Seat No.
2014 ___ ___ 1100
MT -
MATHEMATICS (71) GEOMETRY - PRELIM II - PAPER - 5 (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)
O is the centre of the circle. AB is the longest chord of the circle. If
AB = 8.6 cm, what is the radius of the circle ?
(ii)
A cylinder and a cone have equal radii and equal heights. If the
volume of the cylinder is 300 cm3, what is the volume of the cone ?
(iii)
If sec  = 2, what is the value of tan2  ?
(iv)
What is the equation of a line whose slope is – 2 and y-intercept is 3 ?
(v)
The dimensions of a cuboid are 5 cm, 4 cm and 3 cm. Find its
volume.
(vi)
If x co-ordinate of point A is negative and y co-ordinate is positive,
then in which quadrant point A lies ?
Q.2. Solve ANY FOUR of the following :
5
8
(i)
A ladder 10 m long reaches a window 8 m above the ground. Find the
distance of the foot of the ladder from the base of the wall.
(ii)
Sides of a triangle are 40, 20 and 30. Determine whether they are
sides of a right angled triangle.
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PAPER - 5
Q
•
B
(iii)
In the adjoining figure,
if m (arc APC) = 60º
and m BAC = 80º
Find (a) ABC (b) m (arc BQC).
•P
A
(iv)
C
80º
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)
1
1
If tan A + tan A = 2, show that tan2 A + tan 2 A = 2.
(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)
9
In the adjoining figure,
F
ABCD is a square. The BCE on
side BC and ACF on the diagonal AC
are similar to each other. Then show
1
A (ΔACF)
that A (BCE) =
2
C
D
E
A
(ii)
Find the radius of the circle passing
through the vertices of a right
angled triangle when lengths of
perpendicular sides are 6 and 8.
Q
P
B
6
•Y
8
R
(iii)
Construct the incircle of SRN, such that RN = 5.9 cm, RS = 4.9 cm,
R = 95º.
(iv)
Find the value of k, if (– 3, 11), (6, 2) and (k, 4) are collinear points.
(v)
A building has 8 right cylindrical pillars whose cross sectional diameter
is 1 m and whose height is 4.2 m. Find the expenditure to paint those
pillars at the rate of Rs. 24 per m2.
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Q.4.
(i)
PAPER - 5
Solve ANY TWO of the following :
8
In the adjoining figure,
BC is a diameter of the circle
with centre M. PA is a tangent
at A from P which is a point on
line BC. AD  BC.
Prove that DP2 = BP × CP – BD × CD.
A
B
M
D
C
P
(ii)
Lines x = 5 and y = 4 form a rectangle with co-ordinate axes. Find the
equation of the diagonals.
(iii)
If tan  + sin = m and tan – sin  = n, show that m2 – n2 = 4 mn .
Q.5. Solve ANY TWO of the following :
10
(i)
Prove : In a triangle, the angle bisector divides the side opposite to the
angle in the ratio of the remaining sides.
(ii)
SHR ~ SVU, In SHR, SH = 4.5 cm, HR = 5.2 cm, SR = 5.8 cm and
SH
3
=
; construct SVU.
SV
5
(iii)
The diameter of the base of metallic cone is 2 cm and height is 10 cm.
900 such cones are molten to form 1 right circular cylinder whose radius
is 10 cm. Find total surface area of the right circular cylinder so formed.
(Given  = 3.14)
Best Of Luck

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Seat No.
2014 ___ ___ 1100
MT -
MATHEMATICS (71) GEOMETRY - PRELIM II - PAPER - 5 (E)
Time : 2 Hours
A.1.
(i)
Prelim - II Model Answer Paper
Attempt ANY FIVE of the following :
AB is the longest chord of the circle
But, diameter is the longest chord of the circle
 seg AB is the diameter of the circle
Diameter = 8.6 cm
Max. Marks : 40
[Given]
 Radius = 4.3 cm
(ii)
1
A cylinder and cone have equal height and equal radii
1
 volume of cylinder
 Volume of cone =
3
1
 300
=
3
= 100 cm3
 Volume of the cone is 100 cm3.
(iii)
(iv)
sec  = 2
But, sec 60 = 2
 sec  = sec 60
  = 60º
½
½
Slope (m) = – 2
y intercept (c) = 3
 Equation of the line by slope-intercept form is y = mx + c

y
= – 2x + 3
Length of a cuboid (l)
Its breadth (b)
Its height (h)
Volume of a cuboid
½
[Given]
 2x + y – 3 = 0
(v)
½
=
=
=
=
=
=
 Volume of the cuboid is
½
½
5 cm
4 cm
3 cm
l×b×h
5×4×3
60 cm3
60 cm3.
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(vi)
A.2.
(i)
(ii)
PAPER - 5
If x co-ordinate of point A is negative and y co-ordinate is positive.
Then, point A lies in the II quadrant.
Solve ANY FOUR of the following :
In the adjoining figure,
A
seg AB represents the wall
seg AC represents the ladder
seg BC represents the distance of the
foot of the ladder from the base of the wall
AC = 10 m
C
B
AB = 8 m
(½ mark for figure)
In ABC,
[Given]
m ABC = 90º
 AC 2
= AB2 + BC2
[By Pythagoras theorem]
 (10) 2 = (8)2 + BC2
 100
= 64 + BC2
 BC 2 = 100 – 64
 BC 2 = 36
 BC
= 6m
[Taking square roots]
 The distance of the foot of the ladder from the base of the wall
is 6 m.
1
½
½
½
½
(40) 2
= 1600
......(i)
(20) + (30)2 = 400 + 900
= 1300
......(ii)

(40) 2
 (20)2 + (30)2
[From (i) and (ii)]
 The given sides do not form a
[By Converse of
right angled triangle.
Pythagoras theorem]
2
(iii)
(a) m ABC
 m ABC
 m ABC
1
m(arc APC)
2
[Inscribed angle theorem]
1
=
× 60
2
=
=
30º
=

=
80
 m(arc BQC)
 m(arc BQC)
=
=
½
Q
•
B
½
C
80º
A
(b) m BAC
½
½
•P
1
m(arc BQC) [Inscribed angle theorem]
2
1
m(arc BQC)
2
80 × 2
160º
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
(iv)
PAPER - 5

(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
1
tan A + tan A
(v)
= 2
2

1 

 tan A  tan A 
1
1
 tan2 A + 2 tan A . tan A  tan2 A
1

tan2 A + 2 + tan2 A
1

tan2 A + tan2 A
1

tan2 A + tan 2 A
(vi)
x
=

1
sin 
=

cosec 
=
y
=
= 4
[Squaring both sides] ½
= 4
½
= 4
= 4–2
½
= 2
½
a sin 
a
x
a
x
b tan 
......(i)

1 
 cos ec   sin  


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
1
tan 
=

cot 
=
b
y
b
y
PAPER - 5
.....(ii)
We know,
1 + cot2  = cosec2 
2
 cosec  – cot2  = 1
2


A.3.
(i)
 b
 a
  –  
x
 y
2
a
b2
–
x2
y2

1 
 cot   ta n  


½
½
2
= 1
[From (i) and (ii)]
½
= 1
Solve ANY THREE of the following :
F
C
D
E








(ii)
BCE ~ ACF
A (BCE)
BC2
A (ACF) = AC2
ABCD is a square
AB = BC = CD = AD
In ABC,
m ABC
= 90º
AC 2 = AB2 + BC2
AC 2 = BC2 + BC2
AC 2 = 2BC2
A (BCE)
BC2
=
A (ACF)
2BC2
A (BCE)
1
A (ACF) = 2
1
A (BCE)
=
A (ACF)
2
[Given]
A
B
......(i) [Areas of similar triangles]
......(ii)
½
[Given]
[Sides of a square]
½
[Angle of a square]
[By Pythagoras theorem]
[From (i)]
½
½
[From (i) and (iii)]
½
.....(iii)
In PQR,
[Given]
m PQR = 90º
 PR²
= PQ² + QR²
[By Pythagoras theorem]
 PR²
= 6² + 8²
 PR²
= 36 + 64
 PR²
= 100
½
P
Q
6
•Y ½
8
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






PR
m PQR
= 10 units
= 90º
m PQR
=
1
m(arc PYR)
2
PAPER - 5
[Taking square roots]
[Given]
½
[Inscribed angle theorem]
½
1
m(arc PYR)
2
m (arc PYR)= 180º
½
arc PYR is a semicircle
seg PR is the diameter.
Diameter = 10 units.
½
Radius = 5 units
[ Radius is half of the diameter]
90º
=
 Radius of the circle is 5 units.
(iii)
½
(Rough Figure) S
S
4.9 cm
95º
R
5.9 cm
4.9 cm
O
R
×
×
•• 95º
5.9 cm
N
½ mark for rough figure
½ mark for drawing SRN
1 mark for drawing the angle bisectors
1 mark for drawing the incircle
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(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 =
 (x1, y1)
 (x2, y2)
 (x3, y3)
are collinear
Slope of line BC
y3 – y2
x3 – x2

2 – 11
6 – (– 3)
4–2
= k – 6

–9
63
2
= k –6

–9
9
2
= k –6







(v)
PAPER - 5
½
½
½
½
2
= k –6
– (k – 6) = 2
–k+6
= 2
–k
= 2–6
–k
= –4
k
= 4
The value of k is 4
½
–1
½
Diameter of a pillar = 1 m

Its radius (r)
Its height (h)
Curved surface area of a pillar
=
 1
  m
2
= 4.2 m
= 2rh
22
1
×
× 4.2
7
2
= 13.2 m2
Curved surface area of 8 pillars = 8 × 13.2
= 105.6 m2
Rate of painting
= Rs. 24 per m2
 Total expenditure
=
Area to be painted × Rate of painting
=
105.6 × 24
=
2534.40
½
½
= 2×
 Total expenditure to paint the pillars is Rs. 2534.40.
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(ii)
Solve ANY TWO of the following :
Construction : Draw seg AB and seg AC
Proof : In ADP,
A
[Given]
m ADP = 90º
 AP2 = AD2 + DP2
[By Pythagoras theorem]
P
B
D
C
2
2
M
 DP = AP – AD2
......(i)
Line AP is a tangent to the circle at
point A and line PCB is a secant to the
circle at points C and B
 AP2 = BP × CP
......(ii)
In BAC,
[Angle subtended by semicircle]
m BAC = 90º
seg AD  hypotenuse BC
 AD2 = BD × CD
......(iii) [Property of geometric mean]
 DP2 = BP × CP – BD × CD
[From (i), (ii) and (iii)]
Let line y = 4 intersect the Y-axis at point A
 A  (0, 4)
Let line x = 5 intersect the X-axis at point C
 C  (5, 0)
Let line y = 4 and x = 5 intersect at point B
 B  (5, 4)
X
The origin O  (0, 0)
ABCO is a rectangle
seg AC and seg BO are the diagonals
Equation of line AC by two point form,


x – x1
x1 – x 2
y – y1
= y – y
1
2
x –0
0–5
y –4
= 4–0
x
–5
=
y=4
A
½
1
1
1
½
½
Y
B
½
x=5
A.4.
(i)
PAPER - 5
O
C
Y
X
½
½
½
y –4
4

4x = – 5 (y – 4)

– 5y + 20 = 4x
 4x + 5y – 20 = 0
 Equation of diagonal AC is 4x + 5y – 20 = 0.
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PAPER - 5
Equation of line BD by two point from,

x – x1
x1 – x 2



½
x –5
y –4
=
5–0
4–0


y – y1
= y – y
1
2
x –5
5
4 (x – 5)
5y – 20
4x – 5y
y –4
4
= 5 (y – 4)
= 4x – 20
= 0
=
 Equation of diagonal BD is 4x – 5y = 0.
(iii)
tan  + sin  = m
tan – sin  = n
m2 – n2 = (tan  + sin )2 – (tan  – sin )2
= tan2  + 2 tan .sin  + sin2  – [tan2  – 2 tan  sin +
sin2 ]
= tan2  + 2 tan  .sin  + sin2  – tan2  + 2 tan .sin  –
sin2 
= 4 tan .sin 
.......(i)
4 mn =
4
tan   sin   tan 
½
½
½
– sin  
=
4 tan2  – sin2 
=
4
=
 1

4 sin2  
– 1
2
 cos 

½
=
4 sin2  sec 2  – 1

½
=
sin2 
– sin2 
cos 2 
2

2
4 ta n  . sin 
= 4 × sin  × tan 
From (i) and (ii),
m2 – n2
½
=
1  tan2   sec2  


2
2
 tan   sec  – 1
½
.....(ii)
½
4 mn
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A.5.
PAPER - 5
Solve ANY TWO of the following :
E
(i)
Given : In ABC, ray AD is the bisector
of BAC such that B - D - C.
To Prove :
BD
AB
=
DC
AC
Construction : Draw a line passing
A
xx
B
through C, parallel to line AD and
C
D
(½ mark for figure)
½
½
intersecting line BA at point E, B - A - E.
Proof : In BEC,
line AD || side CE
[Construction]
BD
AB
=
.........(i) [By B.P.T.]
DC
AE
line CE || line AD
[Construction]
 On transversal BE,
BAD  AEC
........(ii) [Converse of corresponding
angles test]
Also, On transversal AC,
DAC ACE
........(iii) [Converse of alternate
angles test]
But, BAD  DAC ........(iv) [ ray AD bisects BAC]
In AEC,
AEC  ACE
[From (ii), (iii) and (iv)]
 seg AC  seg AE
[Converse of Isosceles
triangle theorem]
 AC = AE
........(v)

(ii)
BD
AB
=
DC
AC
[From (i) and (v)]
(Rough Figure) U
R
5.
8c
m

S
5.
2
4.5 cm
cm
H
V
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10 / MT
PAPER - 5
1 mark for SHR
U
1 mark for constructing 5 congruent parts
1 mark for constructing VS5S  HS3S
1 mark for constructing UVS  RHS
1 mark for required SVU
R
5.8
cm
cm
5.2
×
S
4.5 cm
×
H
V
S1
S2
•
S3
S4
•
S5
(iii)
Diameter of the base of metallic cone = 2 cm
2
 Its radius (r) =
= 1 cm
2
Its height (h) = 10 cm
1 2
r h
Volume of a metallic cone =
3
1
=
× × 1 × 1 × 10
3
10
=
cm 3
3
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 Volume of 900 metallic cones
=
=
PAPER - 5
10
3
3000 cm3
900 
½
900 cones are melted to form a right circular cylinder

Volume of a cylinder
=
3000 
For a cylinder, Radius (r2)
=
10 cm and height be h2
Volume of a cylinder
=
r12 h1

3000
=  × 10 × 10 h2

h1
= 30 cm
½
½
½
Total surface area of cylinder
=
2r1 (r1 + h1)
=
2 × 3.14 × 10 (10 + 30)
=
6.28 × 10 × 40
=
62.8 × 40
=
2512 cm2
 Total surface area of the right circular cylinder is 2512 cm2.

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½
½
½