Time : 2 Hours (Pages 3) Max. Marks : 40 Q.1. Solve ANY ...ssc.maheshtutorials.com/images/school/geomatry/paper-two/...Solve ANY FIVE of the following : 5 (i) Lines PM and PN are tangents

Q.1. Solve ANY FIVE of the following : 5

(i) Lines PM and PN are tangents tothe circle with centre O.If PM = 7 cm, find PN.

(ii) Using Euler’s formula, find F,if V = 6 and E = 12.

(iii) If x-coordinate of point A is negative and y-coordinate is positive, then inwhich quadrant point A lies ?

(iv) If m = 5 and c = – 3, then write the equation of the line.

(v) The area of a circle is 314 cm2 and the area of its minor sector is31.4 cm2. Find the area of its major sector.

(vi) Find the value of 3sin2 + 3cos2 .

Q.2. Solve ANY FOUR of the following : 8(i) In the adjoining figure,

line l || line m ||line n.Lines p and q are transversals.From given informationfind ST.

Note :

(i) All questions are compulsory.

(ii) Use of calculator is not allowed.

M

O

N

P

p q

R

TC

B S

Al

m

n

8

10

12

Time : 2 Hours (Pages 3) Max. Marks : 40

MT - MATHEMATICS (71) GEOMETRY - PRELIM II - PAPER - 1 (E)

Seat No.2014 ___ ___ 1100

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PAPER - 12 / MT

(ii) ABC is a right angled at B.D is any point on AB.DE AC. If AD = 6 cm,AB = 12 cm,AC = 18 cm, find AE.

(iii) In the adjoining figure,seg AB and seg AD are chordsof the circle. C be a point ontangent to the circle at point A.If m (arc APB) = 80º and BAD = 30º,then find (i) BAC (ii) m (arc BQD)

(iv) Draw a tangent at any point ‘M’ on the circle of radius 2.9 cm and centre ‘O’.

(v) Eliminate , if, x = a sec , y = b tan

(vi) If sin + sin2 = 1, prove that cos2 + cos4 = 1.

Q.3. Solve ANY THREE of the following : 9

(i) In the adjoining figure,LMN = 90º and LKN = 90º,seg MK seg LN. Prove thatR is the midpoint of seg MK.

(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 thecircles touching at M and N.Prove that CM = CN.

(iii) Construct the incircle of SRN, such that RN = 5.9 cm, RS = 4.9 cm,R = 95º.

B C

E

D

A

A

Q

B

P

C

D

M

L R N

KA

BM N

C




PAPER - 1

Best Of Luck

(iv) Write down the equation of a the line whose slope is 3

2 and which passes

through P where P divides the line segment joining A (– 2, 6) andB (3, – 4) in the ratio 2 : 3.

(v) The radius of a circle is 3.5 cm and area of the sector is 3.85 cm2. Find

the length of the corresponding arc and the measure of arc.

Q.4. Solve ANY TWO of the following : 8

(i) Suppose AB and AC are equal chords of a circle and a line parallel tothe tangent at A intersects the chords at D and E. Prove that AD = AE.

(ii) Find the equation of the straight line passing through the origin and thepoint of intersection of the lines x + 2y = 7 and x – y = 4.

(iii) Eliminate , if x = 2 cos – 3 sin , y = cos + 2 sin

Q.5. Solve ANY TWO of the following : 10

(i) Prove : The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

(ii) SHR ~ SVU, In SHR, SH = 4.5 cm, HR = 5.2 cm, SR = 5.8 cm and

SH

SV =

3

5; construct SVU.

(iii) In the adjoining figure,PR and QS are two diameters of the circle.If PR = 28 cm and PS = 14 3 cm, find(i) Area of triangle OPS(ii) The total area of two shaded segments.

( 3 = 1.73)

3 / MT

P S

Q R

O

120º




A.1. Attempt ANY FIVE of the following :(i)

PM = PN[Length of the two tangent segments from an ½external point to a circle are equal]But, PM = 7 cm [Given]

PN = 7 cm ½

(ii) F + V = E + 2 F + 6 = 12 + 2 ½ F + 6 = 14 F = 14 – 6

F = 8 ½

(iii) = – 30º [Given]sin = sin (– 30) ½

= – sin 30

=1

– 2

sin (– 30) = 1

–2

½

(iv) m = 5, c = – 3 By slope point form, the equation of line is

y = mx + c ½ y = 5x – 3

5x – y – 3 = 0 ½

M

O

N

P

Time : 2 Hours Prelim - II Model Answer Paper Max. Marks : 40

MT - MATHEMATICS (71) GEOMETRY - PRELIM II - PAPER - 1 (E)

Seat No.2014 ___ ___ 1100




PAPER - 1

(v) Area of major sector = Area of circle – Area of minor sector ½= 314 – 31.4= 282.6 cm2

The area of the major sector is 282.6 cm2. ½

(vi) 3sin2 + 3cos2 = 3 (sin2 + cos2 ) ½= 3 (1) [ sin2 + cos2 = 1]= 3 ½

A.2. Solve ANY FOUR of the following :(i)

line l || line m || line n [Given] On transversals p and q,

AB

BC=

RS

ST

[By Property of Intercepts made

by threeparallel lines] 1

8

10=

12

ST[Given]

ST =12 ×10

8½

ST = 15 units ½

(ii) In ABC and AED,BAC DAE [Common angle]ABC AED [ Each is 90º]

ABC ~ AED [By AA test of similarity] 1

AB

AE=

AC

AD[c.s.s.t.] ½

12

AE=

18

6[Given]

AE =12 × 6

18 AE = 4 units ½

2 / MT

B C

E

D

A

p q

R

TC

B S

Al

m

n

8

10

12




PAPER - 13 / MT

(iii) m BAC =1

2 m(arc APB)

[Tangent secant theorem] ½

m BAC =1

2× 80

m BAC = 40º ½

m BAD =1

2m (arc BQD) [Inscribed angle theorem] ½

30 =1

2 m (arc BQD)

m (arc BQD) = 30 × 2

m (arc BQD) = 60º ½

(iv)

½ mark for rough figure½ mark for circle1 mark for drawing perpendicular

(v) x = a sec

sec =x

a......(i) ½

y = b tan

tan =y

b......(ii) ½

MO 2.9 cm

MO 2.9 cm

(Rough Figure)

A

Q

B

P

C

D




PAPER - 14 / MT

1 + tan2 = sec2

1 + y

b

2

=x

a

2

[From (i) and (ii)] ½

1 + y

b

2

2 =x

a

2

2

2

2

x

a –

2

2

y

b= 1 ½

(vi) sin + sin² = 1 [Given] sin = 1 – sin² ½

sin = cos2 sin + cos = 1

1 – sin = cos

2 2

2 2

½

sin2 = cos4 [Squaring both sides] ½

1 – cos2 = cos4 sin + cos = 1

1 – cos = sin

2 2

2 2

½

cos² + cos4 = 1

A.3. Solve ANY THREE of the following :(i) In LMN,

m LMN = 90º [Given]seg MR hypotenuse LN [Given]

MR2 = LR × RN.....(i)[By property of geometric mean] 1

In LKN,m LKN = 90º [Given]seg KR hypotenuse LN [Given]

KR2 = LR × RN .....(ii) [By property of geometric mean] 1 MR2 = KR2 [From (i) and (ii)] MR = KR [Taking square roots] R is the midpoint of seg MK 1

(ii) Line CBA is a secant intersecting thecircle at points B and A and line CM isa tangent to the circle at point M. CM² = CB × CA .......(i)

[Tangent secant property] 1Line CBA is a secant intersecting thecircle at points B and A and line CN isa tangent to the circle at point N.

M

L R N

K

A

BM N

C




PAPER - 15 / MT

CN² = CB × CA ......(ii) [Tangent secant property] 1 CM² = CN² [From (i) and (ii)] CM = CN [Taking square roots] 1

(iii)

½ mark for rough figure½ mark for drawing SRN1 mark for drawing the angle bisectors1 mark for drawing the incircle

(iv) A (– 2, 6), B (3, – 4)Point P divides seg AB internally in the ratio 2 : 3 ½Let, P (x, y)By section formula for internal division,

x =mx nx

m + n

2 1y =

my + ny

m + n2 1

=2 (3) + 3 (–2)

2 + 3 =2 (– 4) 3 6

2 3

=6 – 6

5=

–8 + 18

5

=0

5=

10

5= 0 = 2

P (0, 2) 1

The line having slope 3

2 passes through the point P (0, 2)

The equation of the line by slope point form is,

5.9 cm NR

4.9 cm

S

95º××•

•

O

(Rough Figure)

5.9 cm NR

4.9 cm

S

95º ××••

O




PAPER - 16 / MT

(y – y1) = m (x – x1) ½

(y – 2) =3

2 (x – 0)

2 (y – 2) = 3x 2y – 4 = 3x ½ 3x – 2y + 4 = 0

The equation of the required line is 3x – 2y + 4 = 0 ½

(v) Radius of a circle (r) = 3.5 cmArea of the sector = 3.85 cm2

Area of sector =r

2 × l ½

3.85 =3.5

2 × l

3.85 × 2

3.5= l

385 × 2 ×10

100 × 35 = l ½

l =22

10 l = 2.2 cm ½

Area of sector = r360

2 ½

3.85 =22

3.5 3.5360 7

3.85 =22 35 35

360 7 10 10

½

385

100=

3511

360 10

385 × 360 ×10

100 ×11 × 35 =

= 36º ½

Length of arc is 2.2 cm and measure of an arc is 36º.

A.4. Solve ANY TWO of the following :(i) Construction : Draw seg BC. ½

Proof : Take points R and S on the tangentat A as shown in the figureline DE || line RS [Given]

On transversal AD,EDA DAR[Converse of alternate angles test] ½

B C

D E

R A S•

(½ marks for figure)




PAPER - 17 / MT

EDA BAR .......(i) [ B - D - A]BAR BCA .......(ii) [Angles in alternate segment] ½

EDA BCA ......(iii) [From (i) and (ii)]Similarly, we can prove thatDEA CBA ......(iv) ½In ABC,seg AB seg AC [Given]

BCA CBA ........(v) [Isosceles triangle theorem] ½In DEA,EDA DEA [From (iii), (iv) and (v)]

seg AD seg AE [Converse of isoscelestriangle theorem] 1

AD = AE

(ii) Let line x + 2y = 7 and x – y = 4 intersect at point Ax + 2y = 7 .......(i) ½x – y = 4 ......(ii)Subtracting (ii) from (i),x + 2y = 7x – y = 4(–) (+) (–)

3y = 3 y = 1 ½

Substituting y = 1 in equation (ii),x – 1 = 4

x = 4 + 1 x = 5 ½ A (5, 1) ½

The straight line passes through A (5, 1) and O (0, 0) The equation of the line by two point form,

x – x

x – x1

1 2=

y – y

y – y1

1 2½

x – 5

5 – 0 =y – 1

1 – 0

x – 5

5=

y – 1

1½

x – 5 = 5 (y – 1) x – 5 = 5y – 5 x – 5y – 5 + 5 = 0 ½ x – 5y = 0

The equation of the line passing through the origin and the point

of intersection of the lines x + 2y = 7 and x – y = 4 is x – 5y = 0. ½




PAPER - 1

(iii) x = 2 cos – 3 sin ......(i)y = cos + 2 sin ......(ii)

Multiplying (ii) by 2,2y = 2 cos + 4 sin .....(iii) ½

Subtracting (iii) from (i),x – 2y = 2 cos – 3 sin – (2 cos + 4 sin )

x – 2y = 2 cos – 3 sin – 2 cos – 4 sin x – 2y = – 7 sin

sin =–(x – 2y)

7......(iv) ½

Substituting sin – x 2y

7

in equation (ii)

y = cos + 2 x 2y

7

½

y = cos 2 x 2y

7

y + 2 x 2y

7

= cos

7y 2 x 2y

7

= cos ½

7y 2x 4y

7

= cos

cos =2x 3y

7

½

We know,sin2 + cos2 = 1 ½

– (x – 2y) 2x 3y

7 7

2 2

= 1

(x – 2y) (2x 3y)

49 49

2 2

= 1 ½

Multiplying throughout by 49,

(x – 2y)2 + (2x + 3y)2 = 49 ½

A.5. Solve ANY TWO of the following :(i) Given : ABC ~ PQR.

To Prove : A ( ABC)

A ( PQR)

=

BC

QR

2

2 = AB

PQ

2

2 = AC

PR

2

2 ½

8 / MT




PAPER - 19 / MT

Construction : (i) Draw seg AD side BC, ½

B - D - C(ii) Draw seg PS side QR, Q - S - R

Proof : A ( ABC)

A ( PQR)

=

BC AD

QR PS

......(i)

[ The ratio of the areas of two triangles is equal to ratio

of the products of a base and its corresponding height ] ½

ABC ~ PQR [Given]

AB BC AC

PQ QR PR ......(ii) [c.s.s.t.] ½

Also, B Q .....(iii) [c.a.s.t.]

In ADB and PSQ,

ADB PSQ [Each is a right angle]

B Q [From (ii)]

ADB ~ PSQ [By A-A test of similarity] 1

AD BD AB

PS QS PQ .....(iv) [c.s.s.t.]

AD BC

PS QR ......(v) [From (ii) and (iv)] ½

A ( ABC)

A ( PQR)

=

BC AD

QR PS [From (i)]

A ( ABC)

A ( PQR)

=

BC BC

QR QR [From (v)] ½

A ( ABC)

A ( PQR)

=

BC

QR

2

2 .......(vi)

A(ΔABC)

A(ΔPQR) =BC²

QR² = AB²

PQ² =AC²

PR²[From (ii) and (vi)] ½

Q R

P

SB C

A

D

(½ mark for figure)




PAPER - 1

(ii)

1 mark for SHR1 mark for constructing 5 congruent parts1 mark for constructing VS5S HS3S1 mark for constructing UVS RHS1 mark for required SVU

(iii) Draw seg OM side PS ½

OP = 1

PR2 [Radius is half of diameter]

OP = 1

282

OP = 14 cm ½seg OM chord PS [By construction]

PM =1

PS2 [The perpendicular drawn from

the centre of a circle to a chord

bisec ts the chord]

10 / MT

P S

Q R

O

120º

M

U

S H V

R

(Rough Figure) U

R

S H V

5.2

cm

4.5 cm

S1

S2

S3

S4

S5

××

5.8

cm

•

•




PAPER - 111 / MT

PM =1

14 32

PM = 7 3 cm ½In OMP,OMP = 90º [By construction]

OM2 + PM2 = OP2 [By Pythagoras theorem]

OM2 + 7 32

= 142

OM2 = 196 – 147 OM2 = 49 OM = 7 cm [Taking square roots] ½

Area of OPS =1

2 × base × height ½

Area of OPS =1

2 × PS × OM

=1

2 × 14 3 × 7

= 49 3= 49 (1.73) ½

Area of OPS = 84.77 cm2

Area of sector OPS =360

× r2 ½

=120 22

14 14360 7

=616

3= 205.33 cm2 ½

Area of segment PS = Area of sector OPS – Area of OPS= 205.33 – 84.77= 120.56 cm2 ½

Similarly we can prove, Area of segment QR = 120.56 cm2

Total area of two shaded segments = 120.56 + 120.56 = 241.12 cm2

Area of OPS is 84.77 cm2 and total area of two shaded ½segments is 241.12 cm2.




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Time : 2 Hours (Pages 3) Max. Marks : 40 Q.1. Solve ANY ...ssc.maheshtutorials.com/images/school/geomatry/paper-two/...Solve ANY FIVE of the following : 5 (i) Lines PM and PN are tangents