Paper - I
MAHESH TUTORIALS
S.S.C.
Test - III
Batch : SB
GEOMETRY
Chapter : 1, 2 , 6
Date :
Q.1.
Marks : 30
Time : 1 hr. 15 min.
Solve the following :
3
(i)
Find the total surface area of a cube with side 1 cm.
(ii)
If ABC ~ DEF and DEF ~ XYZ and m A = 70º, then find m X.
(iii)
O is the centre of the circle.
If m ABC = 80º, the find
m (arc AC) and m (arc ABC).
B
O
C
A
Q.2.
Solve the following :
6
(i)
An arc of length 4 cm subtends an angle of measure 40º at the centre.
Find the radius and the area of the sector formed by this arc.
(ii)
In the adjoining figure,
AB2 + AC2 = 122, BC = 10.
Find the length of the median
on side BC.
(iii)
Q.3.
(i)
B
C
Q
Curved surface area of a cone with base radius 40 cm is 1640 sq.cm.
Find the height of the cone.
Solve the following :
9
The curved surface area of a hemisphere is 905
( =
(ii)
A
22
)
7
In the adjoining figure, given two
concentric circles of radii 5 and 3,
A
find the length of a chord of larger
circle which touches the smaller one.
If BD = 5, find BC.
1
cm2, what is its volume?
7
D
O
E
C
B
Paper - I
... 2 ...
M
(iii)
In the adjoining figure,
LMN = 90º and LKN = 90º,
seg MK  seg LN. Prove that
L
R is the midpoint of seg MK.
N
R
K
Q.4.
(i)
Solve the following :
12
A
ABCD is cyclic quadrilateral, lines AB and DC
intersect in the point F and lines AD and BC
B
intersect in the point E. Show that the
circumcircles of BCF and CDE intersect
in a point G on the line EF.
F
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
(a) The curved surface area of the cheese.
(b) The volume of the cheese piece.
22
( =
)
7
(iii)
ABC is a triangle in which AB = AC
and D is any point on BC.
Prove that : AB2 – AD2 = BD. CD.
C
G
6 cm
60º
7 cm
(ii)
D
B
A
D
E
Best Of Luck

C
E
Paper - I
MAHESH TUTORIALS
S.S.C.
Batch : SB
GEOMETRY
Chapter : 1, 2 , 6
Date :
A.1.
(i)
MODEL ANSWER PAPER
Solve the following :
Length of side of cube (i)
=
 Total surface area of a cube =
=
=
=
Test - III
Marks : 30
Time : 1 hr. 15 min.
1 cm
6l 2
6 (l)2
6 (1)
6 cm2
 The total surface area of cube is 6 cm2.
(ii)
ABC ~ DEF
DEF ~ XYZ
 ABC ~ XYZ
A = X
 But, m A = 70º
[Given]
[c.a.c.t.]
[Given]
 m X = 70º
 m (arc AC) =
m (arc ABC) =
=
 m (arc ABC) =
200º

A.2.
(i)
m ABC
=
80º
=
½
½
Solve the following :
Length of arc (l)

measure of arc ()
l

4

4×9
2
r

½
½
1
m (arc AC)
[By Inscribed angle theorem]
2
B
1
m (arc AC)
2
160º
O
C
360º – m (arc AC)
A
360 – 160
(iii)
1
= 4 cm
= 40º

=
× 2r
360
40
=
×2××r
360
½
= r
= 18 cm.
½
Paper - I
... 2 ...
Area of the sector
l ×r
2
4 × 18
=
2
= 36cm2
½
=
½
 Radius of the circle is 18 cm and Area of the sector is 36cm2.
(ii)










(iii)









A.3.
(i)
In ABC,
seg AQ is the median
1
BQ = QC =
× BC
2
1
BQ = QC =
× 10
2
BQ = QC = 5 units ......(i)
AB2 + AC2 = 2AQ2 + 2BQ2
122 = 2AQ2 + 2 (5)2
122 = 2AQ2 + 2 (25)
122 = 2AQ2 + 50
2AQ2 = 122 – 50
2AQ2 = 72
AQ2 = 36
AQ = 6 units
Curved surface area of a cone
its radius (r)
Curved surface area of a cone
1640
1640
40
l
Now,
r 2 + h2
402 + h2
h2
h2
h2
h
Height of a cone is 9 cm.
A
[Given]
[Given]
B
Q
C
[By Appollonius theorem]
[From (i) and given]
½
½
½
[Taking square roots]
=
=
=
=
1640cm2
40 cm
rl
 × 40 × l
=
l
=
=
=
=
=
=
=
41 cm
l2
412
412 – 402
1681 – 1600
81
9 cm
[Taking square roots]
½
½
½
½
½
Solve the following :
Curved surface area of a hemisphere = 905
1
cm2
7
Curved surface area of a hemisphere = 2r 2
1
22

905
= 2×
× r2
7
7
½
½
Paper - I
... 3 ...




6336
7
6336  7
7  2  22
r2
r
= 2×
22
× r2
7
½
= r2
= 144
= 12 cm
[Taking square roots]
2 3
Volume of a hemisphere =
r
3
2
22
=
×
× 12 × 12 × 12
3
7
25344
=
7
= 3620.57 cm3
 Volume of a hemisphere is 3620.57 cm3.
½
½
½
(ii)
D
A













O
Sol. For the smaller circle
Line BE is a tangent to the circle at E and
E
C
line CD is a secant intersecting the circle
at points C and D.
B
BE2 = BC × BD
......(i)
[Tangent secant property]
In OEB,
m OEB = 90º
......(ii)
[Radius is perpendicular to tangent]
OB2 = OE2 + BE2
[By Pythagoras theorem]
5 2 = 32 + BE2
[Given]
2
25 = 9 + BE
25 – 9 = BE2
BE2 = 16
BE = 4 units
.....(iii)
[Taking square roots]
For the larger circle,
seg OE  chord AB
[From (ii)]
AB = 2 × BE
[Perpendicular from the centre of the
circle to the chord bisects the chord]
AB = 2 × 4
AB = 8 units
(4)2 = BC × 5
[From (i) and (iii)]
16 = BC × 5
16
BC =
5
BC = 3.2 units
½
½
½
½
½
½
Paper - I
... 4 ...
(iii)
M





A.4.
(i)
In LMN,
m LMN = 90º
seg MR  hypotenuse LN
MR2 = LR × RN
.....(i)
In LKN,
m LKN = 90º
seg KR  hypotenuse LN
KR2 = LR × RN
.....(ii)
MR2 = KR2
MR = KR
R is the midpoint of seg MK
L
[Given]
[Given]
K
[By property of geometric mean]
[Given]
[Given]
[By property of geometric mean]
[From (i) and (ii)]
[Taking square roots]







1
A
B

1
1
Solve the following :

N
R
D
C
Construction : Draw seg CG.
Proof : Let the circumcircle of BCF and F
E
G
CDE intersect at points C and G.
BCGF is cyclic.
[By definition]
m ABC = m CGF
.....(i) [Exterior angle of a cyclic quadrilateral
is equal to its interior opposite angle]
DCGE is cyclic
[By definition]
m ADC = m CGE
......(ii) [Exterior angle of a cyclic quadrilateral
is equal to its interior opposite angle]
Adding (i) and (ii),
m ABC+ m ADC = m CGF + m CGE .......(iii)
ABCD is cyclic
m ABC + m ADC = 180º ....(iv) [Opposite angle of a cyclic
quadrilateral are supplementary]
m CGF + m CGE = 180º
[From (iii) and (iv)]
Also, CGF and CGE are adjacent angle
CGF and CGE form a linear pair [Converse of linear pair axiom]
ray GF and ray GE form opposite rays
Points F, G, E are collinear.
G lies on line EF
The circumcircle of BCF and CDE intersect in a point on line EF.
½
½
½
½
½
½
½
½
Paper - I
... 5 ...
(ii)
7 cm
6 cm
60º
For a sector,
Measure of arc () = 60º
Radius (r) = 6 cm
(a) Curved surface area of the cheese = Length of arc × height

 2r  h
=
360
60
22
2
67
=
360
7
= 44 cm2
½
½
½
 The curved surface area of the cheese is 44 cm2.
½
(b) Volume of the cheese piece
½
= A (sector) × height

 r 2  h
=
360
60
22

667
=
360
7
= 132 cm3
½
½
 The volume of the cheese piece 132 cm3.
½
A
(iii)






In AEB,
B
D
E
m AEB = 90º
[Given]
2
2
2
AB = AE + BE
......(i)
[By Pythagoras theorem]
In AED,
[Given]
m AED = 90º
AD2 = AE2 + DE2
.....(ii)
[By Pythagoras theorem]
Subtracting equation (ii) from (i),
AB2 – AD2 = AE2 + BE2 – (AE2 + DE2)
AB2 – AD2 = AE2 + BE2 – AE2 – DE2
AB2 – AD2 = BE2 – DE2
AB2 – AD2 = (BE + DE) (BE – DE)
AB2 – AD2 = (BE + DE) × BD
.......(iii) [ B - D - E]
C
½
½
½
½
Paper - I
... 6 ...




In AEB and AEC,
m AEB = m AEC = 90º
hypotenuse AB  hypotenuse AC
seg AE  seg AE
AEB  AEC
seg BE  seg CE
.....(iv)
2
2
AB – AD = (CE + DE) × BD
AB2 – AD2 = CD × BD
[Given]
[Given]
[Common side]
[By hypotenuse side theorem]
[c.s.c.t.]
[From (iii) and (iv)]
[ D - E - C]

1
½
½
Paper - II
S.S.C.
MAHESH TUTORIALS
Batch : SB
Date :
Q.1.
Test - III
Marks : 30
GEOMETRY
Chapter : 1, 2, 6
Time : 1 hr. 15 min.
Solve the following :
3
(i)
The dimensions of a cuboid are 5 cm, 4 cm and 3 cm. Find its volume.
(ii)
A (ΔDEF)
If DEF ~ MNK, DE = 5, MN = 6, find the value of A (ΔMNK) .
(iii)
Q.2.
Line AB is a tangent and line BCD
is a secant. If AB = 6 units, BC = 4 units, A
find BD.
B
Solve the following :
D
C
6
(i)
The dimensions of a cuboid in cm are 16 × 14 × 20. Find its total surface area.
(ii)
Find the perimeter of an isosceles right triangle with each of its congruent
sides is 7 cm.
(iii)
Find the length of the arc of a circle with radius 0.7 m and area of the
sector is 0.49 m2.
Q.3.
(i)
(ii)
Solve the following :
9
22 

The surface area of a sphere is 616cm2. What is its volume ?   =
7 

In the adjoining figure,
in the isosceles triangle PQR,
the vertical P = 50º. The circle
passing through Q and R cuts PQ .
in S and PR in T. ST is joined.
Find PST.
Q
P
50º
S
T
R
Paper - II
... 2 ...
(iii)
Q.4.
(i)
(ii)
Prove that three times the square of any side of an equilateral triangle is
equal to four times the square of an altitude.
Solve the following :
12
Let the circles with centre P and
Q touch each other at point A.
Let the extended chord AB intersect
E
the circle with centre P at point E
and the chord BC touches the circle
with centre P at the point D.Then prove
that ray AD is an angle bisector of the CAE.
P
D
M
Q
C
In the adjoining figure,
m POQ = 30º and radius OP = 12 cm.
Find the following (Given  = 3.14)
O
(a) Area of sector O-PRQ
M
(b) Area of OPQ
12 cm 30º
(c) Area of segment PRQ
Q
P
(iii)
B
A
•
R
45º
In the adjoining figure,
if LK = 6 3 find MK, ML, KN, MN
and the perimeter of MNKL.
M
30º
L
6 3
Best Of Luck

K
N
Paper - II
MAHESH TUTORIALS
S.S.C.
Batch : SB
Date :
A.1.
(i)
GEOMETRY
Chapter : 1, 2, 6
MODEL ANSWER PAPER
Solve the following :
Length of a cuboid (l)
Its breadth (b)
Its height (h)
Volume of a cuboid
=
=
=
=
=
=
Test - III
Marks : 30
Time : 1 hr. 15 min.
5 cm
4 cm
3 cm
l×b×h
5×4×3
60 cm3
½
 Volume of cuboid is 60 cm3.
(ii)
DEF ~ MNK
A (DEF)
DE2
 A (MNK) =
MN2
A (DEF)
52
 A (MNK) = 2
6
A (DEF)
25
 A (MNK) =
36
(iii)




½
[Given]
½
[Area of similar triangles]
[Given]
½
Line BCD is a secant intersecting the circle
at points C and D and line BA is a tangent at A
A
AB2 = BC × BD
62 = 4 × BD
36 = 4 × BD
36
C
B
BD =
4
½
D
 BD = 9 units
A.2.
(i)
Solve the following :
Length of a cuboid (l)
its breadth (b)
its height (h)
Total surface area of a cuboid
½
=
=
=
=
=
16 cm
14 cm
20 cm
2 (lb + bh + lh)
2 (16 × 14 + 14 × 20 + 16 × 20)
½
½
Paper - II
... 2 ...
= 2 (224 + 280 + 320)
= 2 × 824
= 1648 cm2
½
 Total surface area of a cuboid is 1648 cm2.
½
(ii)





Given : In ABC,
A
m ABC = 90º
AB = BC = 7 cm
To find : Perimeter of ABC
7 cm
Sol. In ABC,
m ABC = 90º
[Given]
B
2
2
2
AC = AB + BC
[By Pythagoras theorem]
AC2 = (7)2 + (7)2
AC2 = 49 + 49
AC2 = 98
AC = 49  2
[Taking square roots]
 AC = 7 2 cm
Perimeter of ABC
 Perimeter of ABC
(iii)




=
=
=
AB + BC + AC
7 7  7 2
14  7 2
=
7 2


2 cm
Radius of a circle = 0.7 cm
Area of the sector = 0.49 m2
r
Area of the sector =
×l
2
0.7
0.49 =
×l
2
49
7
=
×l
100
20
49  20
= l
100  7
l = 1.4
 The length of the arc is 1.4 m.
A.3.
(i)
Solve the following :
Surface area of sphere = 616 cm2
Surface area of a sphere = 4r 2
22

616 = 4 ×
× r2
7
616  7

= r2
4  22
7 cm
C
½
½
½
½
½
½
½
½
½
½
Paper - II
... 3 ...


r2
r
= 49
= 7 cm
[Taking square roots]
4 3
Volume of a sphere =
r
3
4 22

777
=
3
7
4312
=
3
= 1437.33 cm3
 The volume is 1437.33 cm3.
(ii)
½
½
½
½
P
50º
S






In PQR
Q
R
[Given]
seg PQ  seg PR
PQR  PRQ
......(i) [Isosceles triangle theorem]
m PQR + m PRQ + m QPR = 180º [Sum of the measures of
angles of a triangle is 180º]
m PRQ + m PRQ + 50 = 180º [From (i) and Given]
2m PRQ = 180º – 50º
2m PRQ = 130º
m PRQ = 65º
......(ii)
SQRT is cyclic
PST TRQ
[An exterior angle of cyclic
quadrilateral is congruent to the
angle opposite to adjacent interior
angle]
PST PRQ
[P - T - R]
 PST = 65º
(iii)
T
½
½
½
½
[From (ii)]
To prove : 3AB2 = 4AD2
Proof : ABC is an equilateral triangle [Given]
In ADB,
[Given]
m ADB = 90º
[Angle of an
m ABD = 60º
equilateral triangle]
 m BAD = 30º
[Remaining angle] B
½
½
A
½
D
C
1
Paper - II
... 4 ...
 ADB is a 30º - 60º - 90º triangle
 By 30º - 60º - 90º triangle theorem,
AD =
3
AB
2
½
[Side opposite to 60º]
½
[Squaring both sides]
½
 2AD =
3 AB
 4AD = 3AB2
 3AB2 = 4AD2
2
A.4.
(i)
Solve the following :
E
In MAD,
MA = MD
P
B
A
D
M
Q
C
½
[The lengths of two tangent
segments from an external point
to a circle are equal]
[Isosceles triangle theorem]
½
 m MAD = m MDA
Let,
m MAD = m MDA = xº ........(i)
CAM ABC
[Angles in alternate segments]
Let,
m CAM = m ABC = yº ........(ii)
m CAD = m CAM + m MAD
[Angles Addition property]
m CAD = (x + y)º
......(iii) [From (i) and (ii)]
DAE is an exterior angle of ADB
 m DAE = m ADB + mABD
[Remote Interior angles theorem]
 m DAE = m ADM + m ABC
[D - M - C - B]
 mDAE = (x + y)º
......(iv) [From (i) and (ii)]
 m CAD = mDAE
[From (iii) and (iv)]
 ray AD is an angle bisector of CAE.
½
½
½
½
½
½
(ii)
Radius of the circle (r)= 12 cm
O
Measure of arc () = 30º
M

12 cm 30º
× r2
Area of sector O - PRQ =
Q
360
•
30
R
P
=
× 3.14 × 12 × 12
360
= 37.68 cm2
½
½
Paper - II
... 5 ...




In OMP,
m OMP = 90º
[Given]
m POM = 30º
[Given and O - M - Q]
m OPM = 60º
[Remaining angle]
OMP is 30º - 60º - 90º triangle
By 30º - 60º - 90º triangle theorem
1
OP
PM =
2
1
=
× 12
2
PM = 6 cm.
OP = OQ
= 12 cm
[Radii of same circle]
1
Area of OPQ
=
× base × height
2
1
× OQ × PM
=
2
1
=
× 12 × 6
2
= 36 cm2
Area of segment PRQ = Area of sector O-PRQ – Area of OPQ
= 37.68 – 36
= 1.68 cm2
1
½
½
½
½
(a) Area of sector O-PRQ is 37.68 cm
(b) Area of OPQ is 36 cm2
(c) Area of segment PRQ is 1.68 cm2
2
(iii)
In MLK,
[Given]
m MLK = 90º
[Given]
m MKL = 30º
M
 m LMK = 60º
[Remaining angle]
 MLK is a 30º - 60º - 90º triangle.
L
 By 30º - 60º - 90º triangle theorem,
LK

=
6 3 =
 MK
 MK
45º
½
30º
6 3
3
MK
2
[Side opposite to 60º]
3
× MK
2
[Given]
6 3 2
3
= 12 units
N
K
½
=
......(i)
½
Paper - II
... 6 ...
1
× MK
[Side opposite to 30º]
2
1
ML
=
× 12
2
ML
= 6 units
......(ii)
In MKN,
m MKN = 90º
[Given]
m MNK = 45º
[Given]
m NMK = 45º
[Remaining angle]
MKN is a 45º - 45º - 90º triangle
By 45º - 45º - 90º triangle theorem,
1
MK = KN =
× MN
......(iii)
2
1
MK =
× MN
[From (iii)]
2
1
12
=
× MN
2
ML







=
 MN = 12 2 units .....(iv)
KN
= 12 units
......(v)
[From (i) and (iii)]
Perimeter of MNKL = MN + KN + KL +ML
= 12 2  12  6 3  6
[From (ii), (iv) and (v) and given]
= 18  12 2  6 3
 Perimeter of MNKL
=

6 32 2 


3 units
½
½
½
1
Paper - III
S.S.C.
MAHESH TUTORIALS
Batch : SB
Date :
Q.1.
Test - III
Marks : 30
GEOMETRY
Chapter : 1, 2, 6
Time : 1 hr. 15 min.
Solve the following :
3
(i)
The radius of the base of a cone is 7 cm and its height is 24 cm. What is
its slant height ?
(ii)
Determine whether 9, 40, 41 are the sides of a right angled triangle.
(iii)
Radius of the circle is 4 cm. What is the length of a chord of the circle ?
Q.2.
Solve the following :
6
(i)
An arc of a circle having measure 36 has length 176 m. Find the
circumference of the circle.
(ii)
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.
(iii)
The perimeter of one face of a cube is 24 cm.
Find (a) the total area of the 6 faces (b) the volume of the cube.
Q.3.
Solve the following :
9
(i)
Two arcs of the same circle have their lengths in the ratio 4:5. Find the
ratio of the areas of the corresponding sectors.
(ii)
Two circles intersect each other in points X and Y. Secants through X and
Y intersect one of the circles in A and D and the other in B and C
respectively. A and B are on opposite sides of line DC. Show that line AD
is parallel to line BC.
(iii)
In the adjoining figure,
PQR = 90º.
T is the mid point of the side QR.
Prove that PR2 = 4PT2 – 3PQ2.
P
Q
T
R
Paper - III
... 2 ...
Q.4.
(i)
Solve the following :
12
Let M be a point of contact of two internally
touching circles. Let line AMB be their
common tangent. The chord CD of the bigger
C
circle touches the smaller circle at point N
and chord CM and chord DM of bigger circle
intersect smaller circle at the points P and R
respectively. Prove that CMN  DMN.
N
P
A
R
B
M
10 cm
10 cm
10 cm
(ii)
D
60 cm
A toy is a combination of a cylinder, hemisphere and a cone, each with
radius 10cm. Height of the conical part is 10 cm and total height is 60cm.
Find the total surface area of the toy. ( = 3.14, 2 = 1.41)
(iii)
In the adjoining figure,
PQRV is a trapezium in
which seg PQ || seg VR.
SR = 4 and PQ = 6. Find VR.
V
Best Of Luck
P
60º
T

6
Q
45º
S
4
R
Paper - III
S.S.C.
Batch : SB
Date :
A.1.
(i)
MAHESH TUTORIALS
Marks : 30
GEOMETRY
Chapter : 1, 2, 6
Time : 1 hr. 15 min.
MODEL ANSWER PAPER
Solve the following :
Radius of base of cone (r)
Its height (h)
l2

l2

l2

l2

l
=
=
=
=
=
=
=
7 cm
24 cm
r 2 + h2
72 + 242
49 + 576
625
25
Test - III
[Taking square roots]
 Slant height of cone is 25 cm.
(ii)
(iii)
(41)2
= 1681
......(i)
(9)2 + (40)2 = 81 + 1600
= 1681
......(ii)
 (41)2
= (9)2 + (40)2
[From (i) and (ii)]
 The given sides form a right angled triangle. [By Converse of
Pythagoras theorem]
1
Radius of the circle is 4 cm
 Diameter of the circle = 4 × 2 = 8 cm
Diameter is the biggest chord of the circle
 The length of the chord of the circle is 8 cm or less than 8 cm.
A.2.
(i)
1
Solve the following :
Length of arc (l)

measure of arc ()
l

176

176


176 × 10
2r
= 176 m
= 36º

=
× 2r
360
36
=
× 2r
360
1
=
× 2r
10
= 2r
= 1760
1
½
½
½
Paper - III
... 2 ...
But, circumference
= 2r
 Circumference of the circle is 1760 m.
(ii)





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
B
AB = 8 m
In ABC,
m ABC = 90º
[Given]
AC2 = AB2 + BC2
[By Pythagoras theorem]
(10)2 = (8)2 + BC2
100 = 64 + BC2
BC2 = 100 – 64
BC2 = 36
[Taking square roots]
BC
= 6m
½
½
C
 The distance of the foot of the ladder from the base of the wall is 6 m.
(iii)
Perimeter of one face of a cube= 24 cm
Perimeter of one face of a cube= 4l

4l = 24
24

l =
4

l = 6 cm
Total surface area of a cube = 6l 2
= 6 (6)2
= 6×6×6
= 216 cm2
Volume of the cube = l 3
= 63
= 216 cm3
½
½
½
½
½
½
½
 Total area of the 6 faces is 216 cm2 and volume of the cube is
216 cm3.
A.3.
(i)
Solve the following :
Ratio of lengths of two arcs is 4 : 5.
Let the common multiple be ‘x’
 Lengths of two arcs are (4x) units and (5x) units respectively
Let the lengths of two arcs be ‘l1’, and ‘l2’ and Areas of their
corresponding sectors be A1 and A2.
 l1 = (4x) units
 l2 = (5x) units
½
½
Paper - III
... 3 ...
Both Arcs are of the same circle.
 Their radii are equal
Now,
r
=
× l1
.......(i)
A1
2
r
A2
=
× l2
......(ii)
2
Dividing (i) and (ii) we get,
A1
l1
=
A2
l2
A1
4x
 A
=
5x
2
 A1 : A2 = 4 : 5
½
½
½
 Ratio of the areas of sectors is 4 : 5.
½
A
(ii)
X
ADYX is cyclic
 XYC DAX
Y
D
......(i)
But, XYC XBC .....(ii)
 DAX XBC
 DAB ABC
 line AD || line BC
C
B
[By definition]
[The exterior angle of a cyclic
quadrilateral is equal to its interior
opposite angle]
[Angles inscribed in the same arc]
[From (i) and (ii)]
[A - X - B]
[By Alternate angles test]
½
1
½
1
P
(iii)
In PQR,
seg PT is the median
 PQ2 + PR2
In PQT,
m PQT

PT 2

QT2
Q
[Given]
T
R
= 2PT2 + 2QT2 ......(i) [By Appollonius theorem]
1
= 90º
= PQ2 + QT2
= PT2 – PQ2
1
[Given]
[By Pythagoras theorem]
.....(ii)
Paper - III
... 4 ...
 PQ2 + PR2
 PQ2 + PR2

PR2

PR2
A.4.
(i)
=
=
=
=
2PT2 + 2 (PT2 – PQ2)
2PT2 + 2PT2 – 2PQ2
4PT2 – 2PQ2 – PQ2
4PT2 – 3PQ2
[From (i) and (ii)]
½
Solve the following :
N
C
P








D
R
Construction : Draw seg NR.
M
A
B
Proof :CMA  CDM
[Angles in alternate segments]
Let,
m CMA = mCDM = xº ........(i)
NMA  NRM
[Angles in alternate segments]
Let,
mNMA  NRM = yº
.......(ii)
m NMC = m NMA – m CMA
[Angle Addition property]
m NMC = (y – x)º
......(iii) [From (i) and (ii)]
NMR  DNR
.......(iv) [Angles in alternate segment]
NRM is an exterior angle of  NDR
m NRM = m NDR + mDNR
[Remote interior angles]
m NRM = CDM + m DNR
[ C - N - D and D - R - M]
y = x + mDNR
[From (i) and (ii)]
m DNR = (y – x)º
......(v)
m NMR = (y – x)º
[From (iv) and (v)]
m NMD = (y – x)º
......(vi) [D - R - M]
CMN  DMN
[From (iii) and (vi)]
½
½
½
½
½
½
½
½
10 cm
10 cm
10 cm
(ii)
½
60 cm
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 = 10cm
Total height of the toy
= 60cm
 Height of the cylindrical part (h1)= 60 – 10 – 10 = 60 – 20 = 40 cm
½
½
Paper - III
... 5 ...


l2
l2
l2
l2
=
=
=
=

l
=
l
= 10 2 cm
Slant height of the conical part (l)
=
=
Total surface area of the toy =
=
=
=
=
=
=
½
r 2 + h2
102 + 102
100 + 100
200
200
[Taking square roots]
= 10 2
½
10 × 1.41
14.1 cm
Curved surface area of the conical ½
part + Curved surface area of the
cylindrical part + Curved surface
area of the hemispherical part
rl + 2rh1 + 2r2
½
r (l + 2h1 + 2r)
3.14 × 10 (14.1 + 2 × 40 + 2 × 10)
31.4 (14.1 + 80 + 20)
31.4 × 114.1
3582.74 cm2
 Total surface area of the toy is 3582.74 cm2.
P
(iii)
60º
V
½
T
½
6
Q
45º
S
4
R
In QSR,
m QSR = 90º
[Given]
m SQR = 45º
[Given]
 m QRS = 45º
[Remaining angle]
 QSR is a 45º - 45º - 90º triangle.
 QS = SR
[Congruent sides of 45º - 45º - 90º
triangle]
But, SR = 4 units
......(i) [Given]
 QS = 4 units
.....(ii)
In PQTS,
seg PQ || seg TS
[Given and V - T - S - R]
[Given]
m T = m S = 90º
 PQTS is a rectangle
 PQ = TS = 6 units
.....(iii) [Opposite sides of a rectangle
QS = PT = 4 units
.....(iv) and from (ii) and given]
½
½
½
... 6 ...





In PTV,
m PTV = 90º
[Given]
m VPT = 60º
[Given]
m PVT = 30º
[Remaining angle]
PTV is a 30º - 60º - 90º triangle.
By 30º - 60º - 90º triangle theorem,
1
× PV
[Side opposite to 30º]
PT =
2
1
4
=
× PV
[From (iv)]
2
PV = 8 units
......(v)
VT
=
3
PV
2
[Side opposite to 60º]
 VT
=
3
×8
2
[From (v)]
 VT
VR
= 4 3 units
.....(vi)
= VT + TS + SR
[V - T - S - R]
 VR
=
4 3 64
 VR
=
4 3  10
 VR
=
2 2 3  5 units

Paper - III
½
½
½
[From (i), (iii) and (vi)]

1

Paper - IV
S.S.C.
MAHESH TUTORIALS
Batch : SB
Date :
Q.1.
Test - III
Marks : 30
GEOMETRY
Chapter : 1, 2, 6
Time : 1 hr. 15 min.
Solve the following :
3
(i)
The radius of the base of a cone is 7 cm and its height is 24 cm. What is
its slant height ?
(ii)
In PQR, m Q = 90º, m P = 30º, m R = 60º. If PR = 8 cm, find QR.
(iii)
In the adjoining figure,
if m APB = 30º, then m AOB ?
P
O
B
A
Q.2.
(i)
Solve the following :
6
Find the measure of the central angle subtended by an arc of length 6.05
m and radius 5.5 m. ( =
(ii)
In ABC, AP is a median.
If AP = 7, AB2 + AC2 = 260
then find BC.
B
(iii)
Q.3.
22
)
7
A
P
C
The volume of a cube is 1000 cm3. Find its total surface area.
Solve the following :
(i)
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.
(ii)
ABCD is a rectangle. Taking AD as a diameter, a semicircle AXD is drawn
which intersects the diagonal BD at X. If AB = 12 cm, AD = 9 cm then find the
values of BD and BX.
9
Paper - IV
... 2 ...
(iii)
Determine the values of x, y and z with
P
the help of information given in the adjoing figure.
4
S
6
X
y
Q
Q.4.
z
Solve the following :
R
12
(i)
Suppose AB and AC are equal chords of a circle and a line parallel to the
tangent at A intersects the chords at D and E. Prove that AD = AE.
(ii)
A test tube has diameter 20 mm and
height is 15 cm. The lower portion is
a hemisphere in the adjoining figure.
Find the capacity of the test tube. ( = 3.14)
(iii)
In the adjoining figure,
PQR = 60º and
ray QT bisects PQR.
seg BA  ray QP and
seg BC  ray QR.
If BC = 8 find the
perimeter of ABCQ. Q
15 cm
P
A
Best Of Luck
T
B
C

R
Paper - IV
MAHESH TUTORIALS
S.S.C.
Batch : SB
A.1.
(i)
Marks : 30
GEOMETRY
Chapter : 1, 2, 6
Date :
MODEL ANSWER PAPER
Solve the following :
Radius of base of cone (r)
its height (h)
l2

l2

l2

l2

l
=
=
=
=
=
=
=
7 cm
24 cm
r 2 + h2
72 + 212
49 + 576
625
25
Test - III
Time : 1 hr. 15 min.
[Taking square roots]
 Slant height of cone is 25 cm
(ii)
1
In PQR,
m P = 30º
m R = 60º
[Given]
m Q = 90º
 PQR is a 30º - 60º - 90º triangle
 By 30º - 60º - 90º triangle theorem,
1
PR
[Side opposite to 30º]
QR =
2
½
½
P
(iii)
1
m (arc AB) [Inscribed angle theorem]
2
O
1
 30º
=
m (arc AB)
2
A
 m (arc AB) = 60º
m AOB = m (arc AB)
[Definition of measure of minor arc]
m APB =
 m AOB = 60º
A.2.
B ½
½
Solve the following :
(i)
l

6.05

605
100

× 2r
360
22

=
×2×
× 5.5
7
360
22

=
×
× 11
7
360
=
½
½
Paper - IV
... 2 ...
605 × 360 × 7
 100 × 22 × 11


= 
= 63º
 Measure of the arc is 63º.
(ii)











(iii)
1
In ABC,
seg AP is the median
AB2 + AC2 = 2AP2 + 2BP2
260 = 2 (7)2 + 2BP2
260 = 2 (49) + 2BP2
260 = 98 + 2BP2
260 – 98 = 2BP2
2BP2 = 162
162
BP2 =
2
BP2 = 81
BP = 9 units
1
BP =
BC
2
1
9 =
BC
2
BC
[Given]
[By Appollonius theorem]
[Given]
A
B
P
½
[ P is the midpoint of seg BC]
½
½
=
=
=
=
=
=
=
=
1000 cm3
l3
1000
10 cm
[Taking cube roots]
6l 2
6 × 102
6 × 10 × 10
600 cm2
 Total surface area of a cube is 600 cm2.
A.3.
(i)
C
[Taking square roots]
= 18 units
Volume of a cube
Volume of a cube

l3

l
Total surface area of a cube
½
Solve the following :
Curved surface area of the frustum of a cone = 180 cm2
Perimeters of circular bases are 18 cm and 6 cm
 2r1 = 18
........(i)
2r2 = 6
........(ii)
Adding (i) and (ii), we get
2r1 + 2r2 = 18 + 6
 2 (r1 + r2) = 24
24
  (r1 + r2) =
2
  (r1 + r2) = 12
.......(iii)
1
1
½
½
½
½
Paper - IV
... 3 ...
Curved surface area of the frustum of a cone

180

180

l
=
=
=
=
(r1 + r2) l
(r1 + r2) l
12 × l [From (iii)]
15 cm
½
9
½
½
 Slant height of the frustum of a cone is 15 cm.
(ii)









In ABD
m BAD
BD2
BD2
BD2
BD2
BD
m BAD
line BA is
A
= 90º
[Angle of a rectangle]
2
2
= AB + AD [By pythagoras theorem]
= 122 + 92 [Given]
= 144 + 81
12
X
= 225
= 15 cm
[Taking square roots]
B
= 90º
[Angle of a rectangle]
a tangent to the circle at point A
[A line perpendicular to the radius at its
outer end is a tangent to the circle]
Line AB is a tangent and line BXD is a secant intersecting at
points X and D
AB2 = BX . BD [Tangent secant property]
122 = BX . 15
144 = BX . 15
BX =

BX = 9.6 cm






In PSQ,
m PSQ = 90º
PQ2
= PS2 + QS2
2
6
= 42 + y 2
36
= 16 + y2
2
y
= 36 – 16
y2
= 20
y
=
½
C
1
½
144
15

(iii)
D
45

½
P
[Given]
[By Pythagoras theorem]
4
S
6
Q
[Taking square roots]
y
= 2 5
In PQR,
m PQR = 90º
[Given]
seg QS  hypotenuse PR [Given]
 QS2
= PS × SR
[By property of geometric mean]

y2
= 4×x
 ( 2 5 )2 = 4 × x
X
y
z
R
1
Paper - IV
... 4 ...

20

x

x







A.4.
(i)
= 4×x
20
=
4
1
= 5
In QSR,
m QSR = 90º
QR2
= QS2 + SR2
z2
= y 2 + x2
z2
= ( 2 5 )2 + (5)2
z2
= 20 + 25
2
z
= 45
z
=
9×5
z
[Given]
[By Pythagoras theorem]
[Taking square roots]
1
= 3 5
Solve the following :
Construction : Draw seg BC
C
Proof : Take points R and S on the tangent B
at A as shown in the figure
line DE || line RS
[Given]
 On transversal AD,
E
D
EDA  DAR
[Converse of
•
alternate angles test]
A
S
R
 EDA  BAR
.......(i)
[ B - D - A]
BAR  BCA
.......(ii)
[Angles in alternate segment]
 EDA  BCA
......(iii)
[From (i) and (ii)]
Similarly, we can prove that
DEA  CBA
......(iv)
In ABC,
seg AB  seg AC
[Given]
 BCA  CBA
........(v)
[Isosceles triangle theorem]
In DEA,
EDA  DEA
[From (iii), (iv) and (v)]
 seg AD  seg AE
[Converse of isosceles triangle theorem]

AD = AE
(ii)
Diameter of a test tube

its radius (r)
Its height (h)
= 20 mm
20
=
2
= 10 mm
= 1 cm
= 15 cm
15 cm
½
½
½
½
½
½
1
½
½
Paper - IV
... 5 ...
Height of hemispherical part (h1) =
=

Height of cylindrical part (h2) =
=
=
Volume of test tube =
=
=
=
=
=
Volume of test tube =
radius of hemisphere
1 cm
h – h1
15 – 1
14 cm
Volume of cylindrical part +
Volume of hemispherical part
2 3
r2h2 +
r
3
2 

r
r2 h2 +
3 

2

3.14 (1) 14 + 
3

44
3.14 ×
3
138.16
3
46.05 cm3
½
½
½
½
½
½
 Capacity of a test tube is 46.05 cm3
(iii)







Ray QT is the angle bisector of PQR [Given]
P
B lies on ray QT
seg BA  ray QP
[Given]
A
T
seg BC  ray QR
B
BA = BC
[Angle bisector theorem]
But, BC = 8 units
.......(i)
[Given]
BA = 8 units
......(ii)
Q
R
C
m PQR = 60º
[Given]
1
m PQT = m RQT =
× 60º [ Ray QT bisects QR]
2
m PQT = m RQT = 30º
......(iii)
In BAQ,
m BAQ = 90º
[Given]
m AQB = 30º
[From (iii), P - A - Q and Q - B - T]
m ABQ = 60º
[Remaining angle]
BAQ is a 30º - 60º - 90º triangle
By 30º - 60º - 90º triangle theorem
 AB =
1
BQ
2
1
BQ
2
 BQ = 16 units
 8
=
½
½
1
[Side opposite to 30º]
[From (ii)]
½
Paper - IV
... 6 ...
AQ =
3
× BQ
2
[Side opposite to 60º]
3
× 16
2
 AQ = 8 3 units
 AQ =
......(iv)
Similarly, QC = 8 3 cm
......(v)
 Perimeter of ABCQ = AB + BC + QC + AQ
= 888 3 8 3
[From (i), (ii), (iv) and (v)]
= 16  16 3
 Perimeter of ABCQ

= 16 1 

3 units

½
½
½
Paper - V
S.S.C.
MAHESH TUTORIALS
Batch : SB
Marks : 30
GEOMETRY
Chapter : 1, 2, 6
Date :
Q.1.
Test - III
Time : 1 hr. 15 min.
Solve the following :
3
(i)
Using Euler’s formula, write the value of V, if E = 30 and F = 12.
(ii)
If PQR XYZ,
(iii)
In the adjoining figure,
chords AB and CD intersect at E.
If DE = 6, BE = 3 and CE = 4, then
find AE.
PR
2
= and PQ = 12, then find XY.
XZ
3
A
C
E
D
Q.2.
B
Solve the following :
6
(i)
An arc of a circle having measure 36 has length 176 m. Find the
circumference of the circle.
(ii)
Find the side of square whose diagonal is 16 2 cm .
(iii)
A cone of height 24 cm has a plane base of surface area 154 cm2. Find its
volume.
In the adjoining figure,
P is the centre of the circle with
radius 18 cm. If the area of the
PQR is 100 cm2 and area of the
segment QXR is 13.04 cm2.
Find the central angle . ( = 3.14)
9
P
cm
(i)
Solve the following :
18
Q.3.
Q
R
X
Paper - V
... 2 ...
(ii)
Two circles intersect each other
C
in points A and B. Secants through
A and B intersects circles in C, D and
M, N. Prove that CM || DN.
A
M
(iii)
(i)
N
B
P
In the adjoining figure,
PQR = 90º.
T is the mid point of the side QR.
Prove that PR2 = 4PT2 – 3PQ2.
Q
Q.4.
D
R
T
Solve the following :
12
Let the circles with centre P and
Q touch each other at point A.
Let the extended chord AB intersect
E
the circle with centre P at point E
and the chord BC touches the circle
with centre P at the point D.Then prove
that ray AD is an angle bisector of the CAE.
P
B
A
D
M
Q
C
(ii)
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 ?
(iii)
ABC is a right angled triangle
C
with A = 90º. A circle is inscribed
in it. The lengths of the sides
containing the right angle are 6 cm
and 8 cm. Find the radius of the circle. 6 cm
N
A
Best Of Luck

M
O
L 8 cm
B
Paper - V
MAHESH TUTORIALS
S.S.C.
Batch : SB
A.1.
(i)
Marks : 30
GEOMETRY
Chapter : 1, 2, 6
Date :
Time : 1 hr. 15 min.
MODEL ANSWER PAPER
Solve the following :
F+V = E+2
 12 + V = 30 + 2
 12 + V = 32

V = 32 – 12

(ii)




1
V = 20
PQR ~ XYZ
PQ
PR
=
XY
XZ
12
2
=
XY
3
12  3
XY =
2
XY = 6 × 3
[Given]
[c.s.c.t.]
 XY = 18 units
(iii)
1
Chords AB and CD intersect each
other at point E inside the circle
 AE × BE = CE × DE
 AE × 3 = 4 × 6
 AE =
46
3
A
D
 AE = 8 units
A.2.
(i)
Test - III
E
B
1
Solve the following :
Length of arc (l)

measure of arc ()
=
=
l
=
176
=

C
176 m
36º

× 2r
360
36
× 2r
360
½
½
Paper - V
... 2 ...

176
=


176 × 10
2r
But, circumference
=
=
=
1
× 2r
10
2r
1760
2r
½
 Circumference of the circle is 1760 m.
(ii)
Given :
ABCD is a square.
½
A
AC = 16 2 cm
To find : Side of a square
x
Sol. ABCD is a square
[Given]
Let the sides of the square be x cm
In ABC,
B
m ABC = 90º
[Angle of a square]
 AC2
= AB2 + BC2
[By Pythagoras theorem]


2
 16 2
 256 × 2
 x2
 x2
 x
= x2 + x2
= 2x2
256  2
=
2
= 256
= 16
x
D
½
16 2 cm
x
C
x
[Taking square roots]
 The side of a square is 16 cm.
(iii)
½
Height of a cone (h) = 24 cm
Surface area of base = 154 cm2
1
Volume of a cone =
× Surface area of base × height
3
1
=
× r2 × h
3
1
=
× 154 × 24
3
= 1232 cm3
½
½
½
 Volume of the cone is 1232 cm3.
Area of sector P-QXR
P
= 18 cm
= 100 cm2
= 13.04 cm2
= Area of PQR +
Area of segment QXR Q
= 100 + 13.04
= 113.04 cm2
½
cm
Solve the following :
Radius of a circle (r)
Area of PQR
Area of the segment QXR
Area of sector P-QXR
½
18
A.3.
(i)
½
½
R
X
½
½
Paper - V
... 3 ...
Area of sector
 113.04
 11304
11304 × 360
 314 × 18 × 18
 

× r2
360

=
× 3.14 × 18 × 18
360

=
× 314 × 18 × 18
360
½
=
½
= 
= 40
 Central angle is 40º.
½
A
C
(ii)
D
M
Construction : Draw seg AB.
½
N
B
Proof : ABMC is cyclic
[By definition]
 m MCA + m MBA = 180º ......(i) [Opposite angles of a cyclic
1
quadrilateral are supplementary]
ABND is cyclic
[By definition]
 MBA ADN
.....(ii) [The exterior angle of a cyclic
1
quadrilateral is equal to its
interior opposite angle]
 m MCA + m ADN = 180º
[From (i) and (ii)]
 m MCD + m CDN = 180º
[C - A - D]
 seg CM || seg DN
[By Interior angles test]
½
(iii)
P
In PQR,
seg PT is the median
 PQ2 + PR2 = 2PT2 + 2QT2 ......(i)
In PQT,
m PQT = 90º
 PT 2
= PQ2 + QT2
2
 QT
= PT2 – PQ2
.....(ii)
2
2
2
2
 PQ + PR = 2PT + 2 (PT – PQ2)
 PQ2 + PR2 = 2PT2 + 2PT2 – 2PQ2
 PR2
= 4PT2 – 2PQ2 – PQ2
2
 PR
= 4PT2 – 3PQ2
[Given]
Q
T
[By Appollonius theorem]
R
1
[Given]
[By Pythagoras theorem]
[From (i) and (ii)]
1
½
½
Paper - V
... 4 ...
A.4.
(i)
Solve the following :
E
In MAD,
MA = MD
P
B
A
D
M
Q
C
[The lengths of two tangent
segments from an external point
to a circle are equal]
[Isosceles triangle theorem]
 m MAD = m MDA
Let,
m MAD = m MDA = xº ........(i)
CAM ABC
[Angles in alternate segments]
Let,
m CAM = m ABC = yº ........(ii)
[Angles Addition property]
m CAD = m CAM + m MAD
m CAD = (x + y)º
......(iii) [From (i) and (ii)]
DAE is an exterior angle of ADB
 m DAE = m ADB + mABD
[Remote Interior angles theorem]
 m DAE = m ADM + m ABC
[D - M - C - B]
 mDAE = (x + y)º
......(iv) [From (i) and (ii)]
 m CAD = mDAE
[From (iii) and (iv)]
 ray AD is an angle bisector of CAE.
½
½
½
½
½
½
½
½
(ii)
6.75 cm
20 cm
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
½
Height of the raised water level (h) = 6.75 m
Volume of water displaced = r 2h
½
=  × 12 × 12 × 6.75 cm3
 Volume of iron ball
=  × 12 × 12 × 6.75 cm3
½
Paper - V
... 5 ...
4 3
r
3
4
 × 12 × 12 × 6.75 =
×  × r3
3
12 × 12 × 6.75 × 3
= r3
4
r3 = 3 × 12 × 6.75 × 3
r3 = 3 × 3 × 3 × 4 × 6.75
r3 = 3 × 3 × 3 × 27
r = 3 3 × 3 × 3 × 3 × 3 × 3 [Taking cube roots]
r =3×3
r =9
But, Volume of iron ball =








½
½
½
½
½
 Radius of the iron ball is 9 cm.
(iii)





C
Construction : Let the sides AB, BC and
½
AC touch the circle at points L, M and N
respectively. Draw seg OA, seg OB,
M
seg OC, seg OL, seg OM and ON.
6 cm
Sol. In ABC,
N
[Given]
m BAC = 90º
O
2
2
2
BC = AC + AB
[By Pythagoras theorem]
½
A
B
2
2
2
BC = 6 + 8
L
8 cm
BC2 = 36 + 64
BC2 = 100
BC = 10 cm
.......(i)
[Taking square roots]
½
Let the radius of the circle be r
½
OL = OM = ON = r
.......(ii)
[Radii of the same circle]
seg OL  side AB
seg OM  side BC
[Radius is perpendicular to the tangent] ½
seg ON  side AC
Area of a triangle
=
1
× base × height
2

A (AOB)
=
1
× AB × OL
2

A (AOB)
=
.....(iii)

A (AOB)
Similarly,
A (BOC)
A (AOC)
= 5r
= 3r
.....(iv)
.....(v)
1
×8×r
2
= 4r
[From (ii)]
½
½
... 6 ...





Adding (iii), (iv) and (v),
A (AOB) + A (BOC) + A (AOC)
A (ABC)
1
× AB × AC
2
1
×8×6
2
24
r
Paper - V
= 4r + 5r + 3r [From (iii), (iv) and (v)] ½
= 12r
[Area addition property]
= 12r
= 12r
= 12r
=2
 The radius of the circle is 2 cm.

½
Paper - VI
S.S.C.
MAHESH TUTORIALS
Batch : SB
Date :
Q.1.
Test - III
Marks : 30
GEOMETRY
Chapter : 1, 2, 6
Time : 1 hr. 15 min.
Solve the following :
3
(i)
Perimeter of one face of a cube is 24 cm. Find the length of its side.
(ii)
In DEF, m D = 90º, m E = 45º, m F = 45º. If EF = 8 2 cm, find DE.
(iii)
If PB = 3, PD = 4, PA = 6, find PC.
A
B
P
D
Q.2.
C
Solve the following :
6
(i)
An arc of length 4 cm subtends an angle of measure 40º at the centre.
Find the radius and the area of the sector formed by this arc.
(ii)
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.
(iii)
The curved surface area of a cone is 4070 cm2 and its diameter is 70 cm.
What is its slant height ?
Q.3.
Solve the following :
9
(i)
The length, breadth and height of a cuboid are in the ratio 5:4:2. If the total
surface area is 1216 cm2, find the dimensions of the solid.
(ii)
ABCD is a parallelogram.
A circle passing through D, A,
B cuts BC in P.
Prove that DC = DP.
D
A
C
P
B
Paper - VI
... 2 ...
(iii)
Q.4.
Adjacent sides of a parallelogram are 11 cm and 17 cm. If the length of one
of its diagonals is 26 cm. Find the length of the other.
Solve the following :
12
(i)
Suppose AB and AC are equal chords of a circle and a line parallel to the
tangent at A intersects the chords at D and E. Prove that AD = AE.
(ii)
Adjoining figure depicts a racing track .
whose left and right ends are semicircular.
The distance between two inner parallel
line segments is 70 m and they are each
105 m long. If the track is 7 m wide, find
the difference in the lengths of the inner
edge and outer edge of the track.
(iii)
Seg AD is the median of ABC,
and AM  BC. Prove that :
70 m
70 m
105 m
7m
2
2
 BC 
(b) AB = AD – BC × DM + 

 2 
2
105 m
A
 BC 
(a) AC = AD + BC × DM + 

 2 
2
7m
2
B
2
Best Of Luck
M

D
C
Paper - VI
MAHESH TUTORIALS
S.S.C.
Batch : SB
Date :
A.1.
(i)
Marks : 30
GEOMETRY
Chapter : 1, 2, 6
Time : 1 hr. 15 min.
MODEL ANSWER PAPER
Solve the following :
Perimeter of one face of a cube
Perimeter of one face of a cube

24

l

l
Test - III
= 24 cm
= 4l
= 4l
24
=
4
= 6
 The length of the side of a cube is 6 cm.
(ii)
1
In DEF,
D = 90º, E = 45º, F = 45º
[Given]
 DEF is 45º - 45º - 90º triangle
1
 EF
 DE =
2
1
8 2
 DE =
2
 DE = 8 cm
(iii)





A.2.
(i)
1
Chords AB and CD intersect each
other at point P outside the circle.
PA × PB = PC × PD
6 × 3 = PC × 4
P
63
PC =
4
9
PC =
2
PC = 4.5 units
Solve the following :
Length of arc (l)

measure of arc ()
l
= 4 cm
= 40º

=
× 2r
360
A
B
D
C
1
½
... 2 ...



4
4×9
2
r
Area of the sector
=
Paper - VI
40
×2××r
360
= r
½
= 18 cm
l ×r
=
2
4 × 18
=
2
= 36cm2
½
 Radius of the circle is 18 cm and Area of the sector is 36cm2.
(ii)





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
B
AB = 8 m
In ABC,
[Given]
m ABC = 90º
2
2
2
AC
= AB + BC
[By Pythagoras theorem]
(10)2 = (8)2 + BC2
100 = 64 + BC2
BC2 = 100 – 64
BC2 = 36
[Taking square roots]
BC
= 6m
½
½
C
 The distance of the foot of the ladder from the base of the wall is 6 m.
(iii)
Diameter of a cone

Its radius (r)
Curved surface area of a cone
Curved surface area of a cone



4070
4070
22 × 5
l
 Slant height of a cone is 37 cm.
½
½
½
= 70 cm.
70
=
2
= 35 cm
= 4070 cm2
= rl
22
 35  l
=
7
½
½
= l
½
= 37
½
Paper - VI
... 3 ...
A.3.
(i)
Solve the following :
Ratio of the length, breadth and height of a cuboid is 5 : 4 : 2
Let the common multiple be ‘x’

Length of a cuboid = 5x cm
its breadth = 4x cm
and its height = 2x cm
Total surface area of a cuboid = 1216 cm2
Total surface area of a cuboid = 2 (lb + bh + lh)

1216 = 2 [(5x) (4x) + (4x) (2x) + (5x) (2x)]
1216

= 20x2 + 8x2 + 10x2
2

608 = 38x2
608

= x2
38

x 2 = 16

x =4
[Taking square roots]
Length of a cuboid = 5x
= 5 (4)
= 20 cm
its Breadth = 4x
= 4 (4)
= 16cm
and its height = 2x
= 2 (4)
= 8 cm
 Dimensions of a cuboid are 20 cm, 16 cm and 8 cm.
(ii)
D
ABPD is cyclic
 DPC DAB
.....(i)
C
P
½
½
½
1
B
[By definition]
A
[An exterior angle of cyclic quadrilateral
is congruent to the angle opposite to
adjacent interior angle]
ABCD is parallelogram
 DCB DAB .......(ii) [Opposite angles of a parallelogram
are congruent]
 DPC DCB
......(iii) [From (i) and (ii)]
In DPC,
DPC DCP
[From (ii) and C - P - B]
 seg DP  seg DC
[Converse of isosceles triangle theorem]
 DP = DC
½
1
1
½
½
Paper - VI
... 4 ...
ABCD is a parallelogram [Given]
1
× BD
.....(i)
OB = OD =
2
(iii)












1
× 26
2
OB = OD = 13 cm
In ADB,
seg AO is the median
AB2 + AD2 = 2AO2 + 2OB2
(11)2 + (17)2 = 2AO2 + 2 (13)2
121 + 289 = 2AO2 + 2 (169)
410 = 2AO2 + 338
410 – 338 = 2AO2
72 = 2AO2
AO2 = 36
AO = 6 cm
1
AO =
× AC
2
OB = OD =
[ Diagonals of paralle log ram
bisec t each other]
½
[Given]
½
[From (i) and by definition]
[By Appollonius theorem]
½
A
11 cm
B
17 cm
D
O
[Taking square roots]
C
½
[ Diagonals of paralle log ram
bisec t each other]
1
× AC
2
AC = 12 cm
6 =
 Length of other diagonal is 12 cm.
A.4.
(i)
½
Solve the following :
C
B
Construction : Draw seg BC.
Proof : Take points R and S on the tangent at A
as shown in the figure
line DE || line RS
[Given]
E
D
 On transversal AD,
•
EDA  DAR
[Converse of
A
S
R
alternate angles test]
 EDA  BAR
.......(i)
[ B - D - A]
BAR  BCA
.......(ii)
[Angles in alternate segment]
 EDA  BCA
......(iii)
[From (i) and (ii)]
Similarly, we can prove that
DEA  CBA
......(iv)
In ABC,
seg AB  seg AC
[Given]
 BCA  CBA
........(v) [Isosceles triangle theorem]
In DEA,
EDA  DEA
[From (iii), (iv) and (v)]
 seg AD  seg AE
[Converse of isosceles triangle theorem]

AD = AE
½
½
½
½
½
½
½
1
Paper - VI
... 5 ...
(ii)
7m
105 m
70 m
70 m
105 m
7m
Diameter of inner circular edge (d1) = 70 m
Width of the track
= 7m
Diameter of outer circular edge (d2) = 70 + 7 + 7
= 84 m
The inner and outer edges of the racing tracks comprises of two
semicircles and parallel segments of length 105 m each
1
1
 Length of outer edge
=
d2 + 105 +
d2 + 105
2
2
= d2 + 210
= (84 + 210) m
Length of inner edge
Difference in the lengths of
inner and outer edge
½
½
½
½
1
1
d1 + 105 +
d1 + 105
2
2
= d1 + 210
= (70 + 210) m
½
=
½
= (84 + 210) – (70 + 210)
½
= 84 + 210 – 70 – 210
= 14
22
= 14 ×
7
= 44 m
½
 The difference in the lengths of inner edge and outer edge of the
track is 44 m.
(iii)
(a)
In AMD,
m AMD = 90º
AD2 = AM2 + DM2
......(i)
In AMC,
m AMC = 90º
 AC2 = AM2 + MC2
 AC2 = AM2 + (DM + DC)2
A
B
M
D
C
[By Pythagoras theorem]
½
[Given]
[By Pythagoras theorem]
[ M - D - C]
½
... 6 ...
Paper - VI
 AC2 = AM2 + DM2 + 2DM × DC + DC2
 AC2 = AM2 + DM2 + 2DC × DM + DC2
 BC 
 AC = AD + BC × DM + 

 2 
2
2
2
[ From (i) and D is the midpo int
of side BC ]
(b) In AMB,
[Given]
m AMB = 90º
 AB2 = AM2 + BM2
[By Pythagoras theorem]
 AB2 = AM2 + (BD – DM)2
 AB2 = AM2 + BD2 – 2BD × DM + DM2
 AB2 = AM2 + DM2 – 2BD × DM + BD2
2
[ From (i) and D is the midpo int
 BC 
 AB2 = AD2 – BC × DM + 

of side BC ]
 2 

½
½
1
½
½