Seat No.
2014 ___ ___ 1100
MT -
MATHEMATICS (71) GEOMETRY - PRELIM II - PAPER - 2 (E)
Time : 2 Hours
(Pages 3)
Max. Marks : 40
Note :
(i)
All questions are compulsory.
(ii)
Use of calculator is not allowed.
Q.1. Solve ANY FIVE of the following :
(i)
5
In the adjoining figure,
if m ABC = 55º, then
what is m AEB.
E
A
•
C
B
(ii)
The dimensions of a cuboid are 3 cm × 9 cm × x cm. The volume of this cuboid
is equal to the volume of a cube with side 6 cm. What is the value of x ?
(iii)
If sin  = 1, what is the value of  ?
(iv)
Find the slope of a line whose inclination is 45º.
(v)
The area of a circle with radius R is equal to the sum of the areas of
circles with radii 6 cm and 8 cm. What is the value of R ?
(vi)
1
What is the value of cot2  – sin2 θ ?
Q.2. Solve ANY FOUR of the following :
(i)
IfABC ~ DEF, and A (ABC) : A (DEF) = 16 : 25, BC = 2.2 cm,
then find EF.
A
(ii)
In the adjoining figure,
AB2 + AC2 = 122, BC = 10.
Find the length of the median
on side BC.
B
Q
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C
8
2 / MT
PAPER - 2
X
(iii)
In the adjoining figure,
m (arc XAZ) = m (arx YBW).
Prove that XY || ZW
Y
A
B
Z
W
(iv)
Draw a circle of radius 2.6 cm. Draw tangent to the circle from any point
on the circle using centre of the circle.
(v)
If the angle  = – 60º, find the value of sin , cos , sec  and tan .
(vi)
a2
b2
–
= 1.
If x = a sin , y = b tan  then prove that 2
x
y2
Q.3. Solve ANY THREE of the following :
(i)
The perpendicular AD on the
base BC of ABC intersects
BC at D so that BD = 3 CD.
Prove that 2AB2 = 2AC2 + BC2.
D
B
(ii)
9
A
In the adjoining figure,
A and B are centers of two
circles touching each other at M.
Line AC and line BD are tangents.
If AD = 6 cm and BC = 9 cm then find
the length of seg AC and seg BD.
C
C
D
A
M
B
(iii)
Construct incircle of SGN such that SG = 6.7 cm, S = 70º,G = 50º and
draw incircle of SGN.
(iv)
Find the value of k if (– 3, 11), (6, 2) and (k, 4) are collinear points.
(v)
A
B
Calculate the area of the shaded
region in the adjoining figure where
ABCD is a square with side 8 cm each.
X
D
8 cm
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3 / MT
Q.4.
(i)
(ii)
PAPER - 2
Solve ANY TWO of the following :
8
In the adjoining figure, in a circle with centre P,
a chord AB is parallel to a tangent and intersects
the radius drawn from a point of contact at
A
the midpoint of the radius. If AB = 12,
find the radius of the circle.
•
C
P
E
F
B
•
D
Y
In the adjoining figure,
two lines are intersecting at point (3, 4).
Find the equation of line PA and line PB.
P (3, 4)
A
60º
45º
X
B
X
Y
(iii)
A pilot in an aeroplane observes that Vashi bridge is on one side of the
plane and Worli sea-link is just on the opposite side. The angle of
depression of Vashi bridge and Worli sea-link are 60º and 30º respectively.
If the aeroplane is at a height of 5500 3 m at that time, what is the
distance between Vashi bridge and Worli sea-link ?
Q.5. Solve ANY TWO of the following :
10
(i)
Prove : If a line parallel to a side of a triangle intersects other sides in
two distinct points, then the line divides those sides in proportion.
(ii)
Draw a triangle ABC, right angled at B such that, AB = 3 cm and
BC = 4 cm. Now construct a triangle similar to ABC, each of whose
7
times the corresponding side of ABC.
sides is
5
(iii)
An ink container of cylindrical shape is filled with ink upto 91%. Ball pen
refills of length 12 cm and inner diameter 2 mm are filled upto 84%. If
the height and radius of the ink container are 14 cm and 6 cm respectively,
find the number of refills that can be filled with this ink.
Best Of Luck

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Seat No.
2014 ___ ___ 1100
MT -
MATHEMATICS (71) GEOMETRY - PRELIM II - PAPER - 2 (E)
Time : 2 Hours
A.1.
(i)
(ii)
Max. Marks : 40
Attempt ANY FIVE of the following :
AEB  ABC
[Angles in alternate segments]
But, ABC = 55º
[Given]
½
 AEB = 55º
½


Volume of cuboid
3×9×x
3×9×x
=
=
=

x
=
Volume of cube
(6) 3
6×6×6
666
39

x
=
8
(iii)


(iv)
Prelim - II Model Answer Paper
sin  = 1
But, sin 90 = 1
sin  = sin 90
[Given]
½
[Given]
½
 = 90º
Inclination of the line ()
 Slope of the line
½
=
=
=
=
45º
tan 
tan 45º
1
 Slope of the line is 1.
(v)




½
According to given condition,
R 2 =  × (6)2 +  × (8)2
R 2 =  [62 + 82]
R2
= 36 + 64
R2
= 100
R
= 10
[Taking square roots]
 The value of R is 10.
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½
½
½
2 / MT
=
1
cot2  – sin2 
cot2  – cosec2 
=
–1
(vi)
A.2.
(i)
PAPER - 2
½
[ 1 + cot2  = cosec2 
 cot2  – cosec2  = – 1]
½
Solve ANY FOUR of the following :
ABC ~ DEF
[Given]

A (ABC)
A (DEF)
=
BC2
EF 2
[Areas of similar triangles]

16
25
=
(2.2)2
EF 2
[Given]

4
5
=
2.2
EF
[Taking square roots]

EF
=
2.2  5
4

EF
=
5.5
2

EF
=
2.75 cm
(ii)










= 6 units
½
½
½
A
In ABC,
seg AQ is the median
[Given]
1
BQ = QC =
× BC
2
1
BQ = QC =
× 10
[Given]
B
Q
C
2
BQ = QC = 5 units
......(i)
AB2 + AC2 = 2AQ2 + 2BQ2
[By Appollonius theorem]
122 = 2AQ2 + 2 (5)2
[From (i) and given]
2
122 = 2AQ + 2 (25)
122 = 2AQ2 + 50
2AQ2 = 122 – 50
2AQ2 = 72
AQ 2 = 36
AQ
½
[Taking square roots]
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½
½
½
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(iii)
PAPER - 2
Construction : Draw seg XW
Proof :
X
1
m (arc XAZ)
2
[Inscribed angle theorem]
m XWZ =
1
m (arc YBW)
2
But, m (arc XAZ) = m (arc YBW)

m XWZ = m WXY

line XY || line ZW
m WXY =
.......(i)
½
Y
A
B
½
.......(ii)
Z
W
.......(iii)
[Given]
[From (i), (ii) and (iii)]
[Alternate angles test]
½
½

(iv)

(Rough Figure)
O 2.6 cm
O
2.6 cm
P
P
½ mark for rough figure
½ mark for circle
1 mark for drawing perpendicular at
point P

 = – 60º
sin (– )
sin (– 60)

sin (– 60)

sec (– )
sec (– 60)
= – sin 
= – sin 60
3
= –
2
= sec 
= sec 60

sec (– 60)
= 2

cos (– )
cos (– 60)
= cos 
= cos 60
(v)
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
cos (– 60)

tan (– )
tan (– 60)
1
2
= – tan 
= – tan 60

tan (– 60)
=
(vi)
a sin 
a
x
=

cosec 
=
a
x
y
=
b tan 
=
b
y
=
b
y
1
tan 
cot 
=
We know,
1 + cot2  =
 cosec2  – cot2  =
2


A.3.
(i)
½
– 3


½
=
x
1
sin 

PAPER - 2
......(i)

1 
 cos ec   sin  


½
.....(ii)

1 
 cot   ta n  


½
cosec2 
1
½
2
 b
 a
  –  
x
 y
2
a
b2
–
x2
y2
=
1
=
1
[From (i) and (ii)]
Solve ANY THREE of the following :
BC = BD + CD
[B - D - C]

BC = 3CD + CD
[ BD = 3CD]

BC = 4CD
.....(i)
In ADB,
[Given]
m ADB = 90º
2
2
2

AB = AD + BD
B
[By Pythagoras theorem]

AB 2 = AD2 + (3CD)2
[ BD = 3CD]
2
2
2

AB = AD + 9CD

AB 2 = AD2 + CD2 + 8CD2 .....(ii)
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½
A
D
½
C
½
½
5 / MT
In ADC,
m ADC = 90º

AC 2 = AD2 + CD2
2
AB = AC2 + 8CD2
 BC 
 AB = AC + 8 

 4 
2
[Given]
[By Pythagoras theorem]
[From (ii) and (iii)]
½
[From (i)]
½
2
2
 AB2 = AC2 + 8 ×
.....(iii)
PAPER - 2
BC2
16
BC2
2
 2AB2 = 2AC2 + BC2
 AB2 = AC2 +
(ii)
[Multiplying throughout by 2]
A-M-B













[If two circles are touching circles then
the common point lies on the line
joining their centres]
C
D
AM = AD = 6 cm........(i)
[Radii of the same circle]
BM = BC = 9 cm
........(ii)
B
A
M
AB = AM + MB
[ A - M - B]
AB = 6 + 9
[From (i) and (ii)]
AB = 15 cm
........(iii)
In ABC,
m ACB = 90º
[Radius is perpendicular to the tangent]
AB² = AC² + BC²
[By Pythagoras theorem]
15² = AC² + 9²
[From (ii) and (iii)]
225 = AC² + 81
AC² = 225 – 81
AC² = 144
AC = 12 cm
[Taking square roots]
In ADB,
m ADB = 90º
[Radius is perpendicular to the tangent]
AB 2 = AD2 + BD2
[By Pythagoras theorem]
152 = 62 + BD2
[From (i) and (iii)]
225 = 36 + BD2
BD 2 = 225 – 36
BD 2 = 189
½
 BD =
½
½
½
½
½
9  21
 BD = 3 21 cm.
[Taking square roots]
 The lengths of seg AC and seg BD are 12 cm and 3 21 cm
respectively.
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PAPER - 2
(Rough Figure)
(iii)
N
N
S
O
S
••70º
50º ××
6.7 cm
70º
50º
6.7 cm
G
G
½ mark for rough figure
½ mark for drawing SGN
1 mark for angle bisectors
1 mark for the incircle
(iv)





Let, A  (– 3, 11)
B  (6, 2)
C (k, 4)
Points A, B and C
Slope of line AB
y 2 – y1
x 2 – x1
2 – 11
6 – (– 3)
–9
63
–9
9



are
=
=
=
=
=

–1
=




– (k – 6)
–k+6
–k
–k
=
=
=
=
(x1, y1)
(x2, y2)
(x3, y3)
collinear
Slope of line BC
y3 – y2
x3 – x2
4–2
k – 6
2
k –6
2
k –6
2
k –6
2
2
2–6
–4
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½
½
½
7 / MT

k
PAPER - 2
= 4
 The value of k is 4
(v)
½
Mark point X as shown in the figure
A
ABCD is a square
[Given]
side = 8 cm
Radius (r) = side of a square
 r = 8 cm
Measure of arc () = 90º
[Angle of a square] D
Area of the segment AXC
8 cm
sin  
 

r2 
2 
 360
=
 3.14  90 sin 90 

82 
2 
 360
=
1.57 1 
 
64 
2
 2
=
=
=
=
=
1.57  1
64 

2

64  0.57
2
36.48
cm2
2
2 × Area of segment AXC
36.48
2×
2
36.48 cm2
 Area of shaded region is 36.48 cm2.
A.4.
(i)
X
=
=
Area of shaded region
B
C
½
½
½
½
½
½
Solve ANY TWO of the following :
Take points E and F as shown in the figure.
m PFC = 90º
......(i)
P
[Radius is perpendicular to the tangent]
½
A
B
E
line CD || chord AB
[Given]
•
•
 On transversal PF,
F
D
C
PFC  PEA ......(ii) [Converse of corresponding angle test] ½
 m PEA = 90º ......(iii) [From (i) and (ii)]
[From (iii)]
seg PE  chord AB
1
 AE = × AB
[The perpendicular drawn from the centre ½
2
of the circle to the chord bisects the chord]
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PAPER - 2
1
 12
[Given]
2
AE = 6 units
Let the radius of the circle be 2x units
PA = PF = 2x units
[Radii of same circle]
1
PE = PF
[ E is the midpoint of seg PF]
2
1
 2x
PE =
2
PE = x units
In PEA,
m PEA = 90º
[From (iii)]
PA2 = PE2 + AE2
[By Pythagoras theorem]
(2x)2 = x2 + 62
4x2 – x2 = 36
3x2 = 36
x2 = 12
x= 4×3
x= 2 3
[Taking square roots]
PA = 2 × 2 3 = 4 3 units
 AE =













 Radius of the circle is 4 3 units.
(ii)





½
½
½
½
½
Inclination of line PA is 45º
Y
Slope of line PA = tan  = tan 45º = 1
P (3, 4)
Line PA passes through point P (3, 4)
The equation of the line by slope point form is,
(y – y1) = m (x – x1)
60º
A 45º
(y – 4) = 1 (x – 3)
X
X
B
y–4 = x–3
x–y–3+4 = 0
x–y+1 = 0
Y
 The equation of line PA is x – y + 1 = 0
½
½
½
½
Inclination of line PB is 60º
 Slope of line PB = tan  = tan 60º = 3
Line PB passes through point P (3, 4)
 The equation of the line by slope point form is,

(y – y1)
= m (x – x1)

(y – 4)
=
3 (x – 3)

y–4
=
3x – 3 3
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9 / MT

3x – y  4 – 3 3
 The equation of PB is
(iii)
PAPER - 2
= 0
½
3x – y  4 – 3 3 = 0
Let A be the point of observer
Let B and C represent the positions of Vashi bridge and Worli
sea-link respectively.
AD represents the height of a plane from the ground
AD = 5500 3 m
EAB and FAC are angles of depression
m EAB = m ABD = 60º
[Converse of alternate angle test]
m FAC = m ACD = 30º
In right angled ADB,
E
F
A
tan 60º =
AD
[By definition]
BD
5500 3

=
3
BD
 BD
= 5500 m
In right angled ADC,





A.5.
(i)
B
AD
DC
1
3
DC
DC
BC
BC
BC
5000 3
DC
5500 × 3
16500 m
BD + DC
5500 + 16500
22000 m
=
=
=
=
=
30º
60º
½
550 3 m
tan 300 =
=
1
60º
D
30º
C
(½ mark for figure)
½
[By definition]
½
[B - D - C]
½
 Distance between Vashi bridge and Worli sea-link is 22 km.
½
Solve ANY TWO of the following :
Given : In ABC,
(i) line DE || side BC
(ii) Line DE intersects sides AB and AC
at points D and E respectively.
½
AD
AE
=
DB
EC
Construction : Draw seg BE and seg CD.
(½ mark for figure)
A
D
E
To Prove :
B
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C
½
10 / MT
PAPER - 2
Proof : ADE and BDE have a common vertex E and
their bases AD and BD lie on the same line AB.
 Their heights are equal
A (ADE)
AD
 A (BDE) =
.......(i) [Triangles having equal heights] ½
DB
ADE and CDE have a common vertex D and
their bases AE and EC lie on the same line AC.
 Their heights are equal.
A (ADE)
AE
 A (CDE) =
.....(ii) [Triangles having equal heights] ½
CE
line DE || side BC
[Given]
BDE and CDE are between the same two parallel lines
½
DE and BC.
 Their heights are equal.
Also, they have same base DE.
½
 A(BDE)
= A(CDE)
[ Areas of two triangles having equal
.....(iii) bases and equal heights are equal ]
A (ADE)
A (ADE)
 A (BDE) = A (CDE) .....(iv) [From (i), (ii) and (iii)]

AE
AD
=
DB
EC
[From (i), (ii) and (iv)]
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½
½
11 / MT - w
(ii)
PAPER - 2
P
(Rough Figure) P
A
A
3 cm
3 cm
B
B
C
4 cm
4 cm
C
R
R
B1
B2
B3
B4
B5
B6
1
1
1
1
1
(iii)
mark
mark
mark
mark
mark
for
for
for
for
for
B7
ABC
constructing 7 congruent parts
constructing CB5B  RB7B
constructing ACB  PRB
required PRB
Height of the cylindrical container (h)
Its radius (r)
Volume of cylindrical container
But, volume of ink filled
in the cylindrical container
=
=
=
=
=
14cm
6 cm
 r 2h
 × 6 × 6 × 14
504 cm3
=
91% of 504 
=

Length of ball pen refill (h1)
its inner diameter
Its radius (r1)
=
=
=
=
Volume of the refill
=
91
× 504 cm 3
100
12m
2 mm
1 mm
1
cm
10
2
r 1 h 1
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½
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½
12 / MT - w
PAPER - 2
1
1
×
× 12
10 10
12 
cm3
=
100
12 

But, volume of ink filled
= 84% of
100
84 12 
×
=
cm 3
100 100
Number of refills that can be filled with ink
Volume of ink filled in the cylindrical container
=
Volume of ink filled in each refill
91 × 504 
100
84 ×12 
=
100 ×100
91 × 504  100 ×100
×
=
100
84 ×12 
= 4550
=
×
 Number of refills that can be filled with this ink is 4550.

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