 21 ½

= 22 cm

Length of arc (MYN) is 22 cm ½

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(iii) Analytical fi gure:

1 mark for ABC 1 mark for constructing BB4C BB7N

1 mark for constructing BCA BNL

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(iv) Proof : In ∆ABC and ∆DAC, A

B D C

½BAC ADC ...(Given) ½BCA ACD ...(Common Angle) ½

∆ABC ∆DAC ...(By AA test of similarity) ½

...(c.s.s.t.) ½

CA2 = CB CD ½

A.5. Solve ANY ONE of the following :

(i)

Q

C

P

A

B

Proof :ABC is an equilateral triangle

mBCA = mCAB = mCBA = 60º ...(i) ½ [Angles of an equilateral triangle]

mCBP = mABP = 12

mCBA [ Ray BP bisects ABC] ½

mCBP = mABP = 12

× 60

mCBP = mABP = 30o ...(ii) ½CAP CBP ...(iii) [Angles inscribed in the same arc

are congruent] ½

mCAP = 30o [From (ii) and (iii)]

mCAQ = 30o ...(iv) [A–P–Q] ½BCA is an exterior angle of CQA

mBCA = mCQA + mCAQ [Remote-interior angles theorem] ½

60 = mCQA + 30 [From (i) and (iv)]

mCQA = 60 – 30

mCQA = 30o ...(v)

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In CAQ,

CAQ CQA [From (iv) and (v)]

seg CQ seg CA [Converse of isosceles triangle theorem] ½

CQ CA ½

(ii) A (PQR) =12

PR×QN P

Q R

N

½

a = 12

PR×QN

a = PR×QN

QN = ...(i) ½

A (PQR) =12

base×height

A (PQR) =12

QR×PQ

a =12

b×PQ ½

2a = b PQ

PQ = ...(ii) ½

In PQR

m PQR = 90o

PR2 = PQ2 = QR2 (by Pythagoras theorem)

PR = 2 2 ½

PR =2

2

PR =2

22

4+

PR =2 2

2

4 +

PR =2 4

...(iii) ½

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QN = 2a2 4

[From (i) and (iii)]

QN = 2a2 4

QN =4 2

½

A.6. Solve ANY ONE of the following :

(i) To Prove : TS2 + TQ2 = TP2 + TR2 P

S

T

Q

R

Construction : Draw a line parallel to side SR,

through point T, intersectingsides PS and QR at point A and B respectively.

Proof :

PQRS is a rectangle ...(Given)

SPQ = PSR = SRQ = PQR = 90° ...(i) (Angles of a rectangle) ½

seg PQ seg SR ...(ii) (Opposite sides of rectangle are parallel)

But, seg AB seg SR ...(iii) [Construction]

seg PQ seg SR seg AB

In ABRS, seg AB seg SR ...[From (iii)]

seg AS seg BR ...(Opposite sides of rectangle are parallel and P - A - S, Q - B - R) ½

In ABRS is parallelogram ...(By definition)

AS = BR ...(v) (Opposite sides of parallelogram are equal)

Similarly by proving ABQP is a parallelogram we get,

AP = BQ ...(vi)

seg PQ seg AB ...[From (ii)]

on transversal PS,

QPS = BAS = 90° ...(vii)

Similarly we can prove, seg BA seg PS ...(viii)

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In TAS, TAS = 90° ...[From (vii)]

TS2 = TA2 + AS2 = 90o ...(ix) (By Pythagoras theorem) ½

In TBQ, TBQ = 90° ...[From (viii)]

TQ2 = TB2 + BQ2 = 90o ...(x) (By Pythagoras theorem) ½

Adding (ix) and (x),

TS2 + TQ2 = TA2 + AS2 + TB2 + BQ2 ...(xi)

In TAP, TAP = 90o ...[From (vii)]

TP2 = TA2 + AP2 ...(xii) (By Pythagoras theorem)

In TBR, TBR = 90o

TR2 = TB2 + BR2 ...(xiii) (By Pythagoras theorem) ½

Adding (xii) and (xiii),

TP2 + TR2 = TA2 + AP2 + TB2 + BR2

TP2 + TR2 = TA2 + BQ2 + TB2 + AS2 ...(xiv) ...[From (vi) and (v)

TS 2 + TQ2 = TP2 + TR2 ...[From (xi) and (xiv) ½

(ii) For the metallic hollow sphere, Outer diameter = 12 cm

Outer radius (r1) = = 6 cm ½

Thickness = 0.01 m

= 0.01 × 100 ...[ 1 m = 100cm]

= 1 cm

Inner radius (r2) = r1 – thickness ½

= 6 – 1

= 5 cm ½

Volume of metal in the hollow metalic sphere

= Volume of the outer sphere – Volume of the inner sphere ½

= r13 – r23

= (r13 – r23) ½

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=

= ½

= ½

= 22 13 ½

V =1144

3 381.33 cm3

Volume of the hollow sphene is 381.33 cm3