A.1. Solve ANY FIVE of the following :

(i)A ( ABE)A ( ABD)

=

BEAD [Triangles with common base] ½

A ( ABE)A ( ABD)

=

69

A ( ABE)A ( ABD)

=

23 ½

(ii) Let r1 = 5 cm and r2 = 3 cmThe circles are touching internally.

Distance between their centres= r1 – r2

= 5 – 3= 2 cm 1

(iii) Given : = – 45ºtan (– ) = – tan ½

tan (– 45º) = – tan 45º

tan (– 45º) = – 1 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

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

MT - wAA.P. SET CODE

2017 ___ __ 1100 - MT - w - MATHEMATICS (71) GEOMETRY- SET - A (E)

BE

A D

O A T

SET - A2 / MT - w

(v) radius = 7 cmCircumference = 2r ½

= 2 ×227

× 7

= 44 cm ½ Circumference is 44 cm.

(vi) Height of an equilateral triangle = 3 side2 ½

=3 62

Height of an equilateral triangle = 3 3 cm ½

A.2. Solve ANY FOUR of the following :

(i)PEEQ =

44.5

PEEQ =

4 104.5 10

PEEQ =

4045 ½

PEEQ =

89 ......(i) ½

PFFR =

89 ......(ii) ½

In PQR,

PEEQ =

PFFR [From (i) and (ii)]

line EF || side QR [By converse of B.P.T.] ½

(ii) Chords AB and CD intersect eachother internally at point P

PA × PB = PC × PD 1 6 × 4 = 8 × PD

PD =248

PD = 3 units 1

P

E F

Q R

4 8

94.5

C

BPA

D

SET - A3 / MT - w

(iii) Curved surface area of a cone = 1640 cm2

its radius (r) = 40 cm.Curved surface area of a cone = r l ½

1640 = × 40 × l

1640

40= l

l = 41 cm ½Now,

r2 + h2 = l2 ½ 402 + h2 = 412

h2 = 412 – 402

h2 = 1681 – 1600 h2 = 81 h = 9 cm [Taking square roots] Height of a cone is 9 cm. ½

(iv) Since the initial arm rotates inclockwise direction and the angle 1is more than – 180º but less than – 270º,the terminal arm lies in II quadrant.

(v) Radius of a right circular cylinder = 3cmits height (h) = 7cm

(a) Curved surface area of a cylinder = 2 rh ½

= 2 ×227

× 3 × 7

Curved surface area of a cylinder= 132 cm2 ½(b) Total Surface area of a cylinder = 2r (r + h) ½

= 2 ×227

× 3 (3 + 7)

= 2 ×227

× 3 × 10

=1320

7= 188.57 cm2 ½

Total surface area of the cylinder is 188.57 cm2.

Y

XX O

Y

220º

(1 mark for figure)

SET - A4 / MT - w

(vi)

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

seg AP is the median [Given] AB2 + AC2 = 2AP2 + 2BP2 [By Apollonius theorem] ½ 260 = 2 (7)2 + 2BP2 [Given] 260 = 2 (49) + 2BP2 ½ 260 = 98 + 2BP2

260 – 98 = 2BP2

2BP2 = 162 ½

BP2 =162

2 BP2 = 81 BP = 9 units [Taking square roots] ½

BP =12 BC [ P is the midpoint of seg BC] ½

9 =12 BC

BC = 18 units ½

A B8.3 cm

1 mark for drawing seg AB1 mark for drawing perpendicular bisector of seg AB

A

B P C

SET - A5 / MT - w

(ii) PA = PB = 5BQ = CQ = 3Let,AS = SD = xCR = DR = y ½PQRS is a parallelogram [Given]

PQ = SR [ Opposite sides of a ½parallelogram are congruent]

PB + BQ = SD + DR [P - B - Q and S - D - R] 5 + 3 = x + y x + y = 8 ......(i) ½

PS = QR [ Opposite sides of aparallelogram are congruent]

PA +AS = QC + CR [ P - A - S and Q - C - R] 5 + x = 3 + y x – y = 3 – 5 x – y = – 2 ......(ii) ½

Adding (i) and (ii)x + y + x – y= 8 + (– 2)

2x = 8 – 2 2x = 6 x = 3 ½

PS = PA + AS [ P - A - S] PS = 5 + x PS = 5 + 3

PS = 8 units ½

(iii)

BP Q

AC

S D R

5 3

[The lengths of the twotangent segments to a circledrawn from an externalpoint are equal]

(Analytical Figure)

X Y

Z

4.6 cm

P

•

• •

SET - A6 / MT - w

(iv) tan = 1

sincos

= 1

sin = cos .......(i) ½1 + tan2 = sec2

1 + (1)2 = sec2 1 + 1 = sec2 2 = sec2

sec = 2 [Taking square roots] ½

cos =1

sec

cos =12

sin =12 [From (i)] ½

X Y•

Z•

• 4.6 cm

P

1 mark for drawing chord XY1 mark for drawing tangent at X1 mark for drawing tangnet at Y

SET - A7 / MT - w

cosec =1

sin

=1

12

½

cosec = 2

sin cos

sec cosec

=

1 12 22 2

½

=2

22 2

=2

2 2 2

=2

2 2

=24

sin cos

sec cosec

=

12

½

(v) Let, A 2 1,5 3 (x1, y1)

B 1 , k2 (x2, y2)

C 4 , 05 (x3, y3)

Points A, B and C are collinearSlope of line AB = Slope of line BC ½

y – yx – x

2 1

2 1=

y – yx – x

3 2

3 2

1k –3

1 2–2 5

=0 – k4 1–5 2

½

3k – 131

10

=– k3

10½

SET - A8 / MT - w

3k – 1 ×10

3 =10– k ×3 ½

3k – 1 = – k 3k + k = 1 4k = 1 ½

k =14

The value of k is14 ½

A.4. Solve ANY TWO of the following :(i) AB represents the height of the tree

The tree breaks at point DAD is the broken part of tree whichthen takes the position of DC ½

AD = DCmDCB = 30ºBC = 30 mIn right angled DBC,

tan 30º =DBBC [By definition] ½

13 =

DB30

DB =303

DB =30 3

3½

DB = 10 3 m

cos 30º =BCDC [By definition] ½

32

=30DC

DC =30 2

3

DC =30 3 2

3 ½

DC = 20 3 m

A

D

B C30 m30º

(½ mark for figure)

SET - A9 / MT - w

AD = DC = 20 3 m

AB = AD + DB [ A - D - B] ½

AB = 20 3 + 10 3

AB = 30 3 m

AB = 30 × 1.73

AB = 51.9 m

The height of tree is 51.9 m. ½

(ii) Given : ABCD is a cyclic

To Prove : m ABC + m ADC = 180º ½ m BAD + m BCD = 180º

Proof :

m ABC =12 m (arc ADC) .....(i)

1m ADC =

12 m (arc ABC) .....(ii)

Adding (i) and (ii), we get

m ABC + m ADC =12 m (arc ADC) +

12 m (arc ABC)

m ABC + m ADC =12 [m (arc ADC) + m (arc ABC)] ½

m ABC + m ADC =12 × 360º [ Measure of a circle is 360º] ½

m ABC + m ADC = 180º .......(iii)

In ABCD,

m BAD + m BCD + m ABC + m ADC = 360º[ Sum of measure of angles of a ½

quadrilateral is 360º]

m BAD + m BCD + 180º = 360º [From (iii)]

m BAD + m BCD = 180º ½

C

DB

A(½ mark for figure)

[Inscribedangletheorem]}

SET - A10 / MT - w

(iii) A (4, 7), B (– 2, 3), C (0, 1)Let, seg AD, seg BE and seg CF be the medians on sides BC,AC and AB respectively.

D, E and F are the midpoints of sides BC, AC andAB respectively.By midpoint formula,

D x + x y + y

,2 2

1 2 1 2 ½

–2 + 0 3 + 1,

2 2

–2 4,2 2

(– 1, 2) ½

E x + x y + y

,2 2

1 2 1 2

4 0 7 1,

2 2

4 8,2 2

(2, 4) ½

F x + x y + y

,2 2

1 2 1 2

4 (–2) 7 3,

2 2

4 – 2 10,

2 2

2 , 52

(1, 5) ½By two point form,The equation of median AD,

x – xx – x

1

1 2=

y – yy – y

1

1 2½

x – 4

4 – (– 1) =y – 77 – 2

x – 44 1 =

y – 75

x – 4

5=

y – 75

D

A (4, 7)

F E

B (– 2, 3) C (0, 1)

SET - A11 / MT - w

x – 4 = y – 7 x – y – 4 + 7 = 0 x – y + 3 = 0 ½

The equation of the median BEx – xx – x

1

1 2=

y – yy – y

1

1 2

x – (–2)–2 – 2 =

y – 33 – 4

x 2– 4

=y – 3

–1 x + 2 = 4 (y – 3) x + 2 = 4y – 12 x – 4y + 2 + 12 = 0 x – 4y + 14 = 0 ½

The equation of the median CFx – xx – x

1

1 2=

y – yy – y

1

1 2

x – 00 – 1 =

y – 11 – 5

x–1 =

y – 1– 4

4x = y – 1 ½ 4x – y + 1 = 0 The equation of the medians of ABC are x – y + 3 = 0,

x – 4y + 14 = 0 and 4x – y + 1 = 0

A.5. Solve ANY TWO of the following :(i) Given : In ABC,

(i) Line l || side BC(ii) Line intersects sides AB and AC at ½

points D and E respectively.A - D - B, A - E - C

To Prove :ADDB

=AEEC

Construction : Draw seg BE and seg CD. ½Proof : ADE and BDE have a common vertex E and theirbases AD and BD lie on the same line AB.

Their heights are equal

A( ADE)A( BDE)

=

ADDB

...(i) [Triangles having equal heights] ½

ADE and CDE have a common vertex D and their bases AEand EC lie on the same line AC. ½

Their heights are equal.

A

CB

D E

(½ mark for figure)

SET - A12 / MT - w

A( ADE)A( CDE)

=

AECE

...(ii) [Triangles having equal heights] ½line DE || side BC [Given]BDE and CDE are between the same two parallel lines DE ½and BC.

Their heights are equal.Also, they have same base DE.

A(BDE) = A(CDE) ...(iii) [Areas of two triangles having ½equal bases and equal heightsare equal]

A( ADE)A( BDE)

=

A( ADE)A( CDE)

...(iv) [From (i), (ii) and (iii)] ½

AD AE=DB EC [From (i), (ii) and (iv)] ½

(ii) Analysis : ABC ~DEF [Given]

ABDE =

BCEF =

ACDF =

23 ......(i) [c.s.s.t.]

B = E = 45º [c.a.s.t.]

ABDE =

23 [From (i)]

BCEF =

23 [From (i)]

5.2DE =

23

4.6EF =

23 2

15.6

2 = DE 13.8

2 = EF

DE = 7.8 cm EF = 6.9 cm

Information for constructing DEF is complete.

1

5.2 cm

4.6 cm45º

B C

A

(Given triangle)

SET - A

2

(iii)

A toy is a combination of cylinder, hemisphere and cone,each with radius 10 cm

r = 10 cm Height of the conical part (h) = 10 cm

Height of the hemispherical part = its radius = 10cmTotal height of the toy = 60cm ½

Height of the cylindrical part (h1) = 60 – 10 – 10= 60 – 20= 40 cm

l2 = r2 + h2 ½ l2 = 102 + 102

l2 = 100 + 100l2 = 200

l = 200 [Taking square roots]

l = 10 2 cm ½

Slant height of the conical part (l) = 10 2= 10 × 1.41= 14.1 cm ½

D

E F6.9 cm45º

7.8 cm

(Required triangle)

13 / MT - w

60

10 cm10 cm

10 cm

SET - A

Total surface area of the toy= Curved surface area of the conical part + Curved surface area 1

of the cylindrical part + Curved surface area of the hemisphericalpart

= rl + 2rh1 + 2r2 ½= r (l + 2h1 + 2r)= 3.14 × 10 (14.1 + 2 × 40 + 2 × 10) ½= 31.4 (14.1 + 80 + 20)= 31.4 × 114.1 ½= 3582.74 cm2

Total surface area of the toy is 3582.74 cm2. ½

14 / MT - w