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Q.1. Attempt ANY SIX of the following : 6

(i) Sides of triangle are given below. Determine whether it is a right

angled triangle : 9, 40, 41.

(ii) In the adjoining figure,

point P is centre of the circle

and line AB is the tangent to the

circle at T. What is the measure ofPTB ?

(ii i) If the angle = 60, find sin .

(iv) Write the slope of each of the line stated below : y + 3 =1

2(x 5).

(v) Using Eulers formula, find V, if E = 30, F = 12.

(vi) In the adjoining figure,if m BAC = 80,Find m (arc BQC)

(vii) If cot =7

24

then find the values of cosec , is in IV quadrant.

Note :

(i) All questions are compulsory.

(ii) Use of calculator is not allowed.

Time : 2 Hours (Pages 4) Max. Marks : 60

Seat No.

Q.P. SET CODE

2011 ___ ___ 1100 - MT - w - MATHEMATICS (71) GEOMETRY - SET - A (E)

P

A T B

B

A

C

Q

P

80

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Q.2. Solve ANY FIVE of the following : 10

(i) In the adjoining figure,

seg DH seg EF and seg GK seg EF.If DH = 12 cm, GK = 20 cm and

A (DEF) = 300 cm2 then find(a) EF (b) A (GEF).

(ii) In the adjoining figure,two circles intersect each otherin two points A and B. Seg AB isthe chord of both circles. Point Cis the exterior point of both thecircles on the line AB. From thepoint C tangents are drawn to the

circles touches at M and N.

Prove that CM = CN.

(ii i) Draw a circle of radius 3.6 cm, take a point M on it. Draw a tangent

to the circle at M without using centre of the circle.

(iv) Prove the following : sec2 + cosec2 = sec2 . cosec2 .

(v) If (1, 2),1

, 32

and (0, k) are collinear points find the value of k.

(vi) The radius of the base of a right circular cylinder is 3 cm, height is

7cm. Find (a) curved surface area (b) volume of the closed right

circular cylinder.22

Given =7

Q.3. Solve ANY FOUR of the following : 12

(i) In the adjoining figure,

if LK = 6 3 find MK, ML, KN, MN

and the perimeter ofMNKL.

D

EH

KF

G

A

BM N

C

N

M

LK

45

30

6 3

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(ii) Construct tangents to the circle from point B with radius 3.5 cm

and centre A. Point B is at a distance 7.3 cm from the centre.

(iii) Prove the following :tan A tan A + sec A + 1

=sec A 1 tan A + sec A 1

(iv) Point (m, 2m 1) lies on the line3x 2y

= 15 3

find m.

(v) Construct GHI such that GI = 5.4 cm, GHI = 75. HR is median.

HR = 3.2 cm.

Q.4. Solve ANY THREE of the following : 12

(i ) Prove that : If a line divides two sides of a triangle in the same ratio,

then the line is parallel to the third side.

(ii) In ABC, midpoints of sides AB,

AC and BC are P, Q and R respectively.

AS BC. Prove that PQRS is a

cyclic quadrilateral.

(iii) AMT ~ AHE, In AMT, MA = 6.3 cm, MAT = 120, AT = 4.9 cm and

MA

HA=

7

5, construct AHE.

(iv) Adjoining figure depicts a racing track

whose left and right ends are

semicircular. The distance between two

inner parallel line segments is 70 m

and they are each 105 m long. If the

track is 7 m wide, find the difference

in the lengths of the inner edge and

outer edge of the track.

A

P Q

BS R

C

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Best Of Luck

Q.5. Solve ANY FOUR of the following : 20

(i) Two poles of height a meters and

b meters are p meters apart.

Prove that the height h drawn from

of the point of intersection N of the

lines joining the top of each pole to

the foot of the opposite pole isab

a + b meters.

(ii) If two circles are internally touching at point P. A line intersect

those two circles in point A, B, C, D respectively then prove thatAPB CPD.

(ii i) A person standing on the bank of a river observes that the angle of

elevation of the top of a tree standing on the opposite bank is

60. When he moves 40 m away from the bank, he finds the angle of

elevation to be 30. Find the height of the tree and the width of the

river. ( 3 = 1.73)

(iv) Find the equation of the line passing through ( 3, 5) and parallel

to x 2y 7 = 0.

(v)

A toy is a combination of a cylinder, hemisphere and a cone, each

with radius 10cm. Height of the conical part is 10 cm and total

height is 60cm. Find the total surface area of the toy. ( = 3.14,

2 = 1.41)

S

Na

b

h

x yTBA

R

p

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A.1. Attempt ANY SIX of the following :

(i) (41)2 = 1681 ......(i)

(9)2

+ (40)2

= 81 + 1600= 1681 ......(ii)

(41)2 = (9)2 + (40)2 [From (i) and (ii)] The given sides form a right angled triangle. [By Converse of

Pythagoras theorem] 1

( ii ) Line AB is the tangent to the circle

at point T and seg PT is the radius.

m PTB = 90 [Radius is perpendicularto the tangent] 1

(iii) = 60sin () = sin

sin ( 60) = sin 60

sin ( 60) = 32

(iv) y + 3 =1

2(x 5)

Comparing with the equation of a line in slope point form,

y y1

= m (x x1)

m = 12

Slope of the line y + 3 = 12

(x 5) is1

2

(v) F + V = E + 2 12 + V = 30 + 2 V = 32 12 V = 20

Time : 2 Hours Prelim I Model Answer Paper Max. Marks : 60

A.P. SET CODE

2011 ___ __ 1100 - MT - w - MATHEMATICS (71) GEOMETRY - SET - A (E)

P

A T B

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(vi) m BAC = 12

m(arc BQC)

80 = 12

m(arc BQC)

m(arc BQC) = 80 2 m(arc BQC) = 160

(vii) cot = 724 1 + cot2 = cosec2

7124

2

= cosec2

491576

= cosec2

576 49576

= cosec2

cosec2 = 625576

is the IV quadrant. cosec = 25

24[Taking square roots]

A.2. Solve ANY FIVE of the following :

(i) (a) Area of triangle =1

2 base height

A (DEF) = 12

EF DH

300 = 12

EF 12 1

300 = EF 6 EF = 300

6

EF = 50 cm

(b)A ( DEF)

A ( GEF)

=

DH

GK[Triangles with common base] 1

300

A ( GEF) =12

20

A (GEF) = 300 2012

A (GEF) = 500 cm2

B

A

C80

Q

P

[Inscribed

angle theorem]

D

EH

KF

G

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(ii)

Line CBA is a secant intersecting

the circle at points B and A and

line CM is a tangent to the circle

at point M.

CM = CB CA .......(i) [Tangent secant property] Line CBA is a secant intersecting the circle at points B and

A and line CN is a tangent to the circle at point N.

CN = CB CA ......(ii) [Tangent secant property] CM = CN [From (i) and (ii)] CM = CN [Taking square roots]

(iii)

A

BM N

C

NM

L

(Rough Figure)

A

N M

L

A

mark for rough figure

1 mark for the circle and tangent

mark for congruent marks

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(iv) L.H.S. = sec2 + cosec2

= 1 1cos sin2 2 1 1sec , coseccos sin

=sin + cos

cos . sin

2 2

2 2

=1

cos . sin2 2 [ sin2 + cos2 = 1] = sec2 . cosec2 = R.H.S.

sec2 + cosec2 = sec2 . cosec2

(v) Let, A (1, 2) (x1, y1)B

1, 3

2 (x

2, y

2)

C (0, k) (x3, y

3)

Points A, B and C are collinear

Slope of line AB= Slope of line BC

y y

x x2 1

2 1=

y y

x x3 2

3 2

3 21

12

=k 3

10

2

11

2

=k 3

1

2 1 = k 3 k = 1 + 3 k = 4 The value of k is 4.

(vi) Radius of a right circular cylinder = 3cm

its height (h) = 7cm

(a) Curved surface area of a cylinder = 2 rh = 2

22

7 3 7

Curved surface area of a cylinder = 132 cm2 (b) Volume of the cylinder = r2h

=22

7 3 3 7

= 198 cm3

Curved surface area is 132 cm2and volume of the cylinder is 198 cm3

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A.3. Solve ANY FOUR of the following :

(i) In MLK,m MLK = 90 [Given]m MKL = 30 [Given]

m LMK = 60 [Remaining angle] MLK is a 30 - 60 - 90 triangle. By 30 - 60 - 90 triangle theorem,

LK =3

2MK [Side opposite to 60]

6 3 = 32

MK [Given]

MK = 6 3 23

MK = 12 units ......(i) ML =

1

2 MK [Side opposite to 30]

ML = 12

12

ML = 6 units ......(ii) In MKN,m MKN = 90 [Given]m MNK = 45 [Given] m NMK = 45 [Remaining angle]

MKN is a 45 - 45 - 90 triangle By 45 - 45 - 90 triangle theorem, MK = KN = 1

2 MN ......(iii)

MK =1

2 MN [From (iii)]

12 = 12

MN

MN = 12 2 units .....(iv) KN = 12 units ......(v) [From (i) and (iii)]

Perimeter ofMNKL = MN + KN + KL +ML

= 12 2 12 6 3 6

[From (ii), (iv) and (v) and given]

= 18 12 2 6 3

Perimeter ofMNKL = 6 3 2 2 3 units

N

M

LK

45

30

6 3

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(ii)

(iii) 1 + tan2 A = sec2 A tan2 A = sec2 A 1 tan A . tan A = (sec A 1) (sec A + 1)

tan A

sec A 1 =sec A 1

tan A

tan A

sec A 1 =sec A 1

tan A

=

tan A sec A 1

sec A 1 tan A

[By theorem on equal ratios]

tan A

sec A 1 =tan A + sec A + 1

tan A + sec A 1

(Rough Figure)

7.3 cmB

C

A

D

M3

.5cm

3.5cm

M7.3 cm

C

A

D

B

3.5cm

3.5cm

mark for rough figure mark for drawing the circle of radius 3.5 cm mark for drawing the perpendicular bisector of seg AB mark for drawing the circle with centre M1 mark for drawing both the tangents from point B

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(iv) Let, A (m, 2m 1)

Point A lies on the line 3x 2y5 3

= 1

Co-ordinates of point A satisfies the equation, 3m 2 (2m 1)

5 3= 1

Multiplying throughout by 15,

3m 2 (2m 1)15 15

5 3

= 15 ( 1)

3 (3m) 5 (4m 2) = 15 9m 20m + 10 = 15 11m = 15 10 11m = 25 m = 25

11

The value of m is 2511

(v)

A.4. Solve ANY THREE of the following :

(i ) Given : In ABC,line l intersects sides AB and AC at points

D and E respectively such that

AD

DB=

AE

ECTo prove : Line l || side BC

Proof : (Indirect method)

mark for rough figure mark for drawing centre O1 mark for drawing the

perpendicular bisector mark for locating point H mark for drawingGHI

RG

H H

I15

753.2cm

5.4 cm

15

O

RG

H

I

75

3

.2cm

5.4 cm

(Rough Figure)

D

C

A

B

E

F

l

( mark for figure)

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Let us suppose that line DE is not parallel to side BC

We can draw a line DF parallel to side BC, such that A-F-C.In ABC,line DF || side BC

ADDB

=AF

FC.........(i) [By B.P.T.]

But,AD

DB=

AE

EC.........(ii) [Given]

AFFC

=AE

EC[From (i) and (ii)]

AF FCFC

=

AE EC

EC

[By componendo]

ACFC

=AC

EC[ A - F - C, A - E - C]

FC = EC F and E are not two different points. Line DF and line DE coincide line DE || side BC line l || side BC

(ii) In ABC,P and Q are midpoint of seg AB and seg AC

seg PQ || seg BC [By midpointtheorem]

seg PQ || seg BR [B - R - C] Similarly, seg QR || seg PB

PBRQ is a parallelogram [By definition]In ASB,m ASB = 90 [Given]seg SP is median to hypotenuse AB [Given]

SP = 12

AB .....(i)

But, PB =1

2AB .....(ii)

In PBS,SP = PB [From (i) and (ii)]

m PBS = m PSB [Isosceles triangle theorem] PBRQ is a parallelogram

A

P Q

BS R

C

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m PBR = m PQR .....(iii) [Opposite angles of a parallelogram are congruent] m PBS = m PQR .....(iv) [B - S - R]

m PSB = m PQR ......(v) [From (iii) and (iv)] But,

m PSB + m PSR = 180 [Linear pair axiom] m PQR + m PSR = 180 [From (v)] PQRS is cyclic [If opposite angles of a

quadrilateral are

supplementary then it is a

cyclic quadrilateral]

(iii)

(iv) Diameter of inner circular edge (d1) = 70 mWidth of the track = 7 m Diameter of outer circular edge (d

2) = 70 + 7 + 7

= 84 m The inner and outer edges of the racing tracks comprises of

two semicircles and parallel segments of length 105 m each

Length of outer edge = 12

d2

+ 105 +1

2d

2+ 105

= d2

+ 210

= (84 + 210) m

T

E

A MH

6.3 cm

4.9cm

120

A1

A2

A3

A4

A5

A6

A7

1 mark for AMT mark for constructing 7 congruent parts

1 mark for constructing HA5A MA

7A

1 mark for constructing EHA TMA mark for required AHE

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Length of inner edge =1

2d

1

+ 105 +1

2d

1

+ 105

= d1

+ 210

= (70 + 210) m Difference in the lengths of

= (84 + 210) (70 + 210) inner and outer edge

= 84 + 210 70 210= 14= 14

22

7= 44 m

The difference in the lengths of inner edge and outer edge ofthe track is 44 m.

A.5. Solve ANY FOUR of the following :

(i )

AB = AT + TB [ A - T - B]

AB = (x + y) = p ......(i) [Given]In ABR and ATN,BAR TAN [Common angle]ABR ATN [ each is 90] 1

ABR ATN [By AA test of similarity] AB

AT=

BR

TN[c.s.s.t]

px

=a

h

[Given and from (i)]

a = hpx

......(ii)

In BAS and BTN,ABS TBN [Common angle]BAS BTN [ each is 90]

BAS BTN [By AA test of similarity] AB

BT=

AS

TN[c.s.s.t]

S

Na

b

h

x yTBA

R

p

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py =

b

h [Given and from (i)]

b = hpy

......(iii) 1

Multiplying (ii) and (iii),

a b =hp

x

hp

y

ab = h pxy

2 2

......(iv)

Adding (ii) and (iii),

a + b =hp

x+

hp

y

a + b = hp 1 1x y

a + b = hp (x y)xy

a + b = hpxy

2

......(v)

Dividing (iv) by (v),

ab

a b =h p hp

xy xy

2 2 2

ab

a b =h p xyxy hp

2 2

2

ab

a b = h

h =ab

a +bmetres

(ii)

Construction : Draw a common tangent

MN at point P.

C

B

P

M

N

A

D

(1 mark for figure)

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Proof : APM ADP [Angles in alternate segment]Let,m APM = m ADP = x ......(i) BPM BCP [Angles in alternate segment]Let,

m BPM = m BCP = y .....(ii) m APB = m BM m APM [Angle addition proerty]

m APB = (y x) .....(iii)[From (i) and (ii)] 1BCP is an exterior angle of CPD,

m BCP = m CPD + m CDP [Remote interior angles theorem] y = m CPD + x [From (i) and A - C - D] m CPD = (y x) .....(iv) [From (iii) and (iv)] m APB = m CPD APB CPD

( ii i) Let seg AB represents the tree

seg BC represents width of river

Let BC = x m

C and D represents the initial and

final positions of the observer

DC = 40 m

ACB and ADB are theangles of elevation

m ACB = 60 and m ADB = 30In right angled ACB,tan 60 =

AB

BC[By definition]

3 = ABx

AB = 3 x m .....(i) 1In right angled ADB,tan 30 =

AB

DB

[By definition]

13

=AB

40 x

AB = 40 x3

m .....(ii) 1

From (i) and (ii) we get,

3 x =40 x

3

3x = 40 + x

A

B

CD

40 m

30 60

( mark for figure)

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3x x = 40

2x = 40 x = 20 BC = 20 m 1 AB = 20 3 m [From (i)] AB = 20 1.73 AB = 34.6 m Height of tree is 34.6 m and width of river is 20 m.

(iv) The equation of the line is x 2y 7 = 0

Slope of line x 2y 7 = 0 iscoefficient of x

coefficient of y=

1

2=

1

2

Slopes of parallel lines are equal,

Slope of line parallel to x 2y 7 = 0 is 12

The line parallel to the line x 2y 7 = 0 whose slope is1

2

passes through ( 3, 5)

The equation of the line by slope point form is,(y y

1) = (x x

1)

y ( 5) = 12

[ x ( 3)]

y + 5 = 12

(x + 3)

2 (y + 5) = x + 3 2y + 10 = x + 3 x 2y + 3 10 = 0 x 2y 7 = 0 The equation of the line passing through ( 3, 5) and parallel to

x 2y 7 = 0 is x 2y 7 = 0.

(v)

A toy is a combination of cylinder, hemisphere and cone, each

with radius 10 cm

r = 10 cm

60 cm

10 cm10 cm

10cm

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