Set - A
Q.P. SET CODE
Seat No.
2013 ___ ___ 1100
- MT - w
Time : 2 Hours
MATHEMATICS (71) GEOMETRY - PAPER A (E)
(Pages 3)
Max. Marks : 40
Q.1. Solve the following : (Any 5)
(i)
In the adjoining figure,
seg BE  seg AB and
E
5
B
seg BA  seg AD.
If BE = 6 and AD = 9
find A (ΔABE)
A (ΔABD)
A
D
A
(ii)
B
What is the relation between
ABC and ADC of cyclic ABCD ?
D
C
(iii)
Find where the angle lies if the terminal arm passes through (5, – 7)
(iv)
Find the slope of a line whose inclination is 45º.
(v)
A (PQR) = 24 cm2, the height QS is 8 cm. What is the length of
side PR ?
(vi)
Volume of a cube is 1000 cm3, find the length of its side.
Q.2. Solve the following : (Any 4)
(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.
8
Set - A
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(ii)
D
In the adjoining figure,
seg AB and seg AD are chords
of the circle. C be a point on
tangent to the circle at point A.
If m (arc APB) = 80º and BAD = 30º,
then find (i) BAC (ii) m (arc BQD)
Q
B
P
A
C
(iii)
Eliminate , if x = a sec , y = b tan 
(iv)
If (– 2, – 3) is a point on the line 2y = mx + 5, find m.
(v)
Find the trigonometric ratios in standard position whose terminal arm
passes through the points (4, 3)
(vi)
Write the equation of a line passing through the origin and the point (– 3, 5)
Q.3.
(i)
Solve the following : (Any 3)
9
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)
If the chord AB of a circle is parallel to the tangent at C, then prove that
AC = BC.
(iii)
Construct tangents to the circle from point B with radius 3.5 cm and
centre A. Point B is at a distance 7.3 cm from the centre.
(iv)
(v)
2 1 1

4

If the points  ,  ,  , k  and  , 0  are collinear then find the value of k.
5
3
2
5

 



Construct LEM such that, LE = 6cm, LM = 7.5 cm, LEM = 90º and
draw its circumcircle.
Q.4. Solve the following : (Any 2)
8
(i)
Prove : The lengths of the two tangent segments to a circle drawn from
an external point are equal.
(ii)
SHR ~ SVU, In SHR, SH = 4.5 cm, HR = 5.2 cm, SR = 5.8 cm and
SH
3
=
; construct SVU.
SV
5
Set - A
3 / MT - w
A piece of cheese is cut in the shape of
the sector of a circle of radius 6 cm. The
thickness of the cheese is 7 cm. Find
(i) The curved surface area of the cheese.
(ii) The volume of the cheese piece.
Q.5. Solve the following : (Any 2)
6 cm
60º
7 cm
(iii)
10
(i)
Prove : In a triangle, the angle bisector divides the side opposite to the
angle in the ratio of the remaining sides.
(ii)
A person standing on the bank of a river observes that the angle of
elevation of the top of a tree standing on the opposite bank is 60º. When
he moves 40 m away from the bank, he finds the angle of elevation to be
30º. Find the height of the tree and the width of the river. ( 3 = 1.73)
(iii)
A 10 m deep well of diameter 1.4 m is dug up in a field and the earth
from digging is spread evenly on the adjoining cuboid field. The length
and breadth of that filled are 55m and 14 m respectively. Find the thickness
of the earth layer spread.
Best Of Luck

Set - A
A.P. SET CODE
2013 ___ ___ 1100
- MT - w
Time : 2 Hours
MATHEMATICS (71) GEOMETRY - PAPER A (E)
Prelim - I Model Answer Paper
Max. Marks : 40
Q.1. Solve the following : (Any 5)
(i)


(ii)
A (ABE)
A (ABD)
=
BE
AD
A (ABE)
A (ABD)
=
6
9
=
2
3
A (ABE)
A (ABD)
½
[Triangles with common base]
½
ABCD is cyclic
[Given]
 m ABC + m ADC = 180º

½
[Opposite angles of quadrilateral
are supplementary]
ABC and ADC are supplementary.
(iii)
(5, – 7)
 x is positive and y is negative
 The terminal arm is in IV quadrant.
(iv)
Inclination of the line () =
 Slope of the line
=
=
=
½
½
½
½
45º
tan 
tan 45º
1
½
 Slope of the line is 1.
(v)
Area of a triangle =

A (PQR) =

24 =
1
 base  height
2
1
× PR × QS
2
½
Q
8 cm
1
 PR  8
2
P
S
R
Set - A
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


(vi)
24 = PR × 4
24
PR =
4
PR = 6 m
½
Volume of a cube
=
Volume of a cube
=

l3
=

l
=
 Length of the side of cube is
1000 cm3
l3
100
10
10 cm.
½
[Taking cube roots]
Q.2. Solve the following : (Any 4)
A
(i)
In the adjoining figure,
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
C
B
AC = 10 m
AB = 8 m
[½ mark for figure]
In ABC,
[Given]
m ABC = 90º
2
2
2
AC = AB + BC
[By Pythagoras Theorem]
 (10) 2 = (8)2 + BC2
 100 = 64 + BC2
 BC 2 = 100 – 64
 BC 2 = 36
[Taking square roots]
 BC = 6 m
 The distance of the foot of the ladder from the base of the
wall is 6 m.
1
m(arc APB)
2
1
m BAC =
× 80
2
m BAC = 40º
1
m (arc BQD)
m BAD =
2
m BAC =
(ii)


 m (arc BQD)
1
m (arc BQD)
2
= 30 × 2
 m (arc BQD)
= 60º

[Tangent secant theorem]
½
½
½
½
½
½
½
[Inscribed angle theorem]
½
30 =
½
Set - A
3 / MT - w
x = a sec 
(iii)


x
a
y = b tan 
y
tan  =
b
1 + tan2  = sec 2 
sec  =

 y
1+  
 b

1+







(v)
y2
b2
x2
y2
–
a2
b2

(iv)
2
 x
=  
 a
=
½
......(i)
½
......(ii)
2
[From (i) and (ii)]
x2
a2
½
½
= 1
The equation of the line is 2y = mx + 5
Let P (– 2, – 3)
Point P lies on the line 2y = mx + 5
Co-ordinates of point P satisfies the equation of the line
2 (– 3) = m (– 2) + 5
– 6 = – 2m + 5
– 2m = – 6 – 5
– 2m = – 11
11
m =
2
11
The value of m is
2
½
½
½
½
The terminal arm passes through P (4, 3)
 x = 4 and y = 3
r =
x2  y2
=
(4)2  (3)2
=
16  9
=
25
 r = 5 units
Let the angle be 
 sin 
=
y
3
=
r
5
½
cosec 
r
5
= y =
3
Set - A
4 / MT - w
cos 
tan 
(vi)





x
4
=
r
5
y
3
=
=
x
4
=
sec 
cot 
r
5
=
x
4
x
4
= y =
3
=
1½
Let O  (0, 0)  (x1, y1)
A  (– 3, 5)  (x2, y2)
The equation of line OA by two point form is,
x – x1
y – y1
=
x1 – x 2
y1 – y 2
x –0
y –0
=
0 – (– 3) 0 – 5
y
x
= –5
3
– 5x = 3y
5x + 3y = 0
½
½
½
 The equation of the line passing through the origin and the
point (– 3, 5) is 5x + 3y = 0.
Q.3.
Solve the following : (Any 3)
(i)





(ii)
½
ABCD is a trapezium
A
side AB || side DC
[Given]
On transversal AC,
O
BAC  DCA
[Converse
of alternate angles test]
D
BAO  DCO ......(i) [ A - O - C]
In AOB and COD,
BAO  DCO
[From (i)]
AOB  COD
[Vertically opposite angles]
AOB ~ COD
[By AA test of similarity]
AO
BO
=
[c.s.s.t.]
CO
DO
AO CO
=
[By Alternendo]
BO DO
D
C
C
1
1
1
B
A
Take a point D on the tangent at
C as shown in the figure.
seg AB || line CD.
[Given]
 On transversal AC,
B
[½ mark for figure]
½
Set - A
5 / MT - w
BAC  ACD
But, ACD  ABC
........(i) [Converse of alternate angles test]
½
.......(ii) [Angles in alternate segment]
½
In ABC,
BAC  ABC
[From (i) and (ii)]
 seg AC  seg BC

[Converse of Isosceles triangle theorem]
AC = BC
1
(iii)
(Rough Figure)
3.5
cm
C
A
7.3 cm
B
cm
3.5
D
3.5
cm
C
A
M
7.3 cm
B
cm
3.5
D
½
½
½
½
1
mark
mark
mark
mark
mark
for rough figure
for drawing the circle of radius 3.5 cm
for drawing the perpendicular bisector of seg AB
for drawing the circle with centre M
for drawing both the tangents from point B
6 / MT - w
(iv)
Let,
Set - A
 2 1
A   ,   (x1, y1)
5 3
1

B   , k   (x2, y2)
2
½
4

C   , 0  (x3, y3)
5
 Points A, B and C are collinear
Slope of line AB = Slope of line BC
y 2 – y1
y3 – y2

= x – x
x 2 – x1
3
2
1
k –
0–k
3

=
4 1
1 2
–
–
5
2
2 5
3k – 1
–k
3

=
3
1
10
10
3k – 1
10
× 10 = – k ×

3
3

3k – 1 = – k

3k + k = 1

4k = 1
1

k =
4
1
 The value of k is
4
(v)
(Rough Figure)
M
7.5 cm
E
6 cm
L
½
½
½
½
½
Set - A
7 / MT - w
M
7.5 cm
O
E
6 cm
L
½ mark for rough figure
½ mark for drawing LEM
1 mark for drawing the
perpendicular bisectors
1 mark for drawing the
circumcircle
Q.4. Solve the following : (Any 2)
(i)
A
Given : (i) A circle with centre O.
(ii) P is a point in the exterior
of the circle.
O
P
(iii) Points A and B are the
points of contact of the two
B
tangents from P to the circle.
[½ mark for figure]
To Prove : PA = PB
Construction : Draw seg OA, seg OB and seg OP.
Proof : In PAO and PBO,
m PAO = m PBO = 90º[Radius is perpendicular to the tangent]
Hypotenuse OP  Hypotenuse OP [Common side]
[Radii of same circle]
seg OA  seg OB

PAO  PBO
[By hypotenuse - side
theorem]
seg PA  seg PB
[c.s.c.t]

PA
= PB
½
½
½
½
½
½
½
Set - A
8 / MT - w
(Rough Figure) U
(ii)
U
5.
8c
m
R
S
4.5 cm
R
cm
H
V
5.8
cm
cm
5.2
×
S
5.
2
4.5 cm
×
H
V
S1
S2
•
S3
S4
•
½ mark for SHR
S5
1 mark for constructing 5 congruent parts
1 mark for constructing VS5S  HS3S
1 mark for constructing UVS  RHS
½ mark for required SVU
(iii)
For a sector,
Measure of arc () = 60º
Radius (r) = 6 cm
(i) Curved surface area of the cheese = Length of arc × height

 2r  h
=
360
½
½
Set - A
9 / MT - w
60
22
2
67
360
7
= 44 cm2
 The curved surface area of the cheese is 44 cm2.
(ii)
Volume of the cheese piece = A (sector) × height
=

 r2  h
360
60
22

667
=
360
7
= 132 cm3
 The volume of the cheese piece 132 cm3.
=
½
½
½
½
½
½
Q.5. Solve the following : (Any 2)
(i)
E
[½ mark for figure]
Given : In ABC,
A
ray AD is the bisector of BAC
xx
such that B - D - C.
BD
AB
=
To Prove :
DC
AC
B
C
D
Construction : Draw a line passing through C,
parallel to line AD and intersecting line BA at point E, B - A - E.
Proof : In BEC,
line AD || side CE
[Construction]


BD
AB
=
.........(i)
DC
AE
line CE || line AD
On transversal BE,
BADAEC
........(ii)
Also, On transversal AC,
DAC ACE
........(iii)
........(iv)
But, BAD  DAC
In AEC,
AEC  ACE
 seg AC  seg AE

AC

BD
=
DC
= AE
AB
AC
[By B.P.T.]
½
½
½
[Construction]
[Converse of corresponding
angles test]
½
[Converse of alternate angles test]
[ ray AD bisects BAC]
½
½
[From (ii), (iii) and (iv)]
[Converse of Isosceles triangle
theorem]
........(v)
1
[From (i) and (v)]
½
10 / MT - w
(ii)
Set - A
A
Let seg AB represents the tree
seg BC represents width of river
Let BC = x m
C and D represents the initial and
final positions of the observer
DC = 40 m
ACB and ADB are the
30º
60º
B
D
angles of elevation
C
40 m
m ACB = 60º and m ADB = 30º
[½ mark for figure]
In right angled ACB,
AB
[By definition]
tan 60º =
BC
AB

=
3
x
 AB
=
.....(i)
3xm
In right angled ADB,
AB
tan 30º =
DB
AB
1

= 40  x
3





40  x
m .....(ii)
3
From (i) and (ii) we get,
40  x
=
3x
3
3x
= 40 + x
3x – x = 40
2x
= 40
x
= 20
BC
= 20 m




AB
AB
AB
Height
 AB
(iii)

[By definition]
=
=
=
=
of
[From (i)]
20 3 m
20 × 1.73
34.6 m
tree is 34.6 m and width of river is 20 m.
Diameter of well = 1.4 m
1.4
Its radius (r) =
2
= 0.7 m
Its depth (h) = 10 m
Volume of cylindrical well = r 2 h
½
½
½
½
½
½
½
½
½
½
½
½
11 / MT - w
Set - A
22
 0.7  0.7  10
7
22
7
7


 10
=
7
10 10
154
=
10
= 15.4 m3
=

Volume of earth dug is 15.4 m3
Now, Earth dug from the well is spread evenly on the adjoining
cuboid field
Volume of cuboid = Volume of earth dug
= 15.4 m3
Length of a cuboid (l) = 55 m
Its breadth (b) = 14 m
Volume of cuboid = l × b × h

15.4 = 55 × 14 × h
154

10 × 55 × 14 = h
1

h =
m
50

h = 0.02 m

The thickness of the earth layer spread is 0.02 cm.

½
½
½
½
½
½
½