Seat No.
2014 ___ ___ 1100
MT -
MATHEMATICS (71) GEOMETRY - PRELIM II - PAPER - 6 (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
A
Line AB is a tangent and line BCD
is a secant. If AB = 6 units, BC = 4 units,
find BD.
D
B
C
(ii)
If the radius is 2 cm and length of corresponding arc is 3.14 cm, find
the area of a sector.
(iii)
What is the directed angle, whose terminal arm lies along the coordinate
axes, called ?
(iv)
A line has the equation y = 3x – 2. State its y-intercept.
(v)
Find the area of a circle with radius 7 cm.
(vi)
If sec  =
2
, find the value of  ?
3
Q.2. Solve ANY FOUR of the following :
(i)
Find the side of square whose diagonal is 16 2 cm .
(ii)
In PQR, seg PM is the median. If PM = 9 and PQ2 + PR2 = 290. Find QR.
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(iii)
In the adjoining figure,
point P is centre of the circle
and line AB is the tangent to the
circle at T. The radius of the circle
is 6 cm. Find PB ifTPB = 60º.
PAPER - 6
P
A
T
B
(iv)
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.
(v)
Prove : sec2  + cosec2  = sec2  . cosec2 
(vi)
If the angle  = – 60º, find the value of sin , cos , sec  and tan .
Q.3. Solve ANY THREE of the following :
9
(i)
ABCD is a trapezium in which AB || DC and its diagonals intersect
AO CO
=
.
each other at the point O. Show that
BO DO
(ii)
Two circles which are not congruent touch externally. The sum of their
areas is 130cm2 and the distance between their centers is 14 cm. Find
radii of circles.
(iii)
Construct the incircle of RST in which RS = 6 cm, ST = 7 cm and
RT = 6.5 cm.
(iv)
Find the equation of the line which passes through (2, 7) and whose
y-intercept is 3.
(v)
The curved surface area of the frustum of a cone is 180 sq. cm and the
perimeters of its circular bases are 18 cm and 6 cm respectively. Find
the slant height of the frustum of a cone.
Q.4.
(i)
Solve ANY TWO of the following :
8
A
In a right angled triangle ABC, ACB = 90º a
circle is inscribed in the triangle with radius r.
a, b, c are the lengths of the sides BC, AC and
AB respectively. Prove that 2r = a + b – c.
Q
P
C
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B
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PAPER - 6
(ii)
Show that (– 2, 1), (0, 3), (2, 1) and (0, – 1) are the vertices of a parallelogram.
(iii)
Two poles of height 18 metres and 7 metres are erected on the ground. A
wire of length 22 metres tied to the top of the poles. Find the angle made
by the wire with the horizontal.
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)
A cylinder of radius 12 cm contains water upto depth of 20 cm. A spherical
iron ball is dropped into the cylinder and thus water level is raised by
6.75 cm. what is the radius of the ball ?
Best Of Luck

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Seat No.
2014 ___ ___ 1100
MT -
MATHEMATICS (71) GEOMETRY - PRELIM II - PAPER - 6 (E)
Time : 2 Hours
A.1.
Prelim - II Model Answer Paper
Max. Marks : 40
Attempt ANY FIVE of the following :
(i)




Line BCD is a secant intersecting
the circle at points C and D and
line BA is a tangent at A
AB2 = BC × BD
62 = 4 × BD
36 = 4 × BD
36
B
BD =
4
½
A
D
C
 BD = 9 units
(ii)
½
Radius (r)
Length of arc (l)
=
=
Area of sector
=
=
=
2 cm
3.14 cm
r
l
2
2
3.14 ×
2
3.14 cm2
 The area of a sector is 3.14 cm2.
½
½
1
(iii)
If the terminal arm of a directed angle lies along the co-ordinate
axes, then it is called a quadrantal angle.
(iv)
Equation of the line is y = 3x – 2
Comparing the given equation with slope-intercept form
y = mx + c, c = – 2
½
 y intercept of the line is – 2.
½
Radius of circle (r)
 Area of the circle
½
(v)
=
=
7 cm
r 2
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=
=
PAPER - 6
22
77
7
154 cm2
 The area of a circle is 154 cm2.
2
3
sec  =
(vi)
½
[Given]
But, sec 30º =
 sec  = sec 30º
2
3
½
  = 30º
A.2.
½
Solve ANY FOUR of the following :
ABCD is a square.
(i)
x
A
AC = 16 2 cm
16 2 cm
To find : Side of a square
x
ABCD is a square [Given]
Let the sides of the square be x cm
B
In ABC,
m ABC = 90º
[Angle of a square] x

AC 2 = AB2 + BC2
[By Pythagoras theorem]
16 2 
2


256 × 2

x2


x
x
= x2 + x 2
= 2x2
256  2
=
2
= 256
= 16
2
D
x
C
½
½
[Taking square roots]
 The side of a square is 16 cm.
(ii)






In PQR,
seg PM is
PQ2 + PR2
290
290
290
290 – 162
128
the
=
=
=
=
=
=

QM
2
=


QM
QM
=
=
2
median
2PM2 + 2QM2
2 (9)2 + 2QM2
2 (81) + 2QM2
162 + 2QM2
2QM2
2QM2
128
2
64
8 units
½
½
[Given]
[By Appollonius theorem]
[Given]
½
P
9
Q
M
[Taking square roots]
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
QM
=

8
=


8×2
QR
=
=
(iii)





1
QR
2
1
QR
2
QR
16 units
=
½
[ M is midpoint of side QR]
½
In PTB,
m TPB = 60º
[Given]
m PTB = 90º
[Radius is perpendicular to tangent]
m PBT = 30º
[Remaining angle]
PTB is a 30º - 60º - 90º triangle
By 30º - 60º - 90º triangle theorem
A
1
PB
[Side opposite to 30º]
PT =
2
1
6
=
PB
[Given]
2
PB = 6 × 2
 PB
P
1
T
B
½
½
12 cm
L
(iv)
PAPER - 6
(Rough Figure)
•
L
•
• M
N
•
A
•
N
A
M
½ mark for rough figure
1 mark for drawing NMA  NLM
½ mark for tangent at M.
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(v)
(vi)
sec2  + cosec2 
1
1

1
1 
= cos2   sin2 
 sec   cos  , cos ec   sin  


sin2  + cos 2 
= cos 2  . sin2 
1
= cos2  . sin2 
[ sin2  + cos2  = 1]
= sec2  . cosec2 
= R.H.S.
 sec2  + cosec2  = sec2  . cosec2 
L.H.S. =

=
sin (– )
=
 sin (– 60) =
– 60º
– sin 
– sin 60
 sin (– 60) =
–
sec (– ) =
 sec (– 60) =
sec 
sec 60
 sec (– 60) =
2
cos (– )
=
 cos (– 60) =
cos 
cos 60
(i)
3
2
tan (– ) =
 tan (– 60) =
1
2
– tan 
– tan 60
 tan (– 60) =
– 3
 cos (– 60) =
A.3.
PAPER - 6
½
½
½
½
½
½
½
Solve ANY THREE of the following :
ABCD is a trapezium
side AB || side DC
 On transversal AC,
BAC  DCA
 BAO  DCO
......(i)
In AOB and COD,
BAO  DCO
AOB  COD
 AOB ~ COD
½
B
A
O
[Given]
D
C
[Converse of alternate angles test]
[ A - O - C]
1
[From (i)]
[Vertically opposite angles]
[By AA test of similarity]
1
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

(ii)














AO
BO
=
CO
DO
AO CO
=
BO DO
PAPER - 6
[c.s.s.t.]
[By Alternendo]
Let the radius of first circle be r1 and that of second circle be r2
Circles are touching externally
r1 + r2 = 14 cm
r 2 = 14 – r1
........(i)
According to given information
A (I Circle) + A (II Circle) = 130  cm2
r12 + r22 = 130 
 (r12 + r22) = 130 
r12 + r22 = 130
2
r1 + (14 – r1)2 = 130
[From (i)]
2
2
r1 + 196 – 28r1 + r1
= 130
2r12 – 28r1 + 196 – 130 = 0
2r12 – 28r1 + 66 = 0
2 (r12 – 14r1 + 33) = 0
r12 – 14r1 + 33 = 0
2
r1 – 11r1 – 3r1 + 33 = 0
r1 (r1 – 11) – 3 (r1 – 11) = 0
(r1 – 3) (r1 – 11) = 0
r1 – 3 = 0
or
r1 – 11 = 0
r 1 = 3 cm
or
r 1 = 11 cm
If r1 = 3
If r1 = 11
then,r 2 = 14 – 3
then, r2 = 14 – 11
r 2 = 11 cm
r 2 = 3 cm
 The radii of the circles is 11 cm and 3 cm.
(iii)
(Rough Figure)
R
6.5 cm
6 cm
S
7 cm
T
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½
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½
½
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PAPER - 6
R
6 cm
S
6.5 cm
O
×
×
•
•
7 cm
T
½ mark for rough figure
½ mark for drawing DRST
1 mark for drawing the angle bisectors
1 mark for drawing the incircle
(iv)









Let A  (2, 7)
The y intercept of the line is 3
The line intersects the y-axis at point (0, 3)
Let B  (0, 3)
The line passes through point A and B
The equation of the line AB
By two point form
x – x1
y – y1
=
x1 – x 2
y1 – y 2
x –2
y –7
=
2–0
7–3
x –2
y –7
=
2
4
4 (x – 2) = 2 (y – 7)
4x – 8 = 2y – 14
4x – 2y – 8 + 14 = 0
4x – 2y + 6 = 0
2x – y + 3 = 0
[Dividing throughout by 2]
 The required equation of the line is 2x – y + 3 = 0
(v)
Curved surface area of the frustum
Perimeters of circular bases are 18

2r 1
= 18
2r 2
= 6
Adding (i) and (ii), we get
of a cone = 180 cm2
cm and 6 cm
........(i)
........(ii)
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


(ii)
½
2r1 + 2r2
2 (r1 + r2)
=
=
=
=
½
(r1 + r2) l
½
(r1 + r2) l
12 × l [From (iii)]
15 cm
 Slant height of the frustum of a cone is 15 cm.
½



A.4.
(i)
= 18 + 6
= 24
24
 (r1 + r2)
=
2
 (r1 + r2)
= 12
.......(iii)
Curved surface area of the frustum of a cone
180
180
l
PAPER - 6
Solve ANY TWO of the following :
Let the centre of the inscribed circle be ‘O’
Let AP = AQ = x
........(i) [The lengths of the two tangent
CP = CR = y
.......(ii) segments to a circle drawn from 1
BR = BQ = z
......(iii) an external point are equal]
a + b – c = BC + AC – AB
A
 a + b – c = CR + RB + AP + PC – (AQ + QB)
[B - R - C, A - P - C , A - Q - B]
 a + b – c = y + z + x + y – (x + z)
[From (i), (ii) and (iii)]
Q
 a+b–c = y+z+x+y–x–z
O
 a + b – c = 2y
P
 a + b – c = 2y
 a + b – c = 2CP
........(iv) [From (ii)]
1
C
B
R
In  PCRO
m OPC = m ORC = 90º [Radius is perpendicular to tangent]
m PCR = 90º
[Given]
 m POR = 90º
[Remaining angle]
 PCRO is a rectangle
[By definition]
1

CP = OR
........(v) [Opposite sides of a rectangle]
½
 a + b – c = 2 OR
[From (iv) and (v)]
 a + b – c = 2r
½
Let, P  (– 2, 1), Q  (0, 3), R (2, 1), S (0, – 1)
Slope of a line
Slope of line PQ
=
y 2 – y1
x 2 – x1
=
3 –1
0 – (– 2)

2
02

2
2
½
P (– 2, 1)
Q (0, 3)
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S (0, –1)
R (2, 1)
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 Slope of line PQ
Slope of line RS
PAPER - 6
½
= 1
–1 – 1
= 0–2
–2
–2
 Slope of line RS
= 1
 Slope of line PQ
= Slope of line RS
 line PQ || line RS
.......(i)
=
Slope of line QR
=
½
½
1–3
2–0
–2
2
= –1
=
 Slope of line QR
Slope of line PS
=
–1 – 1
0 – (– 2)
=
–2
02
½
–2
2
Slope of line PS
= –1
Slope of line QR
= Slope of line PS
line QR || line PS
......(ii)
In PQRS,
side PQ || side RS
[From (i)]
side QR || side PS
[From (ii)]
PQRS is a parallelogram
[By definition]
The points (–2, 1), (0, 3), (2, 1) and (0, – 1) are the vertices of
parallelogram.
=





(iii)




seg AB and CD represents two poles. AB = 18 m, CD = 7 m
seg AC represent the length of the wire. AC = 22 m
EBDC is a rectangle
A
(½ mark for figure)
EB = CD = 7 m
22
m
[Opposite sides of rectangle]
AB = AE + EB
[ A - E - B]
18 m E
C
18 = AE + 7
18 – 7 = AE
7m
AE = 11 m
B
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D
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



A.5.
(i)
PAPER - 6
In right angled AEC,
AE
sin C
=
[By definition]
AC
11
sin C
=
22
1
sin C
=
......(i)
2
But,
1
......(ii)
sin 30º =
2
sin C
= sin 30º
C
= 30º
½
½
½
½
 The angle made by the wire with horizontal is 30º.
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.
To Prove :
Proof :
A
D
B
E
(½ mark for figure)
C
½
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)
DB
.......(i) [Triangles having equal heights]
½
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
=
.......(ii) [Triangles having equal heights]
A (CDE)
CE
line DE || side BC
[Given]
BDE and CDE are between the same two parallel lines
DE and BC.
 Their heights are equal.
Also, they have same base DE.

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 A(BDE) = A(CDE) ......(iii)
PAPER - 6
[ Areas of two triangles having equal
bases and equal heights are equal ]
½

A (ADE)
A (ADE)
=
A (BDE)
A (CDE) ......(iv) [From (i), (ii) and (iii)]
½

AE
AD
=
DB
EC
½
(ii)
[From (i), (ii) and (iv)]
(Rough Figure) P
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
Radius of the cylinder (r) = 12 cm
A spherical iron ball is dropped into the cylinder
and the water level rises by 6.75 cm
 Volume of water displaced = volume of the iron ball
Height of the raised water level (h) = 6.75 m
Volume of water displaced = r 2 h
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
=  × 12 × 12 × 6.75 cm3
Volume of iron ball =  × 12 × 12 × 6.75 cm3
But, Volume of iron ball =





PAPER - 6
 × 12 × 12 × 6.75 =
12 × 12 × 6.75 × 3
4
r3
r3
r3
½
4 3
r
3
4
×  × r3
3
6.75 cm
20 cm
= r3
= 3 × 12 × 6.75 × 3
= 3 × 3 × 3 × 4 × 6.75
= 3 × 3 × 3 × 27

r =


r = 3×3
r = 9
3
3×3×3×3×3×3
½
½
½
½
[Taking cube roots]
 Radius of the iron ball is 9 cm.

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½