2018 ___ ___ 1100 - MT - x - MATHEMATICS (71) Geometry - SET - B (E)

MT - x

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

Q.P. SET CODE

B 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

SET - B2 / MT - x

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

SET - B3 / MT - x

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

(i) (c) 4 1

(ii) (b) cosec2 – sin2 = 1 1

(iii) (b) 25 cm 1

(iv) (c) 1

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

(i) B(k, –5) = (x1, y1)

C(1, 2) = (x2, y2)

Slope of line BC =

½

7 = ½

7(1 – k) = 2 + 5

7(1 – k) = 7 ½

1 – k =77

1 – k = 1 1 – 1 = k k = 0 ½

(ii) Proof : D

E F

Q

P

X

R

In ∆XDE, PQ DE ...(Given)

XPPD

=XQQE

...(i) (Basic proportionality theorem) ½

In ∆XEF, seg QR side EF ... (Given)

XQQE

=XRRF

...(ii) (Basic proportionality theorem) ½

XPPD

=XRRF

...[From (i) and (ii)] ½

seg PR side DF ...(Converse of Basic Proportionality theorem) ½

SET - B4 / MT - x

(iii) mABE = [m(arc DC) + m(arc AE)] A

BE

CD

½

108 = [m(arc DC) + 95] ½

216 = m(arc DC) + 95°

m(arc DC) = 216 – 95° ½

m(arc DC) = 121° ½

A.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

SET - B5 / MT - x

(ii) Given : In ABC, seg CE bisects ACB

A B

C

D

E

To prove : =

Construction : Through B, draw a line parallel to ray CE, Extend AC to intersect it at point D. Proof : In ABD, seg EC || seg BD (Construction)

= . . .(i) By B.P.T

ray CE || ray BD and AD is transversal

ACE CDB . . .(ii) (Corresponding angles) 2

Now, BC as transversal

ECB CBD . . .(iii) (Alternate angles)

But, ACE ECB . . .(iv) (Given) In CBD,

CBD CDB [from (ii), (iii), (iv)]

seg CB seg CD . . .(v) (converse of isosceles triangle theorem)

= from (i) and (v)

(iii) Given : ABCD is cyclic

A

B

C

D

To prove : A + C = 180o

Proof :

A = 12

m (arc BCD) ...(i) } (Inscribed angletheorem)

C =12

m (arc BAD) ...(ii) 2

Adding (i) and (ii)

A + C = 12

[m (arc BCD) + m (arc BAD)]

A + C =12

× 360

A + C = 180o

SET - B6 / MT - x

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

(i) Analytical fi gure: Radius = 3.3 cm (Given) Chord = 6.6 cm (Given)

Chord is twice of radius. Chord PQ is a diameter.

6.6 cm

3.3 cm

O

P

Q

6.6 cm

3.3 cm

O

P

Q

l

1 mark for drawing circle and diameter 2 marks for drawing tangents at P & Q.

SET - B7 / MT - x

(ii) A(3, 8) = (x1, y1)

B(–9, 3) = (x2, y2)

Let point P(0, a) be a point on Y-axis which divides seg AB in the ratio m : n.

P(0, a) = (x, y)

By Section formula,

½

0 ½

0 (m + n) = –9m + 3n

0 = –9m + 3n ½

9m = 3n

=

=

m : n = 1 : 3

Y-axis divides segment joining pointsA and B in the ratios 1 : 3

½

(iii) LHS = 2 2sec cosecq + q

= ½

=

=

= ½

1tancot

tan cot 1

SET - B8 / MT - x

= ½

= tan + cot ½ = R.H.S.

2 2sec cosecq + q = tan + cot

A.4. Solve ANY THREE of the following :

(i) GD is ground level.

BC is base of the ladder of

the fi re brigade van at a

height of 2 m from ground level. ½

‘T’ is top of ladder of the

fi re brigads van at the maximum height

TBC = 70° ...(Angle of elevation)

BT is the length of the ladder

BT = 20 m, BG = 2 m

BGDC is a rectangle ...(By definition) ½

BG = CD = 2 m ...(Opposite sides of a rectangle)

In BCT, BCT = 90°

sin TBC = ...(By definition) ½

sin 70° =

0.94 =

TC = 0.9420

TC = 18.80 m ½

TD = TC + CD ...(T – C – D) ½

TD = 18.80 + 2

TD = 20.80 m

Other end of the ladder can reach20.80 m above the ground ladder. ½

SET - B9 / MT - x

(ii) For cylindrical wrapper,

Diameter = 14 mm ½

Radius (R) mm = 7 mm

Height (H) = 10 cm

i.e. H = 100 mm

For cylindrical tablet, ½

Radius (r) = 7 mm, Height (h) = 5 mm

Let ‘N’ number of tablets can be wrapped in the given wrapper.

N × Volume of tablet = Volume of wrapper. ½

N × r2h = R 2H ½

N × × 7 × 7 × 5 = × 7 × 7 × 100

N = ½

N = 20

20 tablets can be packed in thegiven wrapper. ½

(iii) Analytical fi gure:

SET - B10 / MT - x

1 mark for PQR 1 mark for constructing RR3Q RR4T 1 mark for constructing RQP RTL

(iv) AR = 5AP ...(Given)

A

P Q

RS

...(i) ½

AS = 5AQ ...(Given)

...(ii) ½

In ∆ASR and ∆AQP,

...[From (i) and (ii)].

SAR QAP ...(Vertically opposite angles) 1 ∆ASR ~ ∆AQP ...(By SAS Test of similarity)

...(c.s.s.t.) ½

...[From (i)]

SR = 5 PQ ½

SET - B11 / MT - x

A.5. Solve ANY ONE of the following :

(i) Proof :

A

Q

B

C

PDDAP PAB [Ray AP bisects DAB]

Let mDAP = mPAB = xo ...(i)DCQ QCB [Ray CQ bisects DCB]

Let mDCQ = mQCB = yo ...(ii) ½ABCD is cyclic

DAB + mDCB =180o

[Opposite angles of cyclic quadrilateral are supplementary] ½

DAP + PAB + DCQ + QCB = 180o

x + x + y + y = 180 [From (i) and (ii)] 2x + 2y = 180 ...(iii)

mDAP =12

m(arc DP) [Inscribed angle theorem] ½

x = 12

m(arc DP) [From (i)]

m(arc DP) = 2x ...(iv)

mDCQ =12

m(arc DAQ) [Inscribed angle theorem] ½

y = 12

m(arc DAQ) [From (ii)]

m(arc DAQ) = 2y ...(v) ½

2x + 2y = 180 ...(iii) m(arc DP) = 2x ...(iv) m(arc DAQ) = 2y ...(v) Adding (iv) and (v),

m(arc DP) + m(arc DAQ) = 2x + 2y ½ m(arc PDQ) = 2x + 2y [Arc addition property]

m(arc PDQ) = 180o [From (iii)] ½ Arc PDQ is a semicircle

Seg PQ is the diameter of the circle ½

SET - B12 / MT - x

(ii) Construction : Draw seg AE side BC ½ such that B – D – E - C Proof : In AEB, mAEB = 90o [Construction]

AB2 = AE2 + BE2 ...(i) [By Pythagoras theorem] ½

In AED, mAED = 90o [Construction]

AD2 = AE2 + BE2 ...(ii) [By Pythagoras theorem] ½

Subtracting equation (ii) from (i),A

B CD

AB2 – AD2 = AE2 + BE2 – (AE2 + DE2) ½

AB2 – AD2 = AE2 + BE2 – AE2 – DE2

AB2 – AD2 = BE2 – DE2

AB2 – AD2 = (BE + DE) (BE – DE)

AB2 – AD2 = (BE + DE) BD ...(iii) [B – D – E] ½

In AEB and AEC,mAEB = m AEC = 90o [Construction]

Hypotenuse AB Hypotenuse AC [Given]

seg AE seg AE [Common side]

AEB AEC [Hypotenuse-side theorem] 1

seg BE seg CE …(iv) [c.s.c.t.]

AB2 – AD2 = (CE + DE) BD [from (iii) and (iv)]

AB2 – AD2 = CD BD [C – E – D] ½

A.6. Solve ANY ONE of the following :

(i) To Prove : 2AB2 = 2AC2 + BC2A

BCD

Proof :

DB = CD ...(i) (Given)

In ADB, ADB = 90o (Given)

AB2 = AD2 + DB2 (By Pythagoras theorem) ½

AB2 = AD2 + (3CD)2 [From (i)] AB2 = AD2 + 9CD2 ...(ii)

In ADC, ADC = 90o ...(Given) AC2 = AD2 + CD2 (By Pythagoras theorem) ½ AD2 = AC2 – CD2 ...(iii)

AB2 = AC2 – CD2 + 9CD2 [From (ii) and (iii) ½ AB2 = AC2 + 8CD2 ...(iv)

But BC = CD + DB ...[C - D - B]

SET - B13 / MT - x

BC = CD + 3CD ...[From (i)] BC = 4CD

CD = ...(v) ½

AB2 = AC2 + 8 ...[From (iv) and (v)]

AB2 = AC2 + 8 × ½

AB2 = AC2 + ...[From (iv) and (v)]

2AB2 = 2AC2 + BC2 (Multiplying throughout by 2) ½

(ii) In PQRPQR = 90º [Tangent Theorem]R = 30º [Given]

P = 60º [Remaing angle] PQR is 30º – 60º – 90º triangle

PQ =12

PR (30º – 60º – 90º triangle theorem)

= 12

× 12

AB

Q

P

R

PQ = 6 cm Radiam of the circle (r) = 6 cm

QR = 32

PR

= 32

× 12

QR = 6 3

A( PQR) = 12

QR × PQ

=12

6 6

= 18 = 18 1.73 = 31.14 cm2 1

Area of a sector = r2

SET - B14 / MT - x

A(P–QAB) = 31.14 62

= 3 14 6 6 = 3.14 6

= 18.84 cm2 1 Area of shaded region = A(PQR) – A(P–QAB) = 31.14 – 18.84 = 12.30 cm2 1 Area of the shaded region = 12.30 cm2