TNPSC 2023 -24

GROUP I,II,IV & VAO

ê˜è£˜ ä.ã.âv Üè£ìI

èEî‹ ¬è«ò´
Ü÷¬õèœ / Mensuration

Anna Nagar West, Chennai - 600012

Phone: 9962600037, 9962600038
www.sarkariasacademy.com
 TNPSC èí‚°
Ü÷¬õèœ
Mensuration

ňFóƒèœ (Formulas)

Area (ðóŠ¹) Perimeter (²Ÿø÷¾)

õ®õ‹
(Sq.units) (Units)
ê¶ó‹ : Square a2 (or)
a 1 2
d
2
a a 4a
Þƒ°, d = a 2
(d = ͬôM†ì‹)(Diagonal)
a
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

ªêšõè‹ : Rectangle lxb

l l = c÷‹
b = Üèô‹ 2(l+b)
b b
d = l2+b2
l
º‚«è£í‹ : Triangle
A
1 bh
AB+BC+CA
h 2

B b C
êñð‚è º‚«è£í‹ :
Equilateral Triangle
3a
A
3 a2 àòó‹ :
a h a 4 3
h= 2 a

B a C
Þ¼êñð‚è º‚«è£í‹ :
Isosceles Triangle
A
h a2-h2 2a + 2 a2-h2
a h a

B C
Üêñð‚è º‚«è£í‹ :
Scalene Triangle
A
S(S-a) (S-b) (S-c) a+b+c (or)
a+b+c 2S
c b ÞF™, S = 2

B a C

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Þ¬íèó‹ :
Parallelogram
b
bh 2(a+b)
a h a

b
꣌ê¶ó‹ : Rhombus
a a
d1 1 4a
d xd
2 1 2
d2
a a

êKõè‹ : Trapezium
b
1 AB+BC+CD+DA
2
h(a+b)
h
a
èó‹ : Quadrilaterals

ê˜è£˜ ä.ã.âv Üè£ìI

C
D h2
1 AB+BC+CD+DA
2
d(h1+h2)
h1 d

A B
õ†ì‹ : Circle

2πr (or)
r πr2 44 r
360o 7

ܬóõ†ì‹ : Semi-Circle
r r
(πr+2r) (or)
1 2
180o πr 36r
2
7

裙õ†ì‹ : Quadrant

( πr2 +2r) (or)

r
90o πr2
4 25 r
r 7
πr
2

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õ†ì«è£íŠð°F :
Sector of a circle θ x πr2 (or) lr
360o 2
Þƒ°,
O θ l + 2r
l = 360o x 2πr
θ l = M™L¡ c÷‹ (length of
P Q arc)
l

ðóŠ¹ (Sq.units) èù Ü÷¾

õ®õ‹ ð‚èŠðóŠ¹ ªñ£ˆîŠðóŠ¹ (Volume)
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

(CSA/LSA) (TSA) (Cu.units)

èùê¶ó‹ : Cube

V= a3
4a2
6a
2 ͬôM†ì‹
a a d= 3a
a
a
a

èùªêšõè‹ : Cuboid
V= lbh
2h(l + b) 2(lb+bh+hl)
h d = l2+b2+h2
b
L

˹ (Cone)
π rl π rl + π r 2
l l2 = h2 + r2
h l= Slant = π r (l + r) 1 2
πr h
r height 3

(i) Circumference of the cone (base) = 2π r

(ii) Area of the cone (base) = π r 2

༬÷
(Cyclinder)

2π rh 2π r (h + r ) π r 2h
h

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«è£÷‹ (Sphere)

4π r2 4π r2 4 3
πr
r 3

ܬó‚«è£÷‹ (Hemi-
sphere)
r
2π r 2 3π r 2 2 3
πr
3

àœkìŸø «è£÷‹
(Hollow Sphere)
4π r 2
R
R - Outer Radius 4π ( R 2 + r 2 ) 4
π ( R3 − r 3 )
r r - Inner Radius 3

ê˜è£˜ ä.ã.âv Üè£ìI

àœkìŸø ༬÷
(HollowCylinder)
R

h 2π ( R + r )
2π ( R + r ) h π (R2 − r 2 ) h
(R − r + h)

àœkìŸø
ܬó‚«è£÷‹ (Hollow
Hemi Sphere)

R 2π ( R 2 + r 2 ) π (3R 2 + r 2 ) 2
r
π ( R3 − r 3 )
3

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Frustum of Right circular

cone
r π ( R + r )l
π ( R + r )l + 1
where, πh
h π R2 + π r 2 3
l= h2 + ( R − r )2 [ R 2 + r 2 + Rr ]
R
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

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ê¶ó‹ ñŸÁ‹ ªêšõè‹

Square and Rectangle

Area (ðóŠð÷¾)

♦ Square (ê¶ó‹)
Area of a square (ê¶óˆF¡ ðóŠð÷¾) = a2
Perimeter of a square (ê¶óˆF¡ ²Ÿø÷¾) = 4a
Diagonal of a square (ê¶óˆF¡ ͬôM†ì‹) d = 2a
♦ Rectangle (ªêšõè‹)
ªêšõèˆF¡ ðóŠð÷¾ = c÷‹ x Üèô‹
Area of Rectangle =lxb
ªêšõèˆF¡ ²Ÿø÷¾ = 2 (c÷‹ + Üèô‹)
Perimeter of Rectangle = 2 (l + b)
Note(°PŠ¹)

ê˜è£˜ ä.ã.âv Üè£ìI

(i) (l + b) 2 = l2 + 2lb + b 2
d - diagonal (ͬôM†ì‹)
(ii) d 2 = l2 + b 2
ªêšõèˆF¡ èùÜ÷¾ V= l x b x h
Area of four walls = 2 (l+b) x h
CSA of Cuboid or
(Lateral surface area of cuboid) = 2h (l+b)

b=h
h  b 
l b l
l = 2πr
 Area of cloth required
= LSA + area of the top
= 2h (l+b) +lb
volume of cube
 Rise in water = (dimension)
Area of base vessel

èùê¶óˆF¡ èùÜ÷¾
î‡a˜ àò¼‹ Ü÷¾ = ð£ˆFóˆF¡ Ü®Šð£èˆF¡
ðKñ£í‹

 Surface area of the Floor = Total Cost2

cost per m
ªñ£ˆî ªêô¾
¬ñî£ùˆF¡ ðóŠð÷¾ = å¼ ê.Ü®‚°
Ý°‹ ªêô¾

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1. 40e c÷‹, 30e. Üèô‹ à¬ìò ªêšõè õ®õ õòL¡ å¼ Í¬ôJ™ 7e c÷ºœ÷
èJÁ å¡P™ °F¬ó è†ìŠð†´œ÷¶. °F¬ó GôˆF™ «ñŒ‰î ÞìˆF¡ ðóŠð÷¾
裇è.
A horse is tethered to one corner of a rectangular field of dimensions 40 m by 30 m by a rope 7 m
long for grazing. How much area can the horse left ungrazed?
land

30m

7 e
c÷ºœ÷ 40m
èJÁ
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

π r2
裙õ†ìˆF¡ ðóŠð÷¾ =
4
22 7x7
= x
7 4
77
=
2

= 38.5m 2
°F¬ó «ñò£î ÞìˆF¡ ðóŠð÷¾
= 40 x 30 = 1200
1200 - 38.5 = 1161.5 m2

2. If the side of a square is increased by 20%. Then its area is increased by

å¼ ê¶óˆF¡ ð‚è‹ 20% ÜFèK‚Aø¶, âQ™ Üî¡ ðóŠ¹ âˆî¬ù êîiî‹ ÜFèK‚°‹.
20%
120
= 10 x
100
= 12
Area = a x a : a x a
10 x 10 : 12 x 12
100 : 144
144 - 100 = 44% ðóŠ¹ ÜFèK‚°‹.

3. If the length of a Rectangle is Increased by 20%. Then its area increased by

å¼ ªêšõèˆF¡ c÷‹ 20% ÜFèK‚Aø¶ âQ™ âî¡ ðóŠ¹ âˆî¬ù % ÜFèK‚°‹ -?
20%
120
= 10 x
100
= 12
Area = l x b : l x b

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10 x 10 : 12 x 10
100 : 120
120 - 100 = 20% ÜFèK‚°‹.
4. 24ªê.e c÷ºœ÷ å¼ è‹H å¼ ê¶ó õ®õˆFŸ° õ¬÷‚èŠð´Aø¶. ܉î ê¶óˆF¡
ðóŠð÷¬õ‚ 致H®‚辋?
A wire of length 24 cm is bent to form a square find the area of the square formed
A) 24ê.ªê.e B) 16ê.ªê.e
C) 36ê.ªê.e D) 49ê.ªê.e
Perimeter of a square(ê¶óˆF¡ ²Ÿø÷¾) = 4a
4a = 24
a = 24
4
a = 6 ªê.e
Area of a square(ê¶óˆF¡ ðóŠð÷¾) = a2
= 6x6
= 36 ê.ªê.e

ê˜è£˜ ä.ã.âv Üè£ìI

5. 60 ªê.e ñŸÁ‹ 40 ªê.e ð‚èƒè¬÷‚ ªè£‡ì å¼ ªêšõè õ®õ ܆¬ìˆ î£O™
Þ¼‰¶ 2 ªê.e ð‚è‹ ªè£‡ì âˆî¬ù ê¶óƒè¬÷ à¼õ£‚è º®»‹?
How many square of 2 cm can be made from the rectangular card sheet whose side are 60 cm
and 40 cm
A) 400 B) 300 C) 600 D) 480
ªêšõèˆF¡ ðóŠð÷¾
«î¬õò£ù ê¶óƒèO¡ } =
ê¶óˆF¡ ðóŠð÷¾
â‡E‚¬è
60 x 40
= 2x2
= 30 x 20
= 600

6. 
å¼ Mõê£J î¡Â¬ìò Gôˆ¬î 3 õK¬êèœ è‹H¬ò‚ ªè£‡´ «õLJì
M¼‹¹Aø£˜. ܉î GôˆF¡ Ü÷¾èœ 120e ñŸÁ‹ 68e âQ™ ªñ£ˆîñ£èˆ
«î¬õŠð´‹ è‹HJ¡ c÷ˆ¬î‚ èí‚A쾋?
A farmer wishes to fence his field with 3 rows of wire if the dimension of the field are 120 m
by 68 m find the total length of wire required
A) 1128e B) 346e C) 1200e D) 498e
ªêšõèˆF¡ ²Ÿø÷¾ = 2 (l + b)
«î¬õò£ù è‹HJ¡ 68e

c÷‹ = 3x2 (l + b) 120e

= 3x2 (120 + 68) l = 120e, b = 68e
= 6x188
= 1128e

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7. 
å¼ ªêšõèˆF¡ ðóŠð÷¾ 72 ê.ªê.e ñŸÁ‹ Üî¡ Üèô‹ 6ªê.e âQ™, ܉î
ªêšõèˆF¡ c÷‹ _______
Find the length of the rectangle whose area and breadth are 72 cm2 and 6 cm
A) 15 ªê.e B) 12 ªê.e C) 18 ªê.e D) 6 ªê.e
A = 72 ê.ªê.e, b = 6ªê.e
Area of a rectangle(ªêšõèˆF¡ ðóŠð÷¾) = l x b
lxb = 72
lx6 = 72
l = 726
l = 12ªê.e
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

8. ê¶óˆF¡ ²Ÿø÷¾ 120e âQ™, Üî¡ ðóŠð÷¬õ‚ 致H®‚辋

The perimeter of the square is 120 m then find its area
A) 800e2 B) 900e2 C) 1000e2 D) 700e2
ê¶óˆF¡ ²Ÿø÷¾ = 4a
4a = 120
120
a = 4
a = 30e
ê¶óˆF¡ ðóŠð÷¾ = a2
= 30 x 30
= 900 ê.e (Ü)
= 900 e2
9. 
5e c÷º‹ 4 e Üèôº‹ ªè£‡ì å¼ Ü¬øJ¡ î¬óŠð°F¬ò ê¶ó æ´è¬÷‚
ªè£‡´ GóŠ¹õ CH M¼‹¹Aø£˜. å¼ ê¶ó 憮¡ ðóŠð÷¾ 1 ê.e âQ™,
2
܉î ܬøJ¡ î¬óŠ ð°F¬ò º¿õ¶ñ£è GóŠ¹õˆ «î¬õŠð´‹ æ´èO¡
â‡E‚¬è¬ò‚ 致H®‚辋.
Sibi wants to cover the floor of a room 5 m long and width 4 m by square tiles. If area of each
square tiles is ½ m2 then find the number of tiles required to cover the floor of a room.
A) 20 B) 30 C) 40 D) 50
1
ê¶óˆF¡ ðóŠð÷¾ = 2 ê.e
ªêšõèˆF¡ ðóŠð÷¾ = l x b
(î¬ó)
«î¬õò£ù æ´èO¡
â‡E‚¬è
} = ªêšõèˆF¡ ðóŠð÷¾
ê¶óˆF¡ ðóŠð÷¾

= 5x14
2
20
= 1
2
= 40
Exercise sums:

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1. The Minimum no. of square size tiles to fill the floor of a room. Measuring 9 m x 6 52 m, is
(A) 1240 (B) 1400 (C) 1440 (D) 1660
å¼ Ü¬øJ¡ î¬ó¬òŠ «ð£´õîŸè£ù e„CÁ â‡E‚¬èJô£ù ê¶ó êô¬õ‚ èŸèœ
2
â‡E‚¬è 裇è. ܬøJ¡ c÷ Üèôƒèœ 9e x 6
5
e
Solution:
Room Size l = 9m
b = 6.4m
Area = 9 x 6.4 = 57.6 m2
HCF (9, 6.4)
2 90, 64
45, 32

HCF = 2
Area of the tiles = 0.2 x 0.2
0.04 m2

ê˜è£˜ ä.ã.âv Üè£ìI

57.6
Minimum no.of square tile =
0.04
Answer = 1440

2.If the side of a square is increased by 5cm, then the area increases by 165 sq.cm. The side of the
square is
å¼ ê¶óˆF¡ ð‚èˆF¡ c÷ˆ¬î 5 ªê.e. ÜFèK‚°‹ «ð£¶ Üî¡ ðóŠ¹ 165 ê.ªê.e.
ÜFèK‚Aø¶ âQ™, ܉î ê¶óˆF¡ ð‚èˆF¡ c÷‹
A) 13 cm B) 14 cm C) 33 cm D) 12 cm
Solution:
New side = a + 5
New area = (a + 5) ( a+ 5)
(a + 5)2 = a2 x 165
a2 + 10a + 25 = a2 x 165
10a = 165 - 25
10a = 140

14
140
a=
10

Answer: a = 14 cm

3. What is the least number of square marbles required for a terrace of 15. 17m long and 9.02 m breadth?
15.17 e c÷º‹ 9.02 e Üèôº‹ à¬ìò å¼ Ã¬óJ¡ «ñŸðóŠH™ ðF‚èˆ «î¬õŠð´‹
°¬ø‰îð†ê â‡E‚¬èJô£ù ê¶ó õ®õ ðOƒ°‚èŸèœ âˆî¬ù-?

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A) 1242 B) 407 C) 814 D) 1000

Solution:
Terrace size,
l = 15.17m = 1517 cm
b = 9.02 m = 902 cm
Area = (15.17 x 9.02) x 100
= 1368334 cm2
HCF, 1517 & 902 = 41
2 902 37 1515
11 451 41 41
41 41 1
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

1
Area of the marbles = 41 x 41 = 1681

1368334
Minimum no. of square tiles =
1681

= 814
Answer = 814

4. A Square and a rectangle have equal areas. If their perimeters are P1 and P2 respectively then
å¼ ê¶ó‹ ñŸÁ‹ ªêšõèˆF¡ ðóŠð÷¾èœ êñ‹. ÜõŸP¡ ²Ÿø÷¾èœ º¬ø«ò P1
ñŸÁ‹ P2 âQ™
A) P1 ≤ P2 B) P1 = P2 C) P1 > P2 D) P1 ≥ P2
Solution:
Area
axa=lxb
a2 = lb
a = lb
Perimeter
p1 = p2
4a = 2 (l + b)
2a = (l + b)

l +b
a=  
 2 
l +b
lb =
2

Answer : p1 ≤ p2

5. A square is inscribed in a circle whose radius is 4 cm. The area of the portion between the circle
and the square is
4 ªê.e Ýó‹ ªè£‡ì å¼ õ†ìˆFŸ°œ å¼ ê¶ó‹ õ¬óòŠð´AøªîQ™ õ†ìˆFŸ°‹

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ê¶óˆFŸ°‹ Þ¬ìŠð†ì ð°FJ¡ ðóŠ¹

(A) 16 π - 32 cm2 (B) 32 π - 27 cm2 (C) 20 π +11 cm2 (D) 12 π - 4 cm2
Solution:
Circle (Area) = π r 2
= π (4)
2

= 16 π
Square (Area) = ½ d2
= ½ x (8)2

1
= x 6432
2
= 32

Answer: 16 π - 32 cm2

6.The perimeters of two squares are 40 cm and 32 cm. Find the perimeter of a third square whose
area is equal to the difference of the areas of the two squares.
Þó‡´ ê¶óƒèO¡ ²Ÿø÷¾èœ º¬ø«ò 40 ªêe ñŸÁ‹ 32 ªêe. ÞšM¼ ê¶óƒèO¡

ê˜è£˜ ä.ã.âv Üè£ìI

ðóŠð÷M¡ MˆFò£êˆ¬îŠ ðóŠð÷õ£è‚ ªè£‡ì Í¡ø£õ¶ ê¶óˆF¡ ²Ÿø÷¾ â¡ù-?
A) 40 cm B) 36 cm C) 12 cm D) 24 cm
Solution
Perimeter of 1st square
4a = 40
a = 10
Perimeter of 2nd square
4a = 32
a=8
Area of 3rd square = Area of 1st square - Area of 2nd square
= 100 - 64
a2 = 36
a=6
Perimeter of 3rd square = 4 x 6
= 24 cm
Answer : 24 cm

7. A drain cover is made from a square metal plate of side 40 cm having 70 circular holes of diameter
1 cm each drilled in it. find the area of the cover.
40 ªê.e ð‚躜÷ å¼ ê¶ó à«ô£èˆ ™ 1 ªê.e M†ìºœ÷ 70 õ†ìˆ ¶¬÷èœ
ÞìŠð†´ å¼ è£™õ£Œ Í® ªêŒòŠð´Aø¶ âQ™ ܉î Í®J¡ ðóŠ¬ð‚ 裇
A) 1380 cm2 B) 1545 cm2 C) 1655 cm2 D) 1820 cm2
Solution
Area of square metal plate = 40 x 40

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= 1600 cm2
Area of each hole = π r 2

2 22
= 7 x (0.5)

10 22
Area of 70 hole = 70 x x 0.25
7
5 11 1
= 10 x 22 x
42

= 5 x 11 = 55
Area of the remaining square plate = 1600 - 55 = 1545 cm2
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

Answer: 1545 cm2

8. A chess board contains 64 equal squares and area of each square is 6.25 sq.cm. A border around
the board is 2 cm wide. Then the length of the side of the chess board is . :
å¼ ê¶óƒè ܆¬ìJ™ 64 êñ Ü÷¾ ê¶óƒèœ àœ÷ù. 嚪õ£¼ ê¶óˆF¡ ðóŠð÷¾‹
6.25 ê.ªê.e. ܆¬ìJ™ ²ŸP½‹ 2 ªê.e ÜèôˆFŸ° è¬ó (Border) MìŠð†´œ÷¶
âQ™ ê¶óƒè ܆¬ìJ¡ ð‚èˆF¡ Ü÷¾ â¡ù?
(A) 20 cm (B) 22 cm (C) 24cm (D) 21 cm
solution:
x=?
equal square Area = 6.25 cm2
a2 = 6.25
a = 6.25
a = 2.5 cm
64 equal squares = 8 x 8 grid
= 8 x 2.5 = 20
Answer = 24 cm
The length of the side of the chess board = 24 cm

9. 400 square metres have to be painted. 1 litre of paint A costs 150 Rs and can be used to paint 4
sq.m. 1 litre of paint B costs Rs. 200 and can paint 6 sq.m. Paint C costs Rs. 250 per litre and each
litre paint C can be used to paint 8 sq.m. least cost of painting 400 sq.m. is
400 ê¶ó e†ì˜ ðóŠð÷¾ õ˜í‹ Ìê «õ‡®»œ÷¶. A õ¬è ªðJ‡† L†ì¼‚° 150
Ïð£Œ; 1 L†ì˜ ªè£‡´ 4 ê¶ó e†ì˜ Ìêô£‹. B õ¬è L†ì˜ 200 Ïð£Œ; L†ì¼‚°
6 ê¶ó e†ì˜ Ìêô£‹. C õ¬è L†ì˜ 250 Ïð£Œ; L†ì¼‚° 8 ê¶ó e†ì˜ Ìêô£‹. Iè
°¬ø‰î M¬ôJ™ 400 ê¶ó e†ì¬ó õ˜í‹ Ìê Ý°‹ ªêô¾
(A) Rs. 10,000 (B) Rs. 12,500 (C) Rs. 15,000 (D) Rs. 16,250
Solution:
Area = 400 sq.metres
A type

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l sq.m
1 4
x 400
B type
l sq.m
1 6
x 400
C type
l sq.m
1 8
x 400

4x = 400 6x = 400 8x = 400

400 400 400

x= = 100 x= = 66.6 x= = 50l
4 6 8

1 l = 150 Rs 1l = 200 Rs 1l = 250 Rs

ê˜è£˜ ä.ã.âv Üè£ìI

100 x 150 = 15000 200 x 66.6 = 13320 50 = 250 = 12500

10. Three solid cubes of sides 1 cm, 6 cm and 8 cm are melted to form a new cube. Find the total
surface area of the cube so formed?
1 ªê.e 6 ªê.e ñŸÁñ 8 ªê.e ð‚èƒè¬÷ à¬ìò Í¡Á èù ê¶óƒè¬÷ ༂A å¼
¹Fò èù ê¶ó‹ ªêŒî£™, Üî¡ ªñ£ˆî ¹ø ðóŠð÷¾ â¡ù-?
A) 384 cm2 B) 456 cm2 C) 486 cm2 D) 430 cm2
Solution:
V = a3
V1 = a13 = 13 = 1 cm3
V2 = a23 = 63 = 216 cm3
V3 = a33 = 83 = 512 cm3
V = V1 + V2 + V3
1 + 216 + 512 = 729 cm3
a3 = 729
a = 729 = 9
Surface area of cube = 6a2
=6x9x9
Answer: = 486 cm2

11. The sides of 14 Square’s are 11 cm, 12 cm, 13 cm, ......... 24 cm, then find the total area of 14
squares.
11 ªê.e, 12 ªê.e, 13 ªê.e............. 24 ªê.e ÝAòùõŸ¬ø º¬ø«ò ð‚è Ü÷¾è÷£è‚
ªè£‡ì 14 ê¶óƒèO¡ ªñ£ˆîŠ ðóŠ¹ è£‡è.

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(A) 3515 cm2 (B) 4515 cm2 (C) 2115 cm2 (D) 3215 cm2
Solution:
112 + 122 + 132 + ............ 242
(12 + 22 + ...... 242) - (12 + 22 + 32 + ........ 102)

= 24 x 25 x 49 −−10
10x11x21
x 11 x 21
6 62
= 4900 - 385
= 4515 cm2
Answer = 4515 cm2
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

12. If the length of the diagonal of a square is 20 cm, then its perimeter must be
å¼ ê¶óˆF¡ ͬôM†ì c÷‹ 20 ªê.e âQ™ ê¶óˆF¡ ²Ÿø÷¾
A) 10 2 cm B) 40 cm C) 40 2 cm D) 200 cm
Solution:
d = 2.a
20 = 2.a

20 2 20 2
a= =
2x 2 2

a = 10 2
Perimeter = 4a
= 4 x 10 2
Answer: = 40 2 cm

13.
Find the cost of fencing a square park of side 250 m at the rate of Rs. 20 per meter.
250 e†ì˜ ð‚è Ü÷¾ ªè£‡ì, ê¶ó õ®Mô£ù ̃è£MŸ° «õL ܬñ‚è e†ì˜
å¡Á‚° Ï. 20 ªêô¾ ªêŒî£™ «õL ܬñŠð Ý°‹ ªñ£ˆî ªêô¾ â¡ù?
A) Rs. 10, 000 B) Rs. 5,000 C) Rs. 15,000 D) Rs. 20,000
Solution:
Perimeter = 4a
= 4 x 250 = 1000
1 m = Rs.20
1000 m = x
x = 1000 x 20 = 20000
Answer = Rs.20000

14. If the diagonal of a square is 10 cm, then the side of the square is

16
 TNPSC èí‚°

å¼ ê¶óˆF¡ ͬôM†ì‹ 10cm âQ™ Üî¡ ð‚è Ü÷¾

(A) 5 2 cm (B) 2 2 cm (C) 3 5 cm (D) 5 3 cm
Solution:
d = 2.a
10 = 2.a

10 x 2 10 2
=a = =5 2
2x 2 2

Answer = 5 2 cm

15. What is the total area of eight squares whose sides are respectively 5cm , 6cm, 7 cm , ........,12cm
5 ªêe, 6 ªêe, 7 ªêe ,,,,,12 ªêe ð‚è Ü÷¾è¬÷‚ ªè£‡ì ↴ ê¶óƒèO¡
ðóŠð÷¾èO¡ Ã´î™ â¡ù?
A) 650 cm2 B) 620 cm2 C) 600 cm2 D) 675 cm2
Solution:

ê˜è£˜ ä.ã.âv Üè£ìI

52 = 25
62 = 36
72 = 49
82 = 64
92 = 81
102 = 100
112 = 121
122 = 144
Add the overall numbers = 620
or
(52 + 62+ 72 + ........ 122)
(12 + 22 + 32 + ....... 122) - (12 + 22 + 32 + 42)

12 x 13 x 25 4 x 5 x 9
−
6 6
= 650 - 30= 620

16. The ratio of the area of a square to that of the square drawn on its diagonal is
å¼ ê¶óˆF¡ ðóŠHŸ°‹, Üî¬ìò ͬôM†ì‹ õNò£è õ¬óòŠð´‹ ê¶óˆF¡
ðóŠHŸ°‹ àœ÷ MAîñ£ù¶
(A) 1:1 (B) 1:2 (C) 1:3 (D) 1:4
Solution:

axa a2 1
= = = 2
2a x 2a 2a 2

17
 TNPSC èí‚°

=1:2
Answer = 1 : 2

17.
If the side of a square be increased by 4 cms, the area increases by 60 sq.cms. The side of the
square is
å¼ ê¶óˆF¡ ð‚è‹ 4 ªê.e ÜFèK‚°‹ «ð£¶ Üî¡ ðóŠð÷¾ 60 ê.ªê.e ÜFèK‚Aø¶
âQ™ ܉î ê¶óˆF¡ ð‚è Ü÷¾ â¡ù?
(A) 12 cm (B) 13 cm (C) 14 cm (D) 5.5 cm
Solution
New side = (a + 4)
New area = (a + 4) (a + 4)
(a + 4)2 = a2 + 60
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

a2 + 8a + 16 = a2 + 60
8a = 60 - 16

44
a=
8
11
a=
2

Answer: a = 5.5 cm

18. What would be the measure of the diagonal of a square whose area is equal to 882 cm2?
882 ªê.e2 ä ðóŠð£è‚ ªè£‡ì ê¶óˆF¡ ͬôM†ìˆF¡ Ü÷¾ â¡ù?
(A) 38 cm (B) 32 cm (C) 42cm (D) 48 cm
Solution:
Area = ½ d2
882 = ½ d2
d2 = 2 x 882
d2 = 1764
d = 42
Answer = 42 cm

19. å¼ ªêšõèˆF¡ c÷‹ 10% ÜFèK‚èŠð´Aø¶ ñŸÁ‹ Üî¡ Üèô‹ 10%

°¬ø‚èŠð´Aø¶ âQ™, ¹Fò ªêšõèˆF¡ ðóŠð÷¾
The length of a rectangle is increased by 10% and its breadth is decreased by 10%. The area of
the new rectangle
(A) ñ£ø£ñL¼‚°‹ / remains same (B) 1% ÜFèK‚°‹ / is increased by 1%
(C) 10% °¬ø»‹ / is decreased by 10% (D) 1% °¬ø»‹ / is decreased by 1%
Solution:
l increased by 10%

18
 TNPSC èí‚°

110
10 x = 11
10 0
b decreased by 10%
90
10 x =9
10 0
lxb=lxb
10 x 10 = 11 x 9
100 = 99
1% decreased

20.
å¼ ªêšõèˆF¡ å¼ ð‚è‹ 6 e ñŸÁ‹ Üî¡ Í¬ôM†ì‹ 10 e âQ™ Üî¡
ðóŠð÷¾?
Area of a rectangle one of whose sides is 6 m and diagonal 10m is
A) 60 B) 48 C) 30 D) 68
Solution:
ªêšõèˆF¡ å¼ ð‚è‹ = 6 m
l (or) b

ê˜è£˜ ä.ã.âv Üè£ìI

ªêšõèˆF¡ ͬôM†ì‹ = 10 m
(d)
d2 = l2 + b2
102 = l2 + 62
l2 = 100 - 36
l2 = 64
l=8
Rectangle area = l x b
=8x6
= 48

21. å¼ ªêšõè î¬ó MKŠH¡ ðóŠð÷¾ 60e2. Üî¡ c÷ñ£ù ð°F»‹, ͬôM†ìº‹
ެ퉶, °ÁAò ð°FJ¡ 5 ñ샰 Ü÷MŸ° êñ‹ âQ™, î¬ó MKŠH¡ c÷‹
�
A rectangular carpet has an area of 60 sq.m. Its diagonal and longer side together equal 5 times
the shorter side. The lengh of the carpet is
(A) 5 m (B) 12 m (C) 13 m (D) 14.5 m
Solution:
ªêšõèˆî¬ó MKŠH¡ ðóŠð÷¾ = 60 m2
Rectangle Carpet area (A) = l x b = 60 m2
d+l=5xb
d = 5b - l
squaring both sides
d2 = (5b - 1)2

19
 TNPSC èí‚°

d2 = 25b2 - 10 lb + l2
l2 + b2 = 25b2 - 10 lb + l2
24b2 = 10 lb
24 b2 = 10(60)

600
b2 =
24

b2 = 25, b = 5
ç lb = 60
l x b = 60
60 60
l= = =12 m
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

b 5

22.
å¼ Ü¬øJ¡ c÷‹. Üèô‹ ñŸÁ‹ àòó‹ º¬ø«ò 8e, 10e, 4e ñŸÁ‹ 3e x
1.5e ðóŠð÷¾ ªè£‡ì å¼ è àœ÷¶. õ‡í‹ Ìê ê¶ó e†ì¼‚° Ï. 200
ªêôõ£°‹ â¡ø£™. Üî¡ ²õ˜èÀ‚° õ˜í‹ Ìê âšõ÷¾ ªî£¬è ªêôõ£°‹?
The length, breath and height of a hall are 8m, 10m, 4m respectively and the hall has one door
of area 3m x 1.5m. Find the cost of painting the walls at the rate of Rs. 200 per square metre.
(A) Rs. 28,800 (B) Rs. 59,900 (C) Rs. 27,900 (D) Rs. 60,800
-Solution:
l=8m; b = 10 m ; h = 4 m
Door area = l x b
= 3m x 1.5m
= 4.5 m2
C.S area of the cuboid = 2 ( l + b) x h
= 2 ( 8 + 10) x 4
= 2 ( 18) x 4 = 144 m2
= 144- 4.5
= 139.5 m2
Cost of painiting per square meter = Rs. 200
= 139.5 x 200
= Rs. 27900
Note :
1m2 = 200
139.5m = x

23. å¼ ªêšõèˆF¡ c÷‹ 50% Ü÷MŸ° °¬øˆ¶‹, Üèôˆ¬î 80% Ü÷MŸ°‹

ÜFèK‚°‹ «ð£¶ Üî¬ìò ðóŠH™ ãŸð´‹ ñ£ŸøˆF¡ êîiî‹ â¡ù?
If the length of a rectangle is decreased by 50% and the breadth is increased by 80%, then the
% change in the area of rectangle is
(A) 10% °¬øõ£è / decreased by 10%

20
 TNPSC èí‚°

(B) 10% ÜFèñ£è / increased by 10%

(C) 20% °¬øõ£è / decreased by 20%
(D) 20% ÜFèñ£è / increased by 20%
Solution:
l →50% decrease

50
10x =5
100

b →80% increase

180
10x = 18
100
l x b : lxb
(10 x10) : (5 x 18)
100 : 90
10% decreased

ê˜è£˜ ä.ã.âv Üè£ìI

24. å¼ Ü¬øò£ù¶ 5 e 40 ªê.e c÷º‹ ñŸÁ‹ 4 e 50 ªê.e Üèôº‹ ªè£‡´œ÷¬õ
âQ™ Üî¡ ðóŠð÷¾
A room is 5 m 40 cm long and 4 m 50 cm broad. Its Area is
A) 23. 4 m2 B) 24.3 m2 C) 25 m2 D) 98.01 m2
Solution:
l = 5 m 40 cm = 5.4 m
b = 4 m 50 cm = 4.5 m
A=lxb
= 5.4 x 4.5
= 24.3 m2

25. å¼ ªêšõèˆF¡ c÷ñ£ù¶ 60% ÜFèK‚èŠð´Aø¶. Üî¡ Üèôñ£ù¶ âˆî¬ù

êîiî‹ °¬ø‰î£™ Üî¡ ðóŠð÷¾ º‰¬îò ðóŠð÷¬õ «ð£ô«õ Þ¼‚°‹?
The length of a rectangle is increased by 60% By what percent would the width have to be
decreased so as to maintain the same area

1
A) 37 % B) 60% C) 75% D) 120%
2
Solution:
l →60% decrease

160
10x = 16
100
x
10x = 62.5%
100
100 - 62.5 = 37.5

21
 TNPSC èí‚°

b →37.5% increase
l x b : lxb
(10 x10) : 16 x x
100 : 16 x
100 = 16 x
x = 6.25
100 : 16 x 6.25
100 = 100 ( No change)

10
26. å¼ ªêšõè õ®õ î£O¡ c÷‹ 14 π ªê.e, Üèô‹ ªê.e î£÷£ù¶ c÷ˆF¡ õN«ò å¼
π
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

º¬ø ²¼†ìŠð´õî¡ Íô‹ å¼ à¼¬÷ à¼õ£‚èŠð†ì£™ ܚ༬÷J¡ èù

Ü÷¾ ò£¶?

10
A rectangular piece of paper has length 14π cm and breadth cm. A cylinder is formed by one
π
rolling of the paper along its length. Then volume of the cylinder is
A) 980 cc / è.ªê.e B) 1960 cc/ è.ªê.e C) 1400 cc/ è.ªê.e D) 490 cc/ è.ªê.e
Solution:
ªêšõè õ®õ î£O¡ c÷‹
Rectangular piece of paper length = 14 π cm
10
Üèô‹ = cm
π
14 π
10/π
10/π =

2πr = 14π
14/π
10
h=
π
22 22
2x x r = 14 x
7 7
14
r= = 7 cm
2
2
Volume of cylinder = πr h
22 7
= x 7 x 7x 10 x
7 22
= 490
= 490 cm 3

27. å¼ ªêšõè GôˆF¡ c÷‹ 80 e Üèô‹ 60 e. GôˆF™ c÷ˆFŸ° Þ¬íò£è¾‹,

ÜèôˆFŸ° Þ¬íò£è¾‹ å«ó Üèôºœ÷ õNŠð£¬î àœ÷¶. õNŠð£¬îèO¡
ðóŠ¹ 675 ê.e âQ™ ð£¬îJ¡ Üèô‹ â¡ù?

22
 TNPSC èí‚°

A rectangular ground is 80 m long and 60 m broad. It has two cross roads of equal width one is
parallel to length and the other parallel to breadth. If the area of these roads is 675 sq.m. Find
the width of each road.
(A) 3m (B) 5 m (C) 7m (D) 10m
Solution:
l = 80 m
b= 60 m
Area of rectangle = 80 x 60 = 4800
õNŠð£¬îJ¡ ðóŠ¹ = 675
E H

A I J B

D L K C

F G
80 m

Area = Area of ABCD + Area of EFGH - Area of IJKL

ê˜è£˜ ä.ã.âv Üè£ìI

675 = 80 x x + 60 x x - x 2
x 2 − 140 x + 675 =
0
=x 135
= or x 5
Ans: 5 m
Note :
675 = 80 x 5 + 60 x 5 - 52
= 400+300-25
= 700-25
= 675

28. å¼ ªêšõèˆF¡ c÷‹ 3 ªê.e. Ü÷¾ ÜFèK‚èŠð†´ Üèô‹ 3 ªê.e. Ü÷¾

°¬ø‚èŠð†ì£™ ðóŠð÷õ£ù¶ 18 ªê.e2. Ü÷¾ °¬ø‰¶ M´‹ . c÷ˆF™ 1 ªê.e
Ü÷¾ °¬ø‚èŠð†´ ÜèôˆF™ 2 ªê.e Ü÷¾ ÜFèK‚èŠð†ì£™ ðóŠð÷õ£ù¶ 14
ªê.e2 Ü÷¾ ÜFèKˆ¶ M´‹. âQ™ Ü„ªêšõèˆF¡ c÷‹ ñŸÁ‹ Üèô‹ º¬ø«ò
In a rectangle if the length is increased and breadth is reduced by 3 cm then area is reduced
by 18 cm2. If the length is reduced by 1 cm and breadth is increased by 2 cm then the area is
increased by 14 cm2. Actual length and breadth of the rectangle are respectively.
A) 13 ªê.e; 10 ªê.e B) 14 ªê.e; 9 ªê.e
C) 12 ªê.e; 12 ªê.e D) 15 ªê.e; 8 ªê.e
Solution:
(l + 3)(b - 3) = lb - 18
lb - 3l + 3b - 9 = lb - 18
lb - 3l + 3b - lb = -18 + 9
- 3l + 3b = -9  (1)
(l - 1)(b + 2) = lb + 14

23
 TNPSC èí‚°

lb + 2l - b - 2 = lb + 14
lb + 2l - b - lb = 14 + 2
2l - b = 16 (mutiply by 3)
6l - 3b = 48  (2)
(2) + (1)
6l - 3b = 48
- 3l + 3b = -9
----------------

39
l= = 13
13
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

l = 13, substitute in eqn (2)

2(13) - b = 16
26 - b = 16
26 -16 = b
b = 10
l = 13 ; b = 10
Another Method
Take option (a) l = 13 cm ; b = 10 cm
l b lb l b lb
13 10 130 16 7 112
13 10 130 12 12 144

29. å¼ ªêšõèˆF¡ c÷‹ ñŸÁ‹ Üèô‹ º¬ø«ò ð£F ñŸÁ‹ Þ¼ñ샰 âù ñ£Á‹
«ð£¶ ðóŠð÷M™ ãŸðì‚îò ñ£Á‹ êîiî‹ ò£¶- ?
If length and breadth of a rectangle became half and double respectively, then what will be the
% increase in resultant ?
Solution:
A) 0% B) 55% C) 75% D) 80%
l x b : lxb
(10 x10) : (5 x 20)
100 : 100
Ans : 0% (No change / ðóŠð÷M™ â‰î ñ£Ÿøº‹ ãŸð죶)

30. ªêšõè õ®õ GôˆF¡ å¼ ð‚è‹ 15 e ñŸÁ‹ Üî¡ å¼ Í¬ô M†ì‹ 17 e

âQ™ GôˆF¡ ðóŠð÷¾
One side of a rectangular field is 15m and one of its diagonals is 17m, then area of the field is
A) 32m2 B) 120 m2 C) 2 m2 D) 60 m2
Solution:

24
 TNPSC èí‚°

d 2 = l2 + b 2
17 2 - 152 = b 2
b = 17 2 - 152
b = 289 - 225
b = 64
b=8
Area = l x b
= 15 x 8 = 120 m2

31. å¼õ˜ c÷‹, Üèô‹ ñŸÁ‹ àòó‹ º¬ø«ò 25 ªêe, 40 ªêe, 45 ªêe Ü÷¾¬ìò
å¼ C.p.u, MŸ° à¬ø ¬î‚è M¼‹¹Aø£˜. à¬øJ¡ M¬ô 1 ê.e†ì¼‚° Ï. 40
âQ™ Üõ¼‚° âšõ÷¾ ªêôõ£°‹?
A man wants to stitch a cover for a c.p.u. Whose length, breadth and height are 25 cm, 40 cm
and 45 cm respectively. Find the amount he has to pay if it costs Rs 40 per sq.m
(A) Rs.27.40 (B) Rs. 26.70 (C) Rs. 22.60 (D) Rs. 25.30
Solution:

ê˜è£˜ ä.ã.âv Üè£ìI

Convert to cm in m
l =0.25 m, b = 0.40 m, h = 0.45 m
Area of cloth required = L.S.A + Area of the top (L.S.A - Lateral Surface Area)
= 2h (l + b) + lb
= 2 x 0.45 (0.25 + 40.0 ) + (0.25 x 0.40)
= 0.90 (0.65) + 0.1000
= 0.6850 m2
= 0.6850 x 40
= Rs. 2740

32. å¼ ªêšõè õ®õ GôˆF¡ ðóŠð÷¾ 836ê.e ªêšõè õ®õ GôˆF¡ å¼ ð‚è
Ü÷¾ 22 e âQ™ ܉î GôˆF¡ ²Ÿø÷¬õ 裇è
The Area of a rectangular field is 836 sq.m . One side of the rectangle is 22m. What is the pe-
rimeter of the field?
A) 100 m / e B) 120 m/ e C) 380 m/ e D) 400 m/ e
Solution:
l x b = 836 m 2
b → 22 m
Perimeter = 2 ( l + b)
Length = 836 = 38m
22

l → 38 m
Perimeter = 2 (38 + 22)
= 2(60) = 120m

25
 TNPSC èí‚°

33.
13e c÷‹, 9e Üèôºœ÷ å¼ Ü¬ø¬ò 75 ªê.e ÜèôŠ «ð£˜¬õJù£™, å¼
e†ì¼‚° Ï. 12.40 iî‹ ªêôM™ ÜôƒèK‚è Ý°‹ ªêô¾ 裇è
Find the cost of carpeting a room 13 m long and 9m broad with a carpet 75cm wide if one metre
of carpet costs Rs. 12.40
A) Rs. 1934.40 B) Rs. 1935. 40
C) Rs. 1936.40 D) Rs. 1937.40
Solution:
(i) Length of the Room = 13 m
Breath of the room = 9 m
Area of the room =lxb
= 13 x 9
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

= 117 m 2

75cm
(ii) 0.75m

Area of the Carpet = x x 0.75

(i) = (ii)
117 = x x 0.75
117
x = = 156 m
0.75

1m = Rs.12.40
156 m = x
= 12.40 x 156
= Rs.1934.40

34. å¼ ªêšõè õ®õŠð£ˆFóˆF™ àœ÷ î‡aK™ º¿õ¶ñ£è Í›A å¼ 15ªê.e

ð‚躜÷ èùê¶ó‹ àœ÷¶. ð£ˆFóˆF¡ Ü®Šð£èˆF¡ ðKñ£í‹ 20ªê.e x 15ªê.e
âù Þ¼‰î£™, î‡a˜ àò¼‹ Ü÷¾ 裇è
A cube of edge 15cm is immersed completely in a rectangular vessel containing water. If the
dimensions of the base vessel are 20cm x 15cm, find the rise in water level
A) 10.25 cm / ªê.e B) 11.25 cm/ ªê.e C) 11 cm/ ªê.e D) 10cm/ ªê.e
Solution:

15 cm

Increase in Volume = Volume of cube

= 15 x15 x 15
= 3375 cm3

26
 TNPSC èí‚°

Volume of cube
Rise in water =
Area (dimention of the base vessel)
3375
=
20 x 15
= 11.25 cm

35. å¼ 6ªê.e x12ªê.e x15ªê.e Ü÷¾œ÷ å¼ ªêšõ舶‡´ eî‹ Þ™ô£ñ™ èù

ê¶ó õ®Mô£ù ¶‡´è÷£è ªõ†ìŠð´Aø¶. °¬ø‰îð†ê â‡E‚¬èJô£ù
èùê¶óƒè¬÷‚ 裇è
A rectangular block 6cm by 12cm by 15cm is cut up into an exact number of equal cubes. Find
the least possible number of cubes
A) 37 B) 38 C) 39 D) 40
Solution:

l = 6 cm 3 6, 12, 15

ê˜è£˜ ä.ã.âv Üè£ìI

b = 12 cm 2 , 4, 5
h = 15 cm HCF=3
V = 6 x 12 x 15
of the block ( Rectangular)
= 1080 cm3
Volume of the cube = a 3
= 3 x3 x 3
= 27 cm3
Volume of the rectangular block
Number of cubes =
Volume of the cube
1080
= = 40
27
36.
c÷‹ 60 e Üèô‹ 3 e àòó‹ 5 e à¬ìò ²õ˜ â¿Šð c÷‹ 30 cm x Üèô‹ 15
cm x àòó‹ 20 cm à¬ìò ªêƒèŸèœ âˆî¬ù «î¬õ?
A wall is to be constructed with length 60 m. breadth 3 m and height 5 m. How many bricks
are required to construct a wall with length 30 cm, breadth 15cm and height 20 cm ?
(A) 1,50,000 (B) 1,25,000 (C) 1,00,000 (D) 1,75,000
Solution:
A wall is to be constructed,
(²õ˜) l = 60m = 6000cm [Note: 1m = 100 cm]
b = 3m = 300 cm
h = 5m = 500 cm
ªêƒè™(bricks)l = 30cm
b = 15cm

27
 TNPSC èí‚°

h = 20cm
How many bricks need?
²õK¡ èùÜ÷¾ = lxbxh
(Volume of the wall) = 6000 x 300 x 500
= 900000000 cm 2
ªêƒèŸèO¡ èùÜ÷¾ = lxbxh
(Volume of the Bricks ) = 30 x 15 x 20
= 9000 cm 2
ªêƒèŸèO¡ â‡E‚¬è 900000000
=
9000
No. of bricks required
= 100000
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

37. 3.78 e†ì˜èœ c÷º‹ 5.25 e Üèôº‹ à¬ìò å¼ Hóè£ó‹ å«ó Ü÷¾œ÷ êKò£ù ê¶ó
æ´è¬÷‚ ªè£‡´ ðóŠðŠðì «õ‡´‹. ܉«ï£‚èˆFŸè£è àð«ò£èŠð´ˆî‚îò
憮¡ I芪ðKò Ü÷¾ â¡ù?
A rectangular courtyard 3.78 metres long and 5.25 metres wide is to be paved exactly with
square tiles, all of the same size. What is the largest size of the tile which could be used for the
purpose?
(A) 14 cms (B) 21 cms (C) 42 cms (D) None of these
Solution:
l = 3.78m = 378 cm
b = 5.25m = 525 cm
21 378, 525
18, 25
HCF = 21
HCF 21
憮¡ I芪ðKò Ü÷¾ = 21 cm
(largest size of the tiles)

38. å¼ ¹™ªõO ¬ñî£ù‹ Üî¡ ð‚èƒèœ 2:3 MAîˆF™ àœ÷ å¼ ªêšõè õ®M™

1
àœ÷¶. Ü‰î ¬ñî£ùˆF¡ ðóŠ¹ ªý‚«ì˜ âQ™, Üî¡ c÷ˆ¬î‚ 裇è.
6
1
A lawn is in the form of a rectangle having its sides in the ratio 2 : 3. The area of the lawn is
6
hectares. Find the
length of the lawn.
(A) 40 m (B) 33½ m (C) 50 m (D) 45m
Solution:
2x

3x
¹™ªõO ¬ñî£ùˆF¡ ð‚èƒèO¡ MAî‹ = 2 : 3
(A lawn is in the form of rectangle having its sides in the ratio = 2 : 3)
¬ñî£ùˆF¡ ðóŠ¹ = 1 ªý‚«ì˜
6
(Area of the lawn = 1 hectores) (1 ªý‚«ì˜ = 10000 m2)
6

28
 TNPSC èí‚°

Area = 1 x 10000 m 2
6
1
2x x 3x = x 10000m 2
6
1
6x 2 = x 10000
6
1 1
x 2 = x 10000 x
100006 6
x=
6x6
100
x=
6
100
Length 3x = 3 x
6
300
= = 50
6
Length = 50m

39.
å¼ ªêšõèˆF¡ c÷ˆFŸ°‹ ÜèôˆFŸè‹ àœ÷ «õÁ𣴠23 e. Üî¬ìò

ê˜è£˜ ä.ã.âv Üè£ìI

²Ÿø÷¾ 206 e âQ™ Ü„ªêšõèˆF¡ ðóŠ¹
The difference between the length and breadth of a rectangle is 23m. If its perimeter is 206 m
then its area is
A) 1795 m2/ ê.e B) 2520 m2/ ê.e C) 2240 m2/ ê.e D) 3250 m2/ ê.e
Solution:
Length = l
Breadth = b
l - b = 23
Perimeter of rectangle = 206
2 (l+b) = 206
206
(l+b) =
2
l+b = 103
l +b = 103
l −b = 23
_________
2l = 126
126
=l = 63
2
l = 63, b = 40
lb = 63 x 40
= 2250 m2

40. 39 e 10 ªê.e c÷‹ ñŸÁ‹ 35e 70 ªê.e Üèô‹ ªè£‡ì å¼ ªêšõè õ®õ
ܬø‚° î¬ó æ´ «ð£†ì£™, ܉î ܬø‚° ªð£¼ˆîñ£ù Iè ÜFèð†ê î¬ó
æ´èœ âšõ÷¾ «î¬õ?
A rectangular hall 39m 10cm long and 35m 70 cm broad is to be paved with square tiles. Find
the largest tile which will exactly fit and the number required.

29
 TNPSC èí‚°

A) 480 B) 485 C) 490 D) 483

Solution:
l = 39m 10 cm = 3910 cm
b = 35 m 70 cm = 3570 cm
170 3910, 3570
23, 21
HCF = 170
Area of the tiles = a 2
= 170 x 170
= 28900 cm 2
Area of Hall = 3910 x 3570
= 13958700
13958700
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

n=
28900
= 483 tiles
41.
å¼ ðœOJ™ àœ÷ è¬ô‚ ÃìˆF¡ c÷‹ 45 e. Üèô‹ 27 e. è¬ô‚ Ã숬î„
²ŸP ªõOŠ¹øñ£è 3 e Üèôºœ÷ õ£ó‹ àœ÷¶. õ£óˆF¡ ðóŠð÷¾ â¡ù?
A school auditorium is 45 m long and 27 m wide. This auditorium is surrounded by a. varandha
of width 3 m on its outside. Find the area of the varandha
(A) 864 m2 (B) 846 m2 (C) 468 m2 (D) 648 m2
Solution:
l = 45 m
b = 27 m
Area of small rectangle = 45 x 27
= 1215 m 2
3m

3m 3m
b = 27 m
27 + 3 x 3 = 33m
l = 45 m
3m
l = 45 + 3 + 3 = 51 m
b = 27 + 3 + 3 = 33 m
Area of big rectangle = 51 x 33
= 1683 m 2
Area of varanda = Area of big rectangle - Area of Small rectangle
= 1683 - 1215
= 468 m 2

42.å¼ ªêšõèˆF¡ c÷‹ Üî¡ Üèôˆ¬î Mì 2 ªê.e2 ÜFè‹. ªêšõèˆF¡ ðóŠ¹

80 ªê.e2 âQ™ Üî¡ c÷ Üèôƒèœ ò£¬õ?
The length of a rectangle is 2 cm greater than the breadth. If the area of the rectangle is 80 cm2.
What are the length and breadth?
(A) 10 cm, 8 cm (B) 8cm, 6 cm (C) 6 cm, 4 cm (D) 12 cm, 10 cm
Solution:
Take option
l x b : lxb
30
 TNPSC èí‚°
10 x 10 : 10 x 80
100 : 80
if Area of rectangle 80
l = 10 cm
b = 8 cm

43. å¼ ªêšõèˆF¡ ðóŠ¹ 144 (ªê.e)2 âQ™ Üî¬ìò ð‚èƒèœ 4:9 â¡ø MAîˆF™
Þ¼‚°ñ£ù£™ Üî¬ìò ²Ÿø÷¾ âšõ÷¾ ?
The perimeter of a rectangle having area equal to 144(cm)2and sides in the ratio 4:9 is
A) 52 cm B) 56 cm C) 60 cm D) 64 cm
Solution:
ªêšõèˆF¡ ðóŠð÷¾ = 144 cm 2
l:b =4:9
4x x 9x = 144
36x 2 = 144
144
x2 = =4
36
x=2

ê˜è£˜ ä.ã.âv Üè£ìI

c÷‹ (l)
4x = 4 x 2 = 8
Üèô‹(b)
9x = 9 x 2 = 18
Primeter = 2 (l + b)
= 2 ( 8 + 18)
= 2 (26)
= 52 cm

44. å¼ ªêšõèˆ ªî£†®J™ 2340 m3 Ü÷¾ î‡a˜ àœ÷¶. ܈ªî£†®J¡ c÷‹ 15

m, Üèô‹ 13 m âQ¡, Ýö‹ â¡ù?
A rectangular tank contains 2340 cu.m of water. What is the depth of water in tank if it is 15
metres long and 13 metre wide
(A) 10 m (B) 15 m (C) 8 m (D) 12 m
Solution:
Volume of rectangular tank = l x b x h
ªêšõè õ®õ ªî£†®J¡ èùÜ÷¾ = 2340 m3
l = 15m b = 13m h=?
2340 = 15 x 13 x h
2340
=
15 x 13
= 12 m

45. å¼ ²õK¡ c÷‹ 8 e. àòó‹ 6 e, î®ñ‡ (Üèô‹) 22.5 ªêe Ü÷¾èœ ªè£‡ì¶.
Þ„²õ¬ó 膴õ 25 ªêe x 11.25 ªêe x 6 ªêe Ü÷¾èœ ªè£‡ì âˆî¬ù
ªêƒè™èœ «î¬õŠð´Aø¶?
How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, it
being given that each brick measures 25 cm x 11.25 cm x 6 cm?
(A) 6400 (B) 6300 (C) 6200 (D) 6500

31
 TNPSC èí‚°
Solution:
c÷‹ (l) = 8m = 800 cm
²õK¡ àòó‹ (h) = 6m = 600 cm
²õK¡ Üèô‹ (b)= 22.5 cm
ªêƒè™L¡ c÷‹ (l) = 25 cm
ªêƒè™L¡ àòó‹ (h) = 11.25 cm
ªêƒè™L¡ Üèô‹ (b) = 6 cm
How many bricks needed?
²õK¡ èùÜ÷¾ = l x b x h
Volume of the wall = 800 x 600 x 22.5
= 10800000 cm3
ªêƒèŸèO¡ èùÜ÷¾ = l x b x h
(Volume of the Bricks) = 22 x 11.25 x 6
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

3
= 1687.5 cm
10800000
ªêƒèŸèO¡ â‡E‚¬è =
1687.5
No. of bricks required
= 6400
46. å¼ ªêšõèˆF¡ c÷‹ ñŸÁ‹ ÜèôˆF¡ MAî‹ º¬ø«ò 3:2. Ü„ªêšõèˆF¡
²Ÿø÷¾ ñŸÁ‹ ðóŠð÷M¡ MAî‹ º¬ø«ò 5:9. âQ™ ªêšõèˆF¡ Üèôˆ¬î
e†ìK™ 裇è.
The ratio of length and breadth of a rectangle is 3:2 respectively. The respective ratio of its
perimeter and area is 5:9. What is the breadth of the rectangle in metres?
(A) 6 mt / e†ì˜ (B) 8mt/ e†ì˜ (C) 9mt / e†ì˜ (D) 13 mt/ e†ì˜
Solution:
ªêšõèˆF¡ c÷‹ (ñ) ÜèôˆF¡ MAî‹ = 3 : 2
The ratio of l and b = 3x, 2x
ªêšõèˆF¡ ²Ÿø÷¾ (ñ) ðóŠð÷M¡ MAî‹ = 5 : 9
The ratio of perimeter and area

2(l + b) 5
=
lxb 9
18 (l + b) = 5 lb
18 (3x + 2x) = 5 (3x x 2x)
18 (5x) = 30x2
90x = 30x2
3
x2 90
=
x 30
x=3
ªêšõèˆF¡ Üèô‹ = 2x
Breadth of rectangle = 6 m
ªêšõèˆF¡ c÷‹ l =30 m

47. å¼ ªêšõè ¬ñî£ùˆ¬î ê¶ó Ü®‚° Ï. 1.25 iî‹ êKªêŒò Ý°‹ ªêô¾ Ï. 900
Ý°‹. Ü‹¬ñî£ùˆF¡ c÷‹ 30 m âQ™ Üî¡ Üèôˆ¬î 致H®.
The cost of levelling a rectangular ground at Rs. 1.25 per sq.m is Rs. 900. If the length of the

32
 TNPSC èí‚°
ground is 30 m, then the width is
(A) 330 m (B) 34 m (C) 24 m (D) 18 m
Solution:
Surface area of the Floor = Total Cost2
cost per m
¬ñî£ùˆF¡ ðóŠð÷¾ = ªñ£ˆî ªêô¾
å¼ ê.Ü®‚°
900 Ý°‹ ªêô¾
= = 720 m 2
1.25
Area = l x b
720 = 30 x b

720
b= =24
30

b = 24m

48. ªêšõèŠ ªð†®J¡, c÷, Üèô, àòóƒèœ º¬ø«ò 30 cm, 50 cm ñŸÁ‹ 100 cm âQ™
Üî¡ èù Ü÷¾M¡ ñFŠ¹ â¡ð¶
If the length, breadth and height of a rectangular box are 90 cm, 50 cm and 100 cm respective-

ê˜è£˜ ä.ã.âv Üè£ìI

ly, then its volume will be
(A) 0.45 m2 (B) 0.45 m3 (C) 4500 cm3 (D) 20 cm3
Solution:
Rectangular Box
ªêšõè ªð†®J¡
c÷‹ (l) = 90 cm
Üèô‹ (b) = 50 cm
àòó‹ (h) = 100 cm
V=lxbxh
= 90 x 50 x 100
V = 450000 cm3
100cm = 1m
450000 cm 2 = x
45 0000
V= = 0.45 m3
100 0000
49. ªêšõè õ®Mô£ù 60 e x 42 e ðKñ£í‹ ªè£‡ì è÷ˆF™ å¼ Í¬ôJ™ å¼
°F¬ó «ñŒõîŸè£è 14 e c÷‹ ªè£‡ì èJŸPù£™ è†ìŠð†´œ÷¶. °F¬ó
«ñò£î è÷ˆF¡ ðóŠ¬ð‚ 裇è
A horse is tethered to one corner of a retangular field of dimensions 60 m by 42 m by a rope 14
m long for grazing. How much area can the horse left ungrazed?
(A) 2366 m2 (B) 1366 m2 (C) 1827m2 (D) 2212 m2
Solution:
42 m

60 m
m
14

Horse

Area of the rectangle = 60 x 42

33
 TNPSC èí‚°
ªêšõèˆF¡ ðóŠð÷¾
A = 2520
Area of the Quarteral
裙õ†ìˆF¡ ðóŠð÷¾ = π r
2

11 2 7 4
22 14 x 14
= x
7 4
2

A = 154
Non grassing area = Area of the rectangle - Area of the Quarteral
°F¬ó «ñò£î ð°F = ªêšõèˆF¡ ðóŠð÷¾ - 裙õ†ìˆF¡ ðóŠð÷¾
= 2525 - 154
= 2366 m 2
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

50. å¼ ªêšõè õ®õ õòL¡ ðóŠð÷¾ 240e2. Üî¬ìò c÷Šð‚èˆFL¼‰¶ 8 ªêe

°¬øˆî£™ ܶ å¼ ê¶óñ£°‹. Üî¡ c÷‹, Üèô‹ º¬ø«ò
The area of a rectangular land is 240m2. If 8m is decreased from its length it will become a
square. Then the length and breadth of the land respectively are...
(A) 12 cm, 20 cm (B) 12 cm, 8 cm
(C) 20 cm, 12 cm (D) 20 cm, 8 cm
Solution:
x

8cm

A = l x b = 240 m 2
The area of the rectangular land is = 240 m 2
ªêšõèˆF¡ ðóŠð÷¾ = 240 m 2
b=x-8
l=x
x (x - 8) = 248
x 2 - x = 240
x 2 - x - 240 = 0
(x - 20) (x + 12) = 0
x = 20 , x = -12
Another method:
Take C option
l x b = lb
20 x 12 = 240
(l - 8 cm)
12 x 12  Square
l = 20m ; b = 20 - 12 = 12m

51. What is the approximate ratio of the length of the sides of a “Golden Rectangle”?
îƒè„ ªêšõèˆF¡ ð‚èƒèœ «î£ó£òñ£è â‰î MAîˆF™ ܬñ‰F¼‚°‹
(A) 1:4 (B) 1:8 (C) 1:1.6 (D) 1:2

34
 TNPSC èí‚°
îƒè„ ªêšõèˆF¡ ð‚èƒèœ «î£ó£òñ£è â‰î MAîˆF™ ܬñ‰F¼‚°‹
The approximate ratio of the length of the sides of a golden rectangle
Ans : 1 : 1 : 6

52. 8 e àòó‹, 6 e c÷‹ ñŸÁ‹ 2.5 e Üèôº¬ìò å¼ ªêšõè î‡a˜ ªî£†®J™

âˆî¬ù L†ì˜ î‡a˜ ªè£œÀ‹?
A rectangular water tank is 8 m high, 6 m long and 2.5 m wide. How many litres of water can
it hold?
(A) 120 litres / L†ì˜ (B) 1200 litres / L†ì˜
(C) 12000 litres / L†ì˜ (D) 120000 litres / L†ì˜
Solution:
h=8m l = 6m b = 2.5m
V = lbh
= (8 x 6 x 2.5) x 1003 cm3
= 120 x 100 cm3
= 120 x 1000 l
V = 120000 litres
Note :
1 l = 1000 cm3

ê˜è£˜ ä.ã.âv Üè£ìI

10l = 10000 cm3
100l = 100000 cm3
1000l = 1003 cm3

53. ªêšõèªñ£¡P¡ Üèôñ£ù¶ Üî¡ c÷ˆ¬îMì 27 ªê.e °¬øõ£è àœ÷¶. ²Ÿø÷¾

3 e 6 ªê.e Ýè Þ¼ŠH¡ Ü„ ªêšõèˆF¡ c÷‹, Üèô‹ º¬ø«ò
Breadth of a rectangle is 27 cm less than its length. If the perimeter is 3 m 6 cm then its length
and breadth are respectively
(A) 100 cm, 73 cm (B) 80 cm, 53 cm (C) 90 cm, 63 cm (D) 103.5 cm, 76.5 cm
Solution:
(x-27) = b

x=l

Perimeter = 3m 6cm
²Ÿø÷¾ 2 ( l +b) = 306 cm
2(x + (x - 27)) = 306
2x + 2x - 54 = 306
4x = 306 + 54
90
360
x= = 90
4
x = 90
x - 27 → 90 - 27 = 63 → Breadth
l = 90 cm
b = 63 cm

35
 TNPSC èí‚°
54. å¼ ªêšõèˆF¡ ͬôM†ìˆF¡ Ü÷¾ 17 ªê.e Ý辋 Üî¡ ²Ÿø÷¾ 46 ªê.e
Ý辋 Þ¼ŠH¡, Üî¡ ðóŠð÷¾ â¡ù?
If the diagonal of a rectangle is 17 cm long and the perimeter of the rectangle is 46 cm, then the
area of the rectangle is
(A) 120 cm2 (B) 152 cm2 (C) 112 cm2 (D) 289 cm2
Solution:
ªêšõèˆF¡ ͬôM†ì‹ = 17 cm
Dialgonal of a rectangle
2 ( l + b) = 46 cm
lxb = ?

y
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

d 2= l 2 + b 2
17 2= l 2 + b 2
289= l 2 + b 2
2(l + b) =46
46
l + b= = 23
2
l +b =23
(l + b) ( 23)
2 2
=
l 2 + b 2 + 2lb =
529
289 + 2lb = 529 − 289
240
lb
= = 120
2
lb = 120cm 2

55. å¼ ªêšõèˆF¡ ²Ÿø÷¾ 60 e†ì˜. Ü„ªêšõèˆF¡ c÷‹, ÜèôˆF¡ Þ¼ñ샰

âQ™ Üî¡ ðóŠð÷¾ â¡ð¶
The perimeter of a rectangle is 60 meters. If its length is twice its breadth, then its area is
(A) 160 m2/ ê¶ó e†ì˜ (B) 180 m2/ ê¶ó e†ì˜
(C) 200 m2/ ê¶ó e†ì˜ (D) 220 m2/ ê¶ó e†ì˜
Solution:
Perimeter of a rectangle = 60m
(ªêšõèˆF¡ ²Ÿø÷¾)

2x

²Ÿø÷¾ = 2 ( l + b)

36
 TNPSC èí‚°
60
= m 2(2 x + x)
6 x = 60
60
x=
6
x = 10 breadth
2 x = 20 length
Area of the Rectangle = l x b
= 10 x 20
= 200 m 2

56. å¼ ªêšõèˆF¡ c÷‹ 12 e Üî¡ Üèô‹ 8 e âQ™ ªêšõèˆF¡ ²Ÿø÷õ£ù¶

If length is 12 m and breath 8 m then perimeter of rectangle is
(A) 96 m / e (B) 40 m / e (C) 48 m / e (D) 20 m / e
Solution:
ªêšõèˆF¡ c÷‹ (l) = 12 m
ªêšõèˆF¡ Üèô‹ (b) = 8m
(ªêšõèˆF¡ ²Ÿø÷¾)
Perimeter of rectangle

ê˜è£˜ ä.ã.âv Üè£ìI

= 2 (l + b)
= 2 (12 + 8) = 2(20)
= 40 m
57. Area of the rectangular region 2 ≤ x ≤ 5,-1 ≤ y ≤ 3 is
2 ≤ x ≤ 5,-1 ≤ y ≤ 3 Ý™ ܬìð´Aø ªêšõè ð°Fò£ù¬õ
(A) 9 Üô°èœ (B) 12 Üô°èœ (C) 15 Üô°èœ (D) 20 Üô°èœ
Solution:
Area of the rectangular regoin 2 ≤ X ≤ 5, -1 ≤ Y ≤ 3 is 2 ≤ X ≤ 5, -1 ≤ Y ≤ 3 Ý™
ܬìð´Aø ªêšõè ð°Fò£ù¬õ

-1
2 5

The area of a rectangle = lx x by

lx
= 5 - 2 =3
b
y = 3 - ( - 1) = 4
The area of the rectangle = 3 x 4
= 12 units

58. è†®ì «õ¬ôò£œ å¼õ˜ ܬøJ¡ ªêšõè õ®õ î¬óJ¡ ðóŠð÷¾ 2x3 + 16 âùˆ
b˜ñ£Qˆ¶‚ ªè£‡ì£˜. Üî¡ c÷‹ 2(x+2) âù‚ °P‚èŠð†ì£™ Üèôñ£ù¶ x& ¡
꣘ð£è H¡õ¼õùõŸÁœ â¶?
A mason uses the expression 2x3 + 16 to represent the area of the rectangular floor of a room.
If he decides that the length of the room will be represented by 2(x+2) then what will the width
of the room be represented in terms of x?

37
 TNPSC èí‚°

(A) 2(x - 2) (B) (x2 - 4) (C) (x2 - 2x + 4) (D) (x2 + 2x - 4)

Solution:
ªêšõè î¬óJ¡ ðóŠð÷¾ = l x b
2 x33 ++ 16
2x 16==lxb
lxb
=l 2( x + 2)
= ??
bb =
ªêšõè õ®õ î¬óJ¡ ðóŠð÷¾ = 2 (x3 + 23)
a3 + b3 = (a + b) (a2 - ab + b2)
a = x, b = 2
A== A22= ((2xx+
A 2)(
+( x2)( xx 2 −
+ 2)(
2
−x 222−xx +
+ 4)+4) 
2 x4)
   
22
llxb b== 2(
x lxb
lxb 2( +x2)(
== xx2(+ xx −
+ 2)(
2)( −x 222−xx +
2 x4)
+ 4)+ 4)
2(
2( xx2(+x2)(
+ + 2)(
2)( xx 22 −
−x 222−xx +
2 x4)
+ 4)+ 4)
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

bb =
=b2(=x + 2)( x − 2 x + 4)
2
b= l
l l
22l 2
2( xx2(
+x2)2)+((2)xx (− −x 22−xx ++ 4)+ 4)
2 x4)
= 2(
= =
+
2(
2( xx2( +x2)
+ 2)+ 2)
22
AnsAns
Ans = xx= −
= −x 222−xx + 2 x44+ 4
+

38
 TNPSC èí‚°

º‚«è£í‹ ñŸÁ‹ ˹

Triangle and Cone

Triangle and Cone Sums:

7 cm 25 cm
1. Find Area ? / ðóŠð÷¾ 裇è.
24 cm

Area = s( s − a )( s − b)( s − c )

a + b + c 7 + 24 + 25
=s =
2 2
56
s
= = 28
2

ê˜è£˜ ä.ã.âv Üè£ìI

Area= 28(28 − 7)(28 − 24)(28 − 25)

= 28(21)(4)(3)
= 2x2x7x3x7x2x2x3
=2x7x3x2
= 84 cm 2

0 cm
=4
2. d 32 cm Find Area ?

1
Sector Area = x r x Arc
2
10
1
= x 20 x 32
2
= 320 cm 2
Note :

d 40
r
= = = 20
2 2

a b
3. Area = 900 3 cm 2 , find a = ?
c

39
 TNPSC èí‚°

3 2
Area = a
4
3 2
900 3 = a
4
4
a 2 = 900 3 x
3

a = 900 x 4
a = 30 x 2
a = 60 cm

4. êñð‚è º‚«è£íˆF¡ ²Ÿø÷¾ 30 cm âQ™ ðóŠð÷¾ 裇è ?

TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

The Perimeter of the equilateral Triangle is 30 cm, find area ?

a a

3a = 30 a

30
a=
3
a = 10 cm

3 2
Area = a
4
5 5
3
x 10 x 10 =
4
2
= 25 3

5. If one side of the equilateral triangle is increased by 20%. Then its area increased by
å¼ êñð‚è º‚«è£íˆF¡ ð‚è‹ 20% ÜFèK‚Aø¶ âQ™ Üî¡ ðóŠ¹ âˆî¬ù %
ÜFèK‚°‹ ?
20%
120
= 10 x
100
= 12
Area = a x a : a x a
10 x 10 : 12 x 12
100 : 144
144 - 100 = 44% ðóŠ¹ ÜFèK‚°‹.

6.  4 ªê.e Ü®Šð‚è‹ ñŸÁ‹ 3 ªê.e àòó‹ ªè£‡ì å¼ ªêƒ«è£í º‚«è£íˆF¡

ðóŠð÷¾

40
 TNPSC èí‚°
Find the area of the right angled triangle whose base and height are 4 cm and 3 cm
A) 12ê.ªê.e B) 6ê.ªê.e C) 8ê-.ªê.e D) 5ê-.ªê.e

b = 4 ªê.e, h = 3 ªê.e
3ªê.e
ªêƒ«è£í º‚«è£íˆF¡
ðóŠð÷¾ = 12 bh 4ªê.e
1
= 2 x4 x3
= 2x3
= 6 ê.ªê.e

7. 
å¼ ªêƒ«è£í º‚«è£íˆF¡ ðóŠð÷¾ 54 ê.ªê.e ñŸÁ‹ Üî¡ àòó‹ 12 ªê.e
âQ™ Üî¡ Ü®Šð‚è Ü÷¾ _____
Area of a right angled triangle is 54 cm2 and its height is 12 cm then find its base
A) 3 ªê.e B) 5 ªê.e C) 8 ªê.e D) 9 ªê.e
A = 54 ê.ªê.e, h = 12 ªê.e
ªêƒ«è£í º‚«è£íˆF¡ = 12 bh
ðóŠð÷¾

ê˜è£˜ ä.ã.âv Üè£ìI

1 bh = 54
2
1 xbx12 = 54
2
x2
b = 5412
6
b = 9 ªê.e

8.  å¼ º‚«è£íˆF¡ Þó‡´ «è£íƒèœ 600 ñŸÁ‹ 900 âQ™ Í¡ø£õ¶ «è£í‹

______
Two angles of a triangle are 60° and 90° then find its third angle?
A) 100 B) 200 C) 300 D) 400
º‚«è£íˆF¡ Í¡Á «è£íƒèO¡ Ã´î™ = 1800
60o + 90o + C = 180o
150o + C = 180o
C = 180 - 150 = 30o
C = 30o

9. 
å¼ «î£†ì‹ º‚«è£í õ®õˆF™ àœ÷¶. Üî¡ Ü®Šð‚è‹ 26e. Üî¡ àòó‹
28e Ü‰îˆ «î£†ìˆ¬î„ êñ¡ð´ˆ¶õ å¼ ê¶ó e†ì¼‚° Ï.5 iî‹ âšõ÷¾
ªêô¾ Ý°‹?
A garden is the form of triangle. Its base is 26 m and height is 28 m. find the cost of levelling
the garden at Rs.5 per m2
A) Ï.1840 B) Ï.1820 C) Ï.1810 D) Ï.1830
b = 26 e, h = 28 e
º‚«è£íˆF¡ ðóŠð÷¾ = 12 bh
13
= 1 x 26 x 28
2

41
 TNPSC èí‚°
= 13 x 28
«î£†ìˆF¡ ðóŠ¾ = 364ê.e
1 ê¶ó e†ì¼‚° ªêô¾ = Ï.5
364 ê.e†ì¼‚° ªêô¾ = 364 x 5 = Ï.1820

10. 13ªê.e, 14ªê.e, 15ªê.e ð‚è Ü÷¾è¬÷ ªè£‡ì º‚«è£íˆF¡ ðóŠð÷¾ 裇è.
Find the area of triangle whose sides are 13 cm, 14 cm and 15 cm
A) 84ªê.e2 B) 68ªê.e2 C) 96ªê.e2 D) 64ªê.e2
a = 13, b = 14, c = 15
a+b+c 13+14+15 42
S = 2 = 2 = 2 = 21
ðóŠ¹ = S(S-a) (S-b) (S-c)
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

= 21(21-13) (21-14) (21-15)

= 21(8) (7) (6)
= 7x3x4x2x7x2x3
= 7x3x4
= 21x4
A = 84ªê.e2

11. å¼ º‚«è£ù õ®õ õòL¡ ð‚è c÷ƒèœ 28e, 15e ñŸÁ‹ 41e âQ™ õòL¡
ðóŠð÷¬õ èí‚A†´, «ñ½‹ õò¬ô„ êñ‹ð´ˆî å¼ ê¶ó e†ì¼‚° Ï.20
ªêôõ£°‹ âQ™, õò¬ô„ êñ‹ð´ˆî Ý°‹ ªñ£ˆî ªêô¬õ‚ èí‚A´è.
The lengths of sides of a triangular field are 28 m, 15 m and 41 m. Calculate the area of the
field. Find the cost of levelling the field at the rate of Rs.20 per sq.m.
A) Ï.2420 B) Ï.2520 C) Ï,2440 D) Ï.2540
a = 28e, b = 15e, c = 41e
a+b+c 28+15+41 84
S = 2 = 2 = 2 = 42e
º‚«è£í õòL¡ ðóŠð÷¾ = S(S-a) (S-b) (S-c)
= 42(42-28) (42-15) (42-41)
= 42(14) (27) (1)
= 2x3x7x2x7x3x3x3x1
= 2x7x3x3 = 126 ê.e
1ê.e õò¬ô„ êñŠð´ˆî Ý°‹ ªêô¾ = Ï.20
126 ê.e õò¬ô„ êñŠð´ˆî Ý°‹ ªêô¾
= 20 x 126 = Ï. 2520

12. 180 ªê.e ²Ÿø÷¾ ªè£‡ì å¼ êñð‚è º‚«è£íˆF¡ ðóŠð÷¬õ‚ 裇è.

Find the area of an equilateral triangle whose perimeter is 180 cm.
A) 1550 ªê.e2 B) 1558 ªê.e2
C) 1560 ªê.e2 D) 1568 ªê.e2
êñð‚è º‚«è£íˆF¡ ²Ÿø÷¾ 3a = 180

42
 TNPSC èí‚°
180
a = 3 = a = 60 ªê.e
êñð‚è º‚«è£íˆF¡ ðóŠð÷¾ = 3 a2ê.e
4
3 15
= 4
x 60 x 60
= 3 x 15 x 60 = 1558.84
= 1558 ªê.e2

13. If the volumes of two cones are in the ratio 1:4 and their diameters are in the ratio 4 : 5 then the
ratio of their height is
Þ¼ ˹èO¡ èùÜ÷¾èO¡ MAî‹ 1:4 ñŸÁ‹ ÜõŸP¡ M†ìƒèO¡ MAî‹ 4:5
âQ™ ÜõŸP¡ àóòƒèO¡ MAî‹ â¡ù ?
The ratio volume of cones (˹èO¡ èùÜ÷¾ MAî‹) = 1 : 4
d1 : d2 = 4 : 5
r1 : r2 = 2 : 5/2
ËH¡ èùÜ÷¾ (Volume of cone)
1 2 1
π r1 h1 = π r22h2
3 3

ê˜è£˜ ä.ã.âv Üè£ìI

5 5
2 x 2 x h1 = x x h 2
2 2
h1 5x5 25
= =
h 2 2 x 2 x 2 x 2 16
h1 : h2 = 25 :16

èùÜ÷¾ MAî‹ = 1 : 4
= 25(1) : 16(4)
h1 : h2 = 25 : 64

14. A cone is 8.4 cm height and the radius of its base is 2.1 cm. It is melted and recast into a sphere.
Find the radius of the sphere.
å¼ Ã‹H¡ Ü®Šð‚èˆF¡ Ýó‹ 2.1 ªê.e., àòó‹ 8.4 ªê.e. ܶ à¼õ£‚èŠð†´
å¼ «è£÷ñ£è õ£˜‚èŠð†ì£™ «è£÷ˆF¡ Ýó‹ â¡ù ?
ËH¡ Ü®Šð‚è Ýó‹ = 2.1 cm
Height (àòó‹) = 8.4 cm
Volume of cone = Volume of sphere
ËH¡ èùÜ÷¾ = «è£÷ˆF¡ èùÜ÷¾

1 2 4
π r h = π r3
3 3
2.1x 2.1 x 8.4 = 4 x r 3

37.044
= r3
4
r 3 = 9.261

43
 TNPSC èí‚°

r = 2.1cm

15. Curved surface area of Frustum of a cone is

å¼ Ã‹H¡ Þ¬ì‚è‡ìˆF¡ õ¬÷ðóŠ¹ â¡ð¶
Ans : π(R+r)l sq.units.

Exercise sums:
1. If the area of an equilateral triangle is 4 3 m2, then its perimeter is
å¼ êñð‚è º‚«è£íˆF¡ ðóŠ¹ 4 3 e2 âQ™ Ü‹º‚«è£íˆF¡ ²Ÿø÷¾
(A) 16m (B) 12 m (C) 18 m (D) 9 m
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

Solution:
3 2
Area of an equilaterat triangle (A )= a
4
The area of equilateral triangle is = 4 3 m2

3 2
4 3= a
4
4
a2 = 4 3 x
3
a2 = 4 x 4
a=4
Perimeter = 3a
= 3 x 4 = 12 m
Answer: 12 m

2. The side of an equilateral triangle is 4cm. Then its altitude is
å¼ êñð‚è º‚«è£íˆF¡ ð‚è‹ 4 ªê.e âQ™ Üî¡ àòó‹
(A) 2 3 (B) 4 3 (C) 8 3 (D) 4
Solution:
a = 4 cm
3
Formula : h = a
2
3 2
= x4
2
Answer: h= 2 3

3. Find the percentage increase in the area of a triangle if its each side is doubled.
å¼ º‚«è£íˆF¡ 嚪õ£¼ ð‚躋 Þ¼ ñìƒè£è ÜFèK‚èŠð†ì£™, ܉î
º‚«è£íˆF¡ ðóŠð÷¾ âˆî¬ù êîiî‹ ÜFèKˆF¼‚°‹ ?
(A) 100% (B) 200% (C) 300% (D) 400%
Solution:
½ bh : ½ bh
a x a : 2a x 2a
44
 TNPSC èí‚°
10 x 10 = 20 x 20
100 = 400
300%  (increase)

4. Three angles of a triangle are x-300,x-450, x+150, find the value of x.
å¼ º‚«è£íˆF¡ Í¡Á «è£íƒèœ x-300,x-450, x+150, âQ™ x- ¡ ñFŠ¹
A) 600 B) 400 C) 800 D) 1000
Solution:
Three angles of a triangle are
x - 30°, x - 45°, x + 15°
x - 30° + x - 45° + x + 15° = 180°
3x - 60° = 180°
3x = 180° + 60°
3x = 240°
x = 240° / 3
x = 80°

ê˜è£˜ ä.ã.âv Üè£ìI

Answer = 80°

5. The base of a triangle is four times its height and its area is 50 m2 . The length of the base is
å¼ º‚«è£íˆF¡ Ü®Šð‚è‹, àòóˆF¡ 4 ñ샰‚° êñ‹ ñŸÁ‹ Üî¡ ðóŠð÷¾
50 e2 âQ™ Üî¡ Ü®Šð‚è Ü÷¾
A) 10 m B) 15 m C) 20 m D) 25 m
Solution:
Area = ½ bh
50 = ½ (4x (x))
x2 = 50 x 2 x ¼ , x2 = 25
x=5
Lenght of the base (4x) = 4 x 5 = 20 m
Answer = 20 m

6. Find the length of the altitude of an equilateral triangle of side 3√3 cm

3 3cm ð‚躜÷ êñð‚è º‚«è£íˆF¡ °ˆ¶òó‹ â¡ù?
(A) 27 cm (B) 9 √3 cm (C) 9 cm (D) 4.5 cm
Solution:
a = 3√3 cm

3
h= a
2
3
= x3 3
2

45
 TNPSC èí‚°

3x3
=
2
9
= = 4.5 cm
2
Answer = 4.5cm

7. The sides ( in c.m) of a right angled triangle are x-1, x, x+1. Then area of the right angled triangle
is
å¼ ªêƒ«è£í º‚«è£íˆF¡ ð‚èƒèœ (ªê.e) x-1, x, x+1 âQ™ Ü„ªêƒ«è£í
º‚«è£íˆF¡ ðóŠð÷¾
A) 12 sq.cm B) 6 sq.cm C) 20sq.cm D) 22 sq.cm
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

Solution:
The side of a right angled triangle are x - 1, x, x + 1
= x + 1 is the longest side Pythagoras theorem
(x + 1)2 = (x - 1)2 + (x)2
x2 + 11 + 2x = x2 - 2x + 1 + x2
x2 - 4x = 0
x (x - 4) = 0
x = 0, x = 4
x = 0 can not be side of a triangle one side is x = 4, x - 1 = 3, x+ 1 = 5
a = 3, b = 4, c = 5 A = s ( s − a)( s − b)( s − c)

3+ 4+5
s= A = 6(6 − 3)(6 − 4)(6 − 5)
2
6
12
s= A = 6(3)(2)(1)
2
s = 6 A = 6 x 6 = 36 = 6 cm2
Answer = 6

8. If the side of an equilateral triangle is decreased by 20% its area is decreased by

å¼ êñð‚è º‚«è£íˆF¡ å¼ ð‚è‹ 20% °¬ø‚èŠð†ì£™ Ü‹º‚«è£íˆF¡
ðóŠð÷M™ ãŸð´‹ °¬ø¾
A) 42% B) 36% C) 34% D) 20%
Solution:
3 2 3 2
a : a
4 4

a x a : a x a (Note : 20% )
80
10 x 10 : 8 x 8 ( 10 x )
10 0
100 : 64
64%  (is decreased by 64%)
ç 100& 64 = 36%
46
 TNPSC èí‚°

9. Which of the following is larger in area?

(A) A triangle with base 10 cm and height 8 cm
(B) A triangle with sides 12 cm, 5 cm and 13 cm
(C) An equilateral triangle whose sides are 10 cm each
(D) A right angled triangle whose sides containing the right angle are 3 cm and 4 cm
H¡õ¼õùõŸÁœ ðóŠð÷M™ ªðKò¶ â¶?
(A) Ü®Šð‚è‹ 10 ªê.e àòó‹ 8 ªêe ªè£‡ì º‚«è£í‹
(B) 12 ªêe, 5 ªêe ñŸÁ‹ 13 ªêe ð‚èƒèœ àœ÷ º‚«è£í‹
(C) 10 ªêe ð‚è‹ àœ÷ êñð‚è º‚«è£í‹
(D) ªêƒ«è£íˆ¬î àœ÷ì‚Aò ð‚èƒèœ 3 ªêe ñŸÁ‹ 4 ªêe àœ÷ º‚«è£í‹
Solution:
Take option A: b = 10, h = 8
1
A = bh
2
1 4
= x 10 x 8
2

ê˜è£˜ ä.ã.âv Üè£ìI

A = 40cm
Take option D: b = 3cm, h = 4 cm

1
A= bh
2
1 2
= x 3 x 4 = 6 cm
2

option B: a = 12cm, b = 5cm, c = 13 cm

= s ( s − a)( s − b)( s − c)
= 15(15 − 12)(15 − 5)(15 − 13)
= 15(3)(10)(2)
= 15 x 60 = 900 = 30 cm
option C:

3 2
A= a
4
3
= x102
4

3 5 5
= x 10 x 10
4 2
= 25 3
= 25 x 1.732 = 43.3 (larger)
Answer : C

47
 TNPSC èí‚°
10. Which of the following is not the area.of triangle ABC?
H¡õ¼õùõŸÁœ ⶠABC º‚«è£íˆF¡ ð󊹂è£ù ňFó‹ Ü™ô?
1 1
(A) ab sin c (B) bh
2 2
1
(C) s ( s  a)( s  b)( s  c) (D) d(h1+h2)
2
1
Answer: (D) d(h1+h2)
2

11. The angles of a triangle are in the ratio 3 : 4 : 5 find the angles
å¼ º‚«è£íˆF™ «è£íƒèœ 3 : 4 : 5 â¡ø MAîˆF™ àœ÷ù, Ü‰î «è£íƒè¬÷
裇è
A) 600, 600, 600 B) 400, 600, 800
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

C) 450, 600, 750 D) 450, 450, 900

Solution:
The angles of a triangle are in the ratio 3 : 4 : 5
3x + 4x + 5x = 180
12x = 180
x = 15
The angles are: 45°, 60°, 75°

12. Angles of a triangle are in the ratio 1 : 2 : 3 then the triangle is

(A) Equilateral triangle (B) Isosceles triangle
(C) Right angle triangle (D) Isosceles right angled triangle
å¼ º‚«è£íˆF¡ «è£íƒèœ 1 : 2 : 3 â¡ø MAîˆF™ Þ¼ŠH¡ Ü‹ º‚«è£í‹
âšõ¬è º‚«è£í‹?
(A) êñð‚è º‚«è£í‹ (B) Þ¼êñð‚è º‚«è£í‹
(C) ªêƒ«è£í º‚«è£í‹ (D) Þ¼êñð‚è ªêƒ«è£í º‚«è£í‹
Solution:
1 : 2: 3
1x : 2x : 3x
1x + 2x + 3x = 180
6x = 180
x = 30
30°, 60°, 90°
Answer: Right angle triangles (because one of the angle is 90°)

13. The area of a right angled triangle is 40 times its base, then its height is
å¼ ªêƒ«è£í º‚«è£íˆF¡ ðóŠð÷¾ Üî¡ Ü®Šð‚è Ü÷¬õ Mì 40 ñ샰
âQ™ Üî¡ àòó‹
A) 80 cm B) 40 cm C) 70 cm D) 90 cm
Solution:

h
48
 TNPSC èí‚°

½ b x h = 40 x b
h = 40 x 2
h = 80 cm
Answer: h = 80 cm

14. A square and an equilateral triangle have equal perimeter. If the diagonal of the square is 12 2
cm, then the area of the triangle is
å¼ ê¶óº‹, å¼ êñ ð‚è º‚«è£íº‹ å«ó ²Ÿø÷¾ ªè£‡ì¬õ. ê¶óˆF¡ ͬô
M†ì‹ 12 2 ªê.e âQ™ êñð‚è º‚«è£íˆF¡ ðóŠ¹
A) 64 3 cm2 B) 60 2 cm2 C) 50 5 cm2 D) 58 3 cm2
Solution:
Square (perimeter) = equilateral triangle (perimeter)
4a = 3a

ê˜è£˜ ä.ã.âv Üè£ìI

A = ½ d2
d = 12 2 cm

1
a2 = x 122 x 2
2
a = 12 for square
equilateral triangle, perimeter = 3a
4 (12) = 3a

4 x 12
a= = 16
3
a = 16 for equilateral triangle

3 2
Area of equilateral triangle = a
4
3
= x (16) 2 = 64 3 cm2
4
Answer : 64 3 cm2

15. The height of an equilateral triangle is 10cm. Its area is

êñð‚è º‚«è£íˆF¡ àòó‹10 ªê.e âQ™ Üî¡ ðóŠð÷¾
100 100
A) cm2 B) 30 cm2 C) 100 cm2 D) cm2
3 3
Solution:
Equilateral triangle (h) height is 10 cm
3 2
Area = a
4
5
3 20 20
= x x
4 3 3
49
 TNPSC èí‚°

100
Answer = cm2
3
Note:
3
h= a
2
3
10 = a
2
2
a = 10x
3
20
a=
3
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

16. Area of a triangle is

º‚«è£íˆF¡ ðóŠð÷¾
1
A) Ü®Šð‚è‹×àòó‹ B) ×Ü®Šð‚è‹ ×àòó‹
2
1
B) 2×Ü®Šð‚è‹×àòó‹ D) (Ü®Šð‚è‹+ àòó‹)
1 2
Answer: (B) ×Ü®Šð‚è‹ ×àòó‹
2

17. Area of a triangular garden is 800 sq.m. The height of the gareden is 40 m. Find the base length
of the garden.
å¼ º‚«è£í õ®õˆF½œ÷ «î£†ìˆF¡ ðóŠð÷¾ 800m2. ÜõŸP¡ àòó‹ 40 m âQ™
܈«î£†ìˆF¡ Ü® àòó‹ ò£¶?
(A) 20 m (B) 40 m (C) 10 m (D) 50 m
Solution:
Area = 800 sq.m , h = 40 m
Area = ½bh
1
800 = x b x 40
2
1
b= x 2 x 800
40
1 20
b= x 2 x 800
40
b = 40m
Answer: b = 40m

18. The perimeter of a equilateral triangle is 18 cm. Then the height of the equilateral triangle is
å¼ êñð‚è º‚«è£íˆF¡ ²Ÿø÷¾ 18 cm âQ™ Ü„êñð‚è º‚«è£íˆF¡ àòó‹
2
(A) 7 (B) 3 cm (C) 2 3 cm (D) 3 3 cm
Solution:
Perimeter of equilateral triangle 3a = 18cm
18
a= =6
3
a = 6cm

50
 TNPSC èí‚°

3
h= a
2
3 3
= x 6 = 3 3 cm
2
Answer: 3 3 cm

19. If the lengths of the sides of a triangle are 11 cm, 60 cm and 61 cm, then its area is
å¼ º‚«è£íˆF¡ ð‚èƒèœ 11 ªê.e, 60 ªê.e ñŸÁ‹ 61 ªê.e âQ™ Üî¡ ðóŠð÷¾
â¡ð¶
(A) 660 ê.ªê.e (B) 330 ê.ªê.e (C) 145 ê.ªê.e (D) 310 ê.ªê.e
Solution:
a = 11cm, b = 60 cm, c = 61 cm

a+b+c 11 + 60 + 61 132
S= = = = 66
2 2 2
s = 66
The area of triangle

ê˜è£˜ ä.ã.âv Üè£ìI

= s ( s − a)( s − b)( s − c)

= 66(66 − 11)(66 − 60)(66 − 61)

= 66(55)(6)(5)
= 11x 6x11x 5 x 6 x 5
= 11 x 6 x 5 = 330 sq.cm
Answer: 330 sq.cm

20. Find the area of equilateral triangle whose circumference is 180 cm.
180 ªê.e ²Ÿø÷¾ ªè£‡ì å¼ êñð‚è º‚«è£íˆF¡ ðóŠð÷¬õ‚ 裇è.
(A) 155.88 ªê.e2 (B) 1588.8 ªê.e2 (C) 900 ªê.e2 (D) 900.8 ªê.e2
Solution:
3a = 180 cm
180
a = 3 = 60 cm

3 2
Area of equilatral triangle = a
4

3 15
= x 60 x 60 = 900 x 3
4
= 900 x 1.732 (Note: 1.732 = 3)
Answer = 1588.8 cm2

21. The sides of a triangle are 8 m, 10 m and 6 m, then the area of the triangle is
å¼ º‚«è£íˆF¡ ð‚è Ü÷¾èœ 8 e, 10e ñŸÁ‹ 6e âQ™ Üî¡ ðóŠð÷¾ âšõ÷¾?

51
 TNPSC èí‚°

(A) 24 e2 (B) 18 e2 (C) 36 e2 (D) 72 e2

Solution:
a = 8m, b = 10m, c = 6m

a+b+c
S=
2
8 + 10 + 6 24
= = = 12 cm
2 2
S = 12 cm
A = s ( s − a)( s − b)( s − c)
= 12(12 − 8)(12 − 10)(12 − 6)
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

= 12(4)(2)(6)
= 12 x 2 x 2 x 12
= 12 x 2 = 24 m2
Answer: 24 m2

22.
In a triangle with area 48 sq. cm, its base is 4 cm greater than its altitude. Then the length of the
base of the triangle is
48 ê.ªê.e. ð󊹬ìò å¼ º‚«è£íˆF™, Üî¡ Ü®Šð‚è‹, Üî¡ °ˆ¶òóˆ¬î Mì
4 ªê.e ÜFè‹ âQ™ Üî¡ Ü®Šð‚èˆF¡ Ü÷¾
(A) 8 ªê.e (B) 12 ªê.e (C) 16 ªê.e (D) 10 ªê.e
Solution:
Area of triangle = ½ bh
Consider , h = x, b = x + 4
48 = ½ (x) (x+4)
x2 + 4x = 96
x2 + 4x - 96 = 0
x = 8, Altitude = 8
h = x = 8 cm
b = x + 4 = 8 + 4 = 12cm
Answer: 12 cm x = 8, x = -12 (Base can not be negative)

23. Find the side of the equilateral triangle if the area of an equi-lateral triangle is 900 3 cm2
å¼ êñð‚è º‚«è£íˆF¡ ðóŠð÷¾ 900 3 ªê.e.2 âQ™ ð‚è Ü÷¬õ‚ 裇è.
(A) 30 ªê.e. (B) 60 ªê.e. (C) 90 ªê.e. (D) 120 ªê.e.
Solution:
The area of equilateral triangle (A) = 900 3 cm2
a=?
3 2
A= a
4

52
 TNPSC èí‚°

3 2
900 3 = a
4
4
a2 = 900 3 x
3
a2 = 3600
a = 60
Answer: a= 60 cm

24. The semi perimeter of a triangle having sides 15 cm, 20 cm and 25 cm is
15 ªê.e., 20 ªê.e. ñŸÁ‹ 25 ªê.e. ð‚è Ü÷¾èœ ªè£‡ì å¼ º‚«è£íˆF¡ ܬó„
²Ÿø÷¾
(A) 15 ªê.e. (B) 45 ªê.e. (C) 30 ªê.e. (D) 60 ªê.e.
Solution:
a = 15 cm, b = 20 cm, c = 25 cm

a+b+c 15 + 20 + 25 60
S= = = = 30 cm

ê˜è£˜ ä.ã.âv Üè£ìI

2 2 2
Answer : S = 30 cm (Semi perimeter)

25. Find the area of triangle whose sides are 25cm, 24cm, and 7cm. (Tri)
å¼ º‚«è£íˆF¡ Í¡Á ð‚èƒèœ º¬ø«ò 25 ªê.e, 24 ªê.e ñŸÁ‹ 7ªê.e âQ™
ðóŠð÷¾ â¡ù?
(A) 84 ê.ªê.e (B) 87.5 ê.ªê.e (C) 90 ê.ªê.e (D) 300 ê.ªê.e
Solution:
a = 25cm, b = 24 cm, c = 7 cm

a + b + c 25 + 24 + 7 56
S= = = = 28
2 2 2
A = s ( s − a)( s − b)( s − c)
= 28(28 − 25)(28 − 24)(28 − 7)

= 28(3)(4)(21) = 7056 = 84
Answer: A = 84 cm2

26. The ratio of the area of a square of side ‘a’ and equilateral triangle of side ‘a’ is
‘a’ ð‚è‹ ªè£‡ì å¼ ê¶óˆF¡ ðóŠð÷¾ ñŸÁ‹ ‘a’ ð‚è‹ ªè£‡ì êñð‚è
º‚«è£íˆF¡ ðóŠð÷¾-èO¡ MAî‹ ________ Ý°‹
(A) 2 : 1 (B) 2 : 3 (C) 4 : 3 (D) 4 : 3
Solution:
3 2
a2 : a
4
3 2
a2 : a
4
53
 TNPSC èí‚°

4: 3
Answer : 4 : 3

The circumference of the base of a 12 cm. high wooden solid cone is 44 cm. Find the volume.
27.
ñóˆFù£ô£ù å¼ F‡ñ‚ ËH¡ Ü®„ ²Ÿø÷¾ 44 ªê.e ñŸÁ‹ Üî¡ àòó‹ 12 ªê.e
âQ™ ܈F‡ñ‚ ËH¡ èù Ü÷¬õ‚ 裇è.
(A) 606 cm3 (B) 610 cm3 (C) 614 cm3 (D) 616 cm3
Solution:
Circumference of the base = 2π r
2π r = 44cm
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

h = 12 cm
2π r = 44

44 7
r= x
2 22

r = 7cm

1
Volume of the cone = π r 2 h

3
(ËH¡ èùÜ÷¾)

1 22
= x x 7 x 7 x 12
3 7
V = 616 cm3
Answer : V = 616 cm3

28. A heap of paddy is in the form of a right circular cone whose diameter is 4.8 m and height 1.8m.
If the heap is to be covered exactly by a canvas to protect it from rain, find the area of the canvas
required
«ï˜õ†ì ˹ õ®M™ °M‚èŠð†ì ªïŸ°MòL¡ M†ì‹ 4.8 e ñŸÁ‹ Üî¡ àòó‹ 1.8e
â¡è. Þ‰ªïŸ°Mò¬ô ñ¬öJL¼‰¶ ð£¶è£‚è Aˆî£¡ ¶Eò£™ Iè„êKò£è ÍìŠð´Aø¶
âQ™, «î¬õò£ù Aˆî£¡ ¶EJ¡ ðóŠ¬ð‚ 裇.
(A) 22.6e2 (B) 27.2e2 (C) 13.6e2 (D) 11.3e2
Solution:
d = 4.8 m
r = 2.4 m
h = 1.8 m
l2 = h2 + r2
=l h2 + r 2

=l (1.8) 2 + (2.4) 2
= 3.24 + 5.76

54
 TNPSC èí‚°

= 3.24 + 5.76 = 9
l =3
curved surface area of the cone (CSA)
(ËH¡ õ¬÷ðóŠ¹) = π rl

22
= x 2.4 x 3
7

158.4
= =22.6 m2
7

29. If the height and the base area of a right circular cone are 5cm and 48 sq.cm respectively, then the
volume of the cone is equal to
5 ªê.e àòóº‹, 48 ê.ªê.e Ü®Šð‚èŠ ð󊹋 ªè£‡ì å¼ «ï˜õ†ì‚ ËH¡ èù
Ü÷¾
A) 240 cm3 B) 120 cm3 C) 80 cm3 D) 480 cm3

ê˜è£˜ ä.ã.âv Üè£ìI

Solution:
h = 5 cm
π r 2 = 48 sq. cm

7
r 2 = 48x = 15.27
22
r = 3.9 cm

1
V = π r 2h
3
1 22
x x (3.9) 2 x 5 ≈ 79.89
3 7
= 80
Answer = 80 cm3

30. The radii of two cones are in the ratio 2 : 1 their volumes are equal. Find the ratio of their heights
èù Ü÷¾èœ êññ£è àœ÷ Þó‡´ Ã‹¹èO¡ ÝóƒèO¡ MAî‹ 2 : 1 ÜõŸP¡
àòóƒèO¡ MAî‹
A) 1 : 8 B) 1 : 4 C) 2 : 1 D) 4 : 1
Solution:
Cone 1 Cone 2
Radius 2x x
h1 = ? h2 = ?

1 2 1 2
= V = π r h1 : V π r h2
3 3

55
 TNPSC èí‚°

1 1
π (2 x) 2 h1 : π x 2 h2
3 3
1 1
π (2) 2 x 2 h1 : π x 2 h2
3 3
4h1 : h2

h1 1
:
h2 4
1 : 4

The radius and the slant height of a cone are respectively ' r ' and ' l ' . What is the volume of the
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

31.
cone?
å¼ Ã‹H¡ Ýó‹ ñŸÁ‹ ꣻòó‹ º¬ø«ò ' r ' ñŸÁ‹ ' l ' âQ™ ËH¡ èùÜ÷¾
â¡ù?

1 1
(A) π r 2 l 2 − r 2 (B) π r 2 l 2 + r 2 (C) π rl (D) π r (l + r )
3 3

Solution:

1
V = π r 2h
3
2
l= h 2 + r 2 = h 2= l 2 − r 2

h
= l2 − r2

1
V = π r2 l2 − r2
3

32. A sector containing an angle of 180° is cut off from a circle of radius 7 cm and folded into a cone.

Find the curved surface area of the cone (Take π  22 7 )

7cm Ýóºœ÷ å¼ õ†ìˆFL¼‰¶ 180° ¬ñò «è£í‹ ªè£‡ì å¼ õ†ì‚
«è£íŠð°F¬ò ªõ†®ªò´ˆ¶. Üî¡ Ýóƒè¬÷ å¡P¬íˆ¶ å¼ Ã‹ð£‚Aù£™,

A¬ì‚°‹ ËH¡ õ¬÷ðóŠ¬ð 裇è ( π  22 7 )

(A) 77 (B) 22 (C) 88 (D) 66
Solution:

π r2
Area = xθ
360°
22 x 7 x 7
= x 180°
7 x 360°

11
22 x 7 x 7
= x 180 ° 9 =11 x 7
7 x 360 18 2 °
56
 TNPSC èí‚°

= 77
Answer = 77

Volume of right circular cone is 1782 cm3. Its height is 21 cm. Find the radius of the cone
33.
å¼ F‡ñ «ï˜õ†ì‚ ËH¡ èù Ü÷¾ 1782 è.ªê.e ñŸÁ‹ Üî¡ àòó‹ 21 ªê.e
âQ™ ܂ËH¡ Ýóˆ¬î‚ 裇è
A) 6 cm B) 7 cm C) 8 cm D) 9 cm
Solution:
Volume of the cone = 1782 cm3
h = 21 cm, r = ?

1
V = π r 2h
3
1 2
π r h = 1782
3
3 7 1
r 2 = 1782x x x
1 22 21 7

ê˜è£˜ ä.ã.âv Üè£ìI

1782
r2 =
22
r2 = 81

r = 9 cm
Answer = 9 cm

34.
The volume of the largest circular cone that can be cut of cube whose edge is 14cm is
14 ªê.e ð‚è Ü÷¾èœ ªè£‡ì å¼ èù ê¶óˆF™ Þ¼‰¶ ªõ†® â´‚èŠð´‹ IèŠ
ªðKò ËH¡ èùÜ÷¾
A) 718.67cm3 B) 718.76cm3 C) 781.67cm3 D) 781. 76cm3
Solution:
a = 14 cm, d = 14 cm, h = 14 cm
r = 7 cm

1
V = π r 2h
3
1 22
= x x 7 x 7 x 14
3 7
2156
=
3
= 718.67 cm3

Answer : 718.67 cm3

57
 TNPSC èí‚°

35. A boy cut a sector containing an angle of 140o from a circle of radius 15 cm and he folded the
22
sector into a cone. What is the curved surface area of the cone (π = )
7
15 ªê.e Ýóºœ÷ å¼ õ†ìˆFL¼‰¶ 140o ¬ñò‚«è£í‹ ªè£‡ì å¼ õ†ì‚«è£íŠ
ð°F¬ò å¼ ñ£íõ¡ ªõ†®ªò´ˆ¶ Üî¡ Ýóƒè¬÷ å¡P¬íˆ¶ å¼

 22 
Ëð£‚Aù£™, A¬ì‚°‹ ËH¡ õ¬÷ðóŠ¹ â¡ù?  π = 
 7 
(A) 572 sq.cm (B) 527 sq.cm
(C) 275 sq.cm (D) 257 sq.cm
Solution:

π r2
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

Area = xθ
360°
22 x 15 x 15
x 140°
7 x 360°

11 5 5
22 x 15 x 15
x 140 ° 7 = 11x 5 x 5 = 275 sq.cm
7 x 360 18 ° 93

An iron right circular cone of diameter 8 cm and height 12 cm is melted and recast into spherical
36.
lead shots each of radius 4 mm. How many lead shots can be made?
8 ªê.e M†ìº‹ 12 ªê.e àòóº‹ ªè£‡ì å¼ «ï˜ õ†ì F‡ñ Þ¼‹¹‚
Ëð£ù¶ ༂èŠð†´ 4 I.e Ýóºœ÷ F‡ñ‚ «è£÷ õ®õ °‡´è÷£è
õ£˜‚èŠð†ì£™ A¬ì‚°‹ «è£÷ õ®õ °‡´èO¡ â‡E‚¬è âšõ÷¾?
(A) 75 (B) 750 (C) 7500 (D) 480
Solution:


12 cm

8 cm

d = 8 cm, r = 4 cm, h = 12 cm
radius of spherical lead shot = 4mm
10 mm = 1 cm

4
4 mm = cm = 0.4 cm
10

Volume of right circular cone = n x (volume of spherical lead shots)

Volume of right circular cone

n=
Spherical lead shots
Cone

58
 TNPSC èí‚°

1
= π 42 x12
3
π x 43 cm3
Spherical lead

4
= π r3
3
4
= x π x (0.4)3
3

π x43 3
n= = 1000x
4 4
x π x (0.4)3
3
n = 750
Answer : 750

37. A girl is 37.5 m away from a tower. Her eye level above the ground is 2.5 m. The angle of elevation
of the tower from her eyes is 45°. What is the height of the tower?

ê˜è£˜ ä.ã.âv Üè£ìI

å¼ «è£¹óˆFL¼‰¶ 37.5 e ÉóˆF™ G¡Á ªè£‡®¼‚°‹ å¼ ñ£íM «è£¹óˆF¡
à„C¬ò 45° ãŸø‚ «è£íˆF™ 裇Aø£œ.ÜõÀ¬ìò A¬ìG¬ôŠ 𣘬õ‚
«è£´ î¬óJL¼‰¶ 2.5 e àòóˆF™ àœ÷¶ âQ™, «è£¹óˆF¡ àòó‹ â¡ù?
(A)35 e (B)37.5 e (C) 40 e (D) 42.5 e
Solution:
B

A 45°
2.5 cm 37.5cm
C

θ= 45°

BC
tan 45° = , In BAC,
AC
x
tan 45° =
37.5

tan 45° = 1
x = 37.5
Hight of the tower = 37.5 + 2.5 = 40 m

38. Two right circular cones have equal radii. If their slant heights are in the ratio 4 : 3, then their
respective curved surface areas are in the ratio
Þó‡´ «ï˜õ†ì Ë¹èœ å«ó êññ£ù Ýóƒè¬÷ ªè£‡´œ÷¶. ܬõèO¡ ꣌

59
 TNPSC èí‚°

àòóƒèœ 4 : 3 â¡ø MAîˆF™ Þ¼ŠH¡ ܬõèO¡ õ¬÷Š¹ø ð‡¹èœ º¬ø«ò

................. MAîˆF™ Þ¼‚°‹.
(A) 16 : 9 (B) 2 : 3 (C) 4 : 3 (D) 3 : 4
Solution:
The ratio of the slant height l1 : l2
4:3
Curved surface are π rl1 : π rl2
4:3
39. What is the volume of a cone having a base of radius 10 cm and height 21 cm?
å¼ Ã‹H¡ Ü® Ýó‹ 10 ªêe ñŸÁ‹ àòó‹ 21 ªê.e âQ™ ËH¡ èù Ü÷¾
A) 2200 cm3 B) 3000 cm3 C) 5600 cm3 D) 6600 cm3
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

Solution:
r = 10 cm
h = 21 cm

1
Volume of the cone = π r 2 h
3
1 22 3
= x x (10) x (10) x 21
3 7
= 2200
Answer : 2200 cm3

40. Curved surface area of the cone is

ËH¡ õ¬÷ ðóŠ¹
1
A) π r h
2
B) π r (l + r ) (C) π r 2 (D) π rl
3
Answer: (D) π rl

If the diameter and height of a right circular cone are respectively 12 cm and 8cm. Then its slant
41.
height is
«ï˜ õ†ì‚ ËH¡ M†ì‹ ñŸÁ‹ àòó‹ º¬ø«ò 12 ªê.e ñŸÁ‹ 8 ªê.e âQ™ Üî¡
ꣻòó‹ â¡ð¶
A) 10 cm B) 20 cm C) 30 cm D) 96 cm
Solution:

h = 8cm

12 cm

d = 12 cm, r = 6 cm
h = 8 cm

60
 TNPSC èí‚°

2
l= h2 + r 2

=l h2 + r 2

=l (8) 2 + (6) 2
=l 64 + 36
l = 100
l = 10 cm
Answer = 10 cm

42. If the base of a cone is not circular then, it is called

A) circular cone B) oblique cone
C) right circular cone D) oblique and circular cone
ËH¡ Ü®Šð£è‹ õ†ìõ®M™ Þ™¬ô âQ™, ܶ................Ý°‹.
A) õ†ì‚ ˹ B) ꣌¾‚ ˹
C) «ï˜õ†ì‚ ˹ D) ꣌¾ ñŸÁ‹ õ†ì‚ ˹

ê˜è£˜ ä.ã.âv Üè£ìI

43.
Find the area of the iron sheet required to prepare a cone 24 cm high with base radius 7 cm
24 ªê.e. àòóº‹ Ü®ð‚è Ýó‹ 7 ªê.e. ªè£‡ì ˹ îò£K‚è «î¬õò£ù Þ¼‹¹
î膮¡ ðóŠ¹ è£‡è
(A) 704 cm2 (B) 702 cm2 (C) 700 cm2 (D) 668 cm2
Solution:
h = 24 cm
r = 7 cm

=l h2 + r 2

=
l (7) 2 + (24) 2
=l 49 + 576

l = 625
l = 25
Area of iron sheet (TSA) = π rl + π r 2
= π r (l + r )

22
=
x7(25 + 7)
7
= 704 cm2

Answer : 704 cm2

44. A circus tent is cylindrical to a height of 3 m and conical above it. If the base radius is 52.5 m and
slant height of the cone is 53 m, find the area of canvas required to make the tent.

61
 TNPSC èí‚°

å¼ ê˜‚èv Ãì£óñ£ù¶ 3 e àòóºœ÷ ༬÷J¡ e¶ ˹ ܬñ‰î£Ÿ «ð£¡ø

õ®õˆF½œ÷¶. Üî¡ Ü®Šð‚è Ýó‹ 52.5 e. ËH¡ ꣻòó‹ 53 e âQ™, Ü‚Ãì£ó‹
ܬñ‚èˆ «î¬õò£ù Aˆî£¡ ¶EJ¡ ðóŠ¬ð‚ èí‚A´è.
(A) 315 π m2 (B) 3097.5 π m2 (C) 2782.5 π m2 (D) 2997.5 π m2
Solution:
Total canvas used = curved surface area of the cylinder + curved surface area of the cone
= 2π rh + π rl
= π r (2h
(2h ++rl)
)

53 m
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

52.5 m

3m

105 m

== π rx(252.5
h + r(6
) + 53)
= π rx(252.5
h + rx) 59
= 3097.5
= π rm (22h + r )
Answer: =
3097.5π rm(22 h + r )

Find the length of the canvas 1.25 m wide required to build a conical tent of base radius 7 metres
45.
and height 24 metres.
7 e†ì˜ Ü® Ýóº‹, 24 e†ì˜ àòóº¬ìò å¼ Ã‹¹ õ®õ Ãì£óˆF¬ù õ®õ¬ñ‚èˆ
«î¬õò£ù 1.25e Üèôº¬ìò ºó†´ˆ¶EJ¡ c÷‹ 裇è
(A) 430 m (B) 440m (C) 445 m (D) 450 m
Solution:
b = 1.25 m
l=?
2
l= h2 + r 2

=l h2 + r 2

=l 49 + 576

l = 625
l = 25 m
Area of canvas = π rl

62
 TNPSC èí‚°

22
= x 7 x 25
7
A = 550 m2

Area 550
Length of Canvas = =
width 1.25

= 440 m
Answer: 440 m

46. The volume of a cone is 216 π cu.cm. If the base radius 9 cm. Find the height of the cone
å¼ «ï˜ õ†ì‚ ËH¡ èù Ü÷¾ 216 π è.ªê.e. ܂ËH¡ Ýó‹ 9 ªê.e . âQ™,
Üî¡ àòóˆ¬î‚ 裇è
(A) 7cm (B) 8cm (C) 6cm (D) 5cm
Solution:
V = 216 π cu.cm
r = 9 cm

ê˜è£˜ ä.ã.âv Üè£ìI

h=?
1 2
π r h = 216 π
3
216 x 3
h=
r2

216 x 3
=
93 x 9
216
=
27
h = 8 cm
Answer: 8 cm

47. The volume of a solid right circular cone is 4928 cu.cm. If its height is 24 cm then find the radius
of the cone, (use π = 22/7)
å¼ F‡ñ «ï˜õ†ì‚ ËH¡ èù Ü÷¾ 4928 è. ªêe Üî¡ àòó‹ 24 ªêe âQ™
ËH¡ Ýó‹ 裇è. Þƒ° (π = 22/7)
(A) 49 cm (B) 28 cm (C) 24 cm (D) 14 cm
Solution:
V = 4928 cu.cm
h = 24 cm
r=?

1
π r 2 h = 4928
3

63
 TNPSC èí‚°

7 1
r 2 = 4928 x 3 x x
22 24 8
34496
r2 =
176
r2 = 196
r = 14 cm
Answer: r = 14 cm

48. Curved Surface area of Frustum of a cone is

(A) 2πr2h sq.units (B) 2πrh sq.units
(C) πrl sq.units (D) π(R+r)l sq.units
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

å¼ Ã‹H¡ Þ¬ì‚è‡ìˆF¡ õ¬÷ðóŠ¹ â¡ð¶

(A) 2πr2h ê.Üô°èœ (B) 2πrh ê. Üô°èœ
(C) πrl ê. Üô°èœ (D) π(R+r)l ê. Üô°èœ
Answer: (D) π(R+r)l sq.units

49. A sector containing an angle of 140º is cut off from a circle of radius 9 cm and folded into a cone.

 22 
Find the curved surface area of the cone.  π = 
 7 

9 ªê.e Ýóºœ÷ å¼ õ†ìˆFL¼‰¶ 140º ¬ñò‚ «è£í‹ ªè£‡ì å¼ õ†ì

«è£íŠð°F¬ò ªõ†®ªò´ˆ¶ Üî¡ Ýóƒè¬÷ å¡P¬í‰¶ å¼ Ã‹ð£‚Aù£™,

 22 
A¬ì‚°‹ ËH¡ õ¬÷ðóŠ¬ð 裇è.  π = 
 7 
(A) 99 sq. cm (B) 254.57 sq. cm
(C) 22 sq. cm (D) 126 sq. cm
Solution:
r = 9 cm
θ 140°
=

π r2
A= xθ
360°

2211 9x 9
= x x 140 ° 20
7 360° 40 2

= 11 x 9 = 99
Answer: A = 99sq. cm

50. A cone of height 24 cm is made up of modeling clay. A child reshapes it in the form of a cylinder
of same radius as cone. Find the height of a cylinder
èOñ‡ªè£‡´ ªêŒòŠð†ì 24 cm àòóºœ÷ å¼ Ã‹¬ð å¼ ° - ö‰¬î Ü«î Ýóºœ÷
64
 TNPSC èí‚°

æ˜ à¼¬÷ò£è ñ£ŸÁAø¶ âQ™ ༬÷J¡ àòó‹ 裇è.

(A) 42 cm (B) 48cm (C) 8 cm (D) 21 cm
Solution:
h1 = 24 cm
Volume of cone = Volume of cylinder

1 2
π r h1 = π r 2 h2
3
1
π r 2 h1 = π r 2 h2
3
1
h1 = h2
3
1
x24 = h2
3
1 8
x 24 = h2
3

ê˜è£˜ ä.ã.âv Üè£ìI

h2 = 8 cm


51.
Using clay, a student made a right circular cone of height 48 cm and base radius 12 cm. Another
student reshapes it in the form of sphere. Find the radius of the sphere.
èOñ‡¬íŠ ðò¡ð´ˆF å¼ ñ£íõ¡ 48 ªêe àòóº‹ 12 ªêe Ýóº‹ ªè£‡ì
«ï˜ õ†ì F‡ñ‚ Ë¬ð„ ªêŒî£˜. ܂ˬð ñŸªø£¼ ñ£íõ˜ å¼ F‡ñ‚ «è£÷ñ£è
ñ£ŸPù£˜. Üšõ£Á ñ£ŸøŠð†ì ¹Fò «è£÷ˆF¡ Ýóˆ¬î‚ 裇è.
(A) 12 cm (B) 15 cm (C) 9 cm (D) 14 cm
Solution:
h = 48 cm
R = 12 cm
Volume of cone = Volume of Sphere

1 4
π R2h = π r 3
3 3
R 2 h = 4r 3

12
3 12 x 12 x 48
r =
4
r = 1728 3

r = 12 cm
Answer : r = 12 cm

52. The radius and height of the cone are 7 cm and 24 cm respectively. Find the surface area of the
cone (CSA)
å¼ Ã‹H¡ Ýó‹ 7 ªê.e. àòó‹ 24 ªê.e, âQ™ õ¬÷ðóŠ¹ è£‡è.
65
 TNPSC èí‚°

(A) 505 cm2 (B) 290 cm2 (C) 360 cm2 (D) 550 cm2
Solution:
r = 7 cm
h = 24 cm
Surface area of the cone
CSA = π rl
2 2 2
l= h + r
=l h2 + r 2

=l (24) 2 + (7) 2

=l 576 + 49
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

l = 625
l = 25

22
CSA = x 7 x 25 = 550 cm2
7
Answer: 550 cm2

53. A cone of height 7 cm and base radius 3 cm is carved a from a rectangular block of wood 10 cm
x 5 cm x 2 cm. The percentage of wood wasted is
10 ªê.e x 5 ªê.e x 2 ªê.e Ü÷¾ ªè£‡ì ªêšõè õ®õ ñóŠðô¬èJL¼‰¶ 7 ªê.e
àòó‹ 3 ªê.e Ýó‹ ªè£‡ì ˹ ªõ†® â´ˆ¶ ªêŒòŠð´Aø¶, âQ™, âˆî¬ù
êîiî‹ ñó‚ 膬ì ií£°‹.
(A) 34% (B) 46% (C) 54% (D) 66%
Solution:
7cm

r = 3cm

l = 10 cm
b =5 cm
h = 2 cm

1
Volume = l x b x h Volume = π r 2 h
3
1 22
= 10 x 5 x 2 = 100 = x x 3 2 x 7 = 66 cm3
3 7
Volume of rectangular block = 100 cm3
Volume of cone = 66 cm3
The percentage of wood wasted is = 100 - 66
= 34
Answer: 34 %

66
 TNPSC èí‚°

54.
If the ratio of volumes of two cones is 2 : 3 and the ratio of the radii of their bases is 1 : 2, then the
ratio of their heights will be ?
(A) 3 : 8 (B) 8 : 3 (C) 9 : 2 (D) 8 : 1
Þ¼ ˹èO¡ èù Ü÷¾èO¡ MAî‹ 2:3 ñŸÁ‹ ÜõŸP¡ Ü®Šð‚è ÝóƒèO¡
MAî‹ 1:2 Ý辋 Þ¼‰î£™ ÜõŸP¡ àòóƒèO¡ MAî‹ ?
Solution:
Cone Cone radius radius
V1 : V2 r1 : r2
2 : 3 1 : 2

1 2
π r1 h1 2
3 =
1 2
π r2 h2 3
3

1  1 
3  π r12 h1  = 2  π r2 2 h2 

ê˜è£˜ ä.ã.âv Üè£ìI

3  3 

1  1 
3  π ( x) 2 h1  = 2  π (2 x) 2 h2 
3  3 

1
π x x 2 x h1 = 2x x π x4 x 2 xh2
3
8
h1 = h2
3
h1 8
=
h2 3

Ratio h1 : h2 = 8 : 3

55.
If the radius of the base of a cone is tripled and height is doubled then the volume is made
å¼ Ã‹H¡ Ýó‹ º‹ñìƒè£è¾‹ àòó‹ Þ¼ñìƒè£è¾‹ ñ£Pù£™ èù Ü÷¾ âˆî¬ù
ñìƒè£è ñ£Á‹ ?
(A) 6 times (B) 12 times (C) 18 times (D) unchanged
Solution:
Cone r : h
3:2

1
V = π r 2h
3
1
V = π (3r ) 2 (2h)
3

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 TNPSC èí‚°

1
V = π x 9r 2 x 2h
3
1 
V = 18  π r 2 h 
3 

V = 18 times (Volume is made)

56. If V is the volume of the cone of radius r and V1 is the volume of the cone when the radius is
doubled then
ËH¡ Ýó‹ ‘r’ â¡ø£™ Üî¡ èùÜ÷¾ ‘V’ Ý°‹. ËH¡ Ýóˆ¬î Þ󆮈 Üî¡
èù Ü÷¾ V1 Ý°‹. Üšõ£ªø¡ø£™.
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

(A) V = 4V1 (B) V= 2V1 (C) V1= 2V (D) V1 = 4V

Solution:

1
V = π r 2h
3
1
V1 = π (2r ) 2 h
3
1
V1 = π x 4r 2 h
3
1 
V1 = 4  π r 2 h 
 3 
V1 = 4 V

57. The base and height of a cylinder and cone are the same. The
volume of the cyclinder is x cm3. The volume of cone is
Ü®ð‚躋 àòóº‹, ༬÷ ñŸÁ‹ ËH™ êññ£è àœ÷ù. ༬÷J¡ èù Ü÷¾
x cm3 âQ™ ËH¡ èùÜ÷¾

1 1 1
(A) x cm3 (B) x cm3 (C) x cm3 (D) x cm3
4 3 2
Solution:

V1 = x cm3 V2 = ?

1
V1 = π r 2 h
V2 = (π r 2 h)
3

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 TNPSC èí‚°

1
x = π r 2 h V2 = x cm3
3
1
Answer : x cm3
3

ê˜è£˜ ä.ã.âv Üè£ìI

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 TNPSC èí‚°
èù„ê¶ó‹ & èù„ªêšõè‹
Cube & Cubiod

Cube and Cubiod Sums:

1. å¼ èù„ê¶óˆF¡ ªñ£ˆîðóŠ¹ 294cm2 âQ™ ÜõŸP¡ ð‚è ðóŠ¹, èùÜ÷¾ 裇è.

If the total surface area of a cube is 294 cm2, then find the lateral surface area and Volume respec-
tively
i) Lateral surface area (ð‚èŠðóŠ¹) = 4a2
6a 2 = 294

294
a2 =
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

6
2  294 
= 4=
  196cm 2
 6 
3 2
= 196 cm

ii) Volume (èùÜ÷¾) = a3

294
a2
= = 49
6
a=7
a3 = 7 x 7 x 7
a 3 = 343cm 3

2. Find the height of a cuboid whose base area is 96 cm2 and volume is 480 cm3.
å¼ èù„ªêšõèˆF¡ Ü®Šð‚è ðóŠð÷¾ 96 cm2 ñŸÁ‹ èùÜ÷¾ 480 cm3 âQ™ àòó‹
â¡ù ?
Base area of cuboid (èù„ªêšõèˆF¡ Ü®Šð‚è‹) = 96 cm3
l x b = 96 cm 2
x b x h 480 cm 3
Volume l =
=
96 x h = 480

480
h=
96

h = 5 cm

3. å¼ èù„ê¶ó õ®õ c˜ˆªî£†®J¡ ªè£œ÷÷¾ 8000 L âQ™ Üî¡ ð‚è‹ ò£¶ ?

A cubical tank can hold 8000 l of water. Find the dimension of its side ?
1000 l = 1m3
1 Hectare = 10000 m2

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.. èù„ê¶óˆF¡ èùÜ÷¾. = a 3 = 8000 l

3
a3 = 8m
a= 38
a = 2m

4. If the total surface area of a cube is 486 cm2. Find the lateral surface area and volume respectively.
å¼ èù„ê¶óˆF¡ ªñ£ˆî ¹øŠðóŠ¹ 486 cm2 âQ™ Üî¡ ð‚èŠ ð󊹋, èù Ü÷¾‹
º¬ø«ò ......, ...... Ý°‹.
6a 2 = 486cm 2
èùê¶óˆF¡ ªñ£ˆî ¹øŠðóŠ¹ (cube)
486
a2 =
6

i) lateral surface (ð‚èŠ ðóŠ¹) = 4a 2 162

2 486
 
= 4 3
 6 
= 324cm 2

ê˜è£˜ ä.ã.âv Üè£ìI

ii) Volume (èùÜ÷¾) = a3

486
a2 =
6
a 2 = 81 ⇒ a = 9
a3 = 9 x 9 x 9
a 3 = 729cm 3

5. Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm3.
å¼ èù„ªêšõèˆF¡ Ü®ŠðóŠ¹ ñŸÁ‹ èùÜ÷¾èœ º¬ø«ò 180 cm2 ñŸÁ‹ 900 cm3
âQ™ Üî¡ àòó‹ â¡ù?
Base area of cuboid (èù„ªêšõèˆF¡ Ü®ŠðóŠ¹) = 180 cm2
l x b = 180 cm2
Volume (èùÜ÷¾) = l x b x h
l x b x h = 900cm 3
180 x h = 900
900
h=
180
h = 5cm

6. The lateral surface area of a cube of side 12cm is

12cm ð‚è Ü÷¾ ªè£‡ì èù„ê¶óˆF¡ ð‚èŠ ðóŠ¹
Side of cube (èù„ê¶óˆF¡ ð‚è Ü÷¾) = 12 cm = a
Lateral surface (ð‚èŠðóŠ¹) = 4a2

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 TNPSC èí‚°
= 4 x 12 x 12
= 576 cm2
7. A cubical tank can hold 27,000 l of water. Find the dimension of its side.
å¼ èù„ê¶ó õ®õ c˜ˆªî£†®J¡ ªè£œ÷÷¾ 27,000L âQ™ Üî¡ ð‚è Ü÷¬õ‚
裇.
èù„ê¶ó õ®õ c˜ˆªî£†®J¡ ªè£œ÷¾ = 27,000 L
Volume of cube a3 = 27000 [1000 l = 1 m3]
a3 = 27 m3
a = 3 27
a = 3m
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

8. The volume of the cube is 729cm3. What is the length of the diagnol.
å¼ èù ê¶óˆF¡ èùÜ÷¾ 729 ªê.e.2 âQ™ Üî¡ Í¬ôM†ì‹ âšõ÷¾?
èù„ê¶óˆF¡ èùÜ÷¾ (Volume of cube a3 = 729 cm3

a = 3 729
a = 9cm
length of diagonal (.....) d = 3a

d = 9 3cm

9. Water flow into a tank 200m x 150m through a rectangular pipe. 1.5x1.25m at 20kmph. Water
level rise by 2 metres in _______ minutes.
200 e c÷º‹, 150 e. Üèô‹ àœ÷ ªî£†®J¡ àœ«÷ 1.5e Üèô‹ àœ÷ ªî£†®J¡
àœ«÷ 1.5e c÷º‹ 1.25 e. Üèôº‹ àœ÷ ªêšõè °ö£Œ Íôñ£è ñE‚° 20A.e./ñ
«õèˆF™ î‡a˜ M¿‰î£™, î‡a˜ 2 e àòó âˆî¬ù GIìƒèœ Ý°‹-?
Volume of water that has to flow
= 200 x 150 x 2 = 60000m3
Speed of water flow = 20 x 5/18
S = V/t
100
= m/s
18
100
Volume of water flowing in 1 second = 1.5 x 1.25 x 18

Number of seconds required to rise the level of water by 2 metres

100
= 60000 ÷ (1.5 x 1.25 x )
18

= 5760 sec.
= 96 min (or) 1 hour 36 minutes.

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 TNPSC èí‚°

1000 5
Note : Speed of water = 20 x 60 x 60 =20 x 18
Another Method:
Volume of the tank = 200 x 150 x 2
= 60000 m3
20x1000
Length of water in 1 min =
60

1000
= m
3

1000
Volume of water flown per min = 1.5 x 1.25 x
3
= 625 m3 / min
60000
Time requried to rise the water by 2 m = = 96 mins
625

10. The lateral surface area and total surface area of the cube having side 24cm are respectively
ð‚èÜ÷¾ 24 ªê.e. à¬ìò å¼ èùê¶óˆF¡ ¹øŠðóŠ¹ ñŸÁ‹ ªñ£ˆîŠ ðóŠ¹ â¡ð¶
º¬ø«ò
èù„ê¶óˆF¡ ð‚èÜ÷¾ (side) a = 24

ê˜è£˜ ä.ã.âv Üè£ìI

¹øŠðóŠ¹ (Lateral surface area) = 4a2
= 4 x 24 x 24
= 2304 cm2
ªñ£ˆîŠðóŠ¹ (Total surface area = 6a2
= 6 x 24 x 24
= 3456 cm2
Ans : 2304 cm2, 3456 cm2

11. A brick measures 20cm x 10cm x 7.5cm. How many bricks will be required for a wall 20m x 2m
x 0.75m?
å¼ ªêƒè™L¡ Ü÷¾ 20ªê.e. x 10. ªê.e. x 7.5 ªê.e. 20e. x 2e. x 0.75 e. Ü÷¾œ÷
²õ˜ 膴õ âˆî¬ù ªêƒèŸèœ «î¬õŠð´‹ ?

Volume of wall
Number of bricks = Volume of 1 brick

²õK¡ èù Ü÷¾
ªêƒèŸèO¡ â‡E‚¬è = å¼ ªêƒè™L¡ èùÜ÷¾
1m = 100 cm
10 10
2000 x 200 x 75
= 20 x 10 x 7.5
= 20,000

12. 3 ªê.e, 4 ªê.e, 5 ªê.e ð‚è Ü÷¾¬ìò Í¡Á èù„ ê¶óƒèœ ༂èŠð†´ å¼
ªðKò èù„ê¶óñ£è ñ£ŸøŠð†ì£™, Üî¡ ªñ£ˆîŠ ¹øŠðóŠ¬ð‚ 裇è.
The sides of a three metallic cubes are 3 cm, 4 cm and 5 cm, it is melted and formed into a
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 TNPSC èí‚°

large cube. Then find its total surface area

A) 208 ê.ªê.e B) 216 ê.ªê.e C) 218 ê.ªê.e D) 220 ê.ªê.e
ªðKò èù„ê¶óˆF¡ CPò èùê¶óƒèO¡
=
èù Ü÷¾ èù Ü÷¾èO¡ ôî™
A3 =a13 + a23 + a33
= 33 + 43 + 53
=27 + 64 + 125
A3 = 216
ªðKò èù„ê¶óˆF¡
ð‚è Ü÷¾ A = 3 216
A = 6ªê.e
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

ªðKò èù„ ê¶óˆF¡

ð‚è Ü÷¾ =6a2
=6x62
=216 ê.ªê.e
13. 15 ªê.e ð‚è Ü÷¾œ÷ å¼ èù„ê¶ó‹ à¼õ£‚è 3 ªê.e ð‚è Ü÷¾œ÷ èù„ê¶óƒèœ
âˆî¬ù «î¬õ?
How many 3 cm side cubes are needed to make a 15 cm side cube?
A) 125 B) 120 C) 115 D) 116
«î¬õò£ù èù„ê¶óƒèO¡ = ªðKò èù„ê¶óˆF¡ èùÜ÷¾

â‡E‚¬è CPò èù„ê¶óˆF¡ èùÜ÷¾

15x15x15
= 3x3x3
= 125 èù ê¶óƒèœ «î¬õ
14. 
å¼ èù„ê¶ó õ®õ c˜ˆ«î‚航®J¡ èùÜ÷¾ 64,000L âQ™, Üî¡ ð‚è
Ü÷¾ 裇è.
A cubical tank can hold 64,000 l of water. Find the length of its side in metres
A) 3e B) 4e C) 5e D) 6e
1000 L†ì˜= 1e3
ªî£†®J¡ èùÜ÷¾ = 64,000L
64,000
= 1,000
V = 64e3
a3 = 64e3
a = 3 64
a = 4e
15. å¼ èù„ê¶óˆF¡ ð‚èŠðóŠ¹ 900 ê.ªê.e âQ™, ð‚è Ü÷¬õ‚ 裇è.
If the lateral surface area of cube is 900 cm2. Then, find its side?
A) 12 ªê.e B) 13 ªê.e C) 14 ªê.e D) 15 ªê.e
èù„ê¶óˆF¡ ð‚èðóŠ¹ = 900 ê.ªê.e
4a2 = 900
900
a2 = 4

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 TNPSC èí‚°

a2 = 225
a = 225
ð‚èÜ÷¾ a= 15 ªê.e
16. 64 è.ªê.e èù Ü÷¾œ÷ Þ¼ èù„ê¶óƒèœ Þ¬í‚èŠð†ì£™ à‡ì£°‹ à¼õˆF¡
ð‚èŠðóŠ¹ è£‡è.
The volume of the two identical cube is 64 cm3 is joined end to end. Find the lateral surface
area of the new resulting shape
A) 94 ê.ªê.e B) 96 ê.ªê.e C) 98- ê.ªê.e D) 100 ê.ªê.e
Þ¼ èù„ ê¶óƒèœ Þ¬í»‹ «ð£¶ èùªêšõè‹ à¼õ£°‹
èù„ê¶óˆF¡ èùÜ÷¾ = 64 è.ªê.e
a3 = 64 4
a = 64
3 4
4 4 4
a = 4 ªê.e
8
ªðKò èù ªêšõèˆF¡ c÷‹ (l) = 4+4 = 8 ªê.e
àòó‹ h = 4 ªê.e, Üèô‹ b = 4 ªê.e
èùªêšõèˆF¡ ð‚èŠðóŠ¹ = 2h (l+b) ê.Ü

ê˜è£˜ ä.ã.âv Üè£ìI

= 2x4 (8+4)
= 8x12
= 96 ê.ªê.e
17. c÷‹ 60e, Üèô‹ 0.3e, àòó‹ 2e à¬ìò ²õ˜ â¿Šð 30ªê.e x 15 ªê.e x 20
ªê.e Ü÷¾ ªè£‡ì ªêƒè™èœ âˆî¬ù «î¬õ?
How many bricks of 30 cm, 15 cm and 20 cm are required to build a wall 60 m long, 0.3 m
wide and 2 m hight?
A) 4000 B) 500 C) 6000 D) 7000
«î¬õŠð´‹ ªêƒèŸèO¡ ²õK¡ èùÜ÷¾
=
â‡E‚¬è å¼ ªêƒè™L¡ èùÜ÷¾

= LBH
lbh
L = 60e = 6000ªê.e, B = 0.3e = 30 ªê.e
H = 2e = 200ªê.e,
l = 30ªê.e, b = 15ªê.e, h = 20 ªê.e
6000x30x200
= 30x15x20
= 4000
18. å¼ èù„ ªêšõèˆF¡ c÷‹ 7.5 e, Üèô‹ 3e, àòó‹ 5e âQ™ Üî¡ ªñ£ˆîŠ
ðóŠ¹ ñŸÁ‹ ð‚èŠ ðóŠ¬ð‚ 裇è.
Find the total surface area and lateral surface area of a cuboid whose length, breadth and height
are 7.5 m, 3 m and 5 m respectively
A) 150e2, 105e2 B) 160e2, 105e2
C) 150e2, 110e2 D) 160e2, 110e2
c÷‹ (l)= 7.5 e, Üèô‹ (b) = 3e
àòó‹ (h) = 5e
ªñ£ˆîŠðóŠ¹ (TSA) = 2 (lb + bh + hl)

75
 TNPSC èí‚°

= 2 ((7.5x3) + (3x5) + (7.5x5))

= 2 (22.5 + 15 + 37.5)

= 2 x 75
T.S.A = 150 e2
ð‚èŠðóŠ¹ (CSA)= 2 (l + b) x h
= 2 (7.5 + 3) x 5
= 2 (10.5) 5
= 105e2

19.  5 ªê.e ð‚è Ü÷¾ ªè£‡ì èù„ ê¶óˆF¡ ªñ£ˆîŠ ðóŠ¹ ñŸÁ‹ ð‚èŠðóŠ¬ð‚
裇è.
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

Find the total and lateral surface area of the cube whose side is 5 cm.
A) 150 ê.ªê.e, 100 ê.ªê.e2 B) 160 ê.ªê.e, 80 ê.ªê.e2
C) 150 ê.ªê.e, 80 ê.ªê.e2 D) 160 ê.ªê.e, 100 ê.ªê.e2
a= 5ªê.e
ªñ£ˆîŠ ðóŠ¹ = 6a2
= 6(52)
= 6x25
= 150 ê.ªê.e2
ð‚èŠðóŠ¹ = 4a2
= 4(52)
= 4x25
= 100 ê.ªê.e2

20. å¼ èù„ê¶óˆF¡ ªñ£ˆîŠ ¹øŠðóŠ¹ (TSA) 486 ªê.e2 âQ™ Üî¡ ð‚èŠ ðóŠ¬ð‚
(CSA) 裇è.
A cube has the total surface area of 486 cm2. Find its lateral surface area
A) 316 ªê.e2 B) 324 ªê.e2
C) 216 ªê.e2 D) 224 ªê.e2

ªñ£ˆîŠ ¹øŠðóŠ¹ (TSA) = 486 ªê.e2
6a2= 486
486
a2 = 6
a2 = 81
a = 9
èù„ê¶óˆF¡ ð‚è Ü÷¾ = 9ªê.e
èù„ê¶óˆF¡ ð‚è ðóŠ¹ (CSA) = 4a2
= 4x92
= 4x81
= 324ªê.e2

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21. å¼ èù„ê¶óˆF¡ ªñ£ˆîŠðóŠ¹ 2400 ªê.e2 âQ™, Üî¡ ð‚èŠðóŠ¬ð‚ 裇è
If the total surface area of the cube is 2400 cm2, then find its lateral surface area
A) 2400 ªê.e2 B) 1600 ªê.e2
C) 2100 ªê.e2 D) 1800 ªê.e2
èù„ê¶óˆF¡ ªñ£ˆîŠðóŠ¹ 6a2 = 2400
a = 400
2

a = 20 ªê.e
ð‚èŠðóŠ¹ = 4a2
= 4x202 = 4 x 400
= 1600 ªê.e2

22. 
7 ªê.e ð‚è Ü÷¾œ÷ å«ó ñ£F£ò£ù Þó‡´ èù„ê¶óƒèœ å¡Áì¡ å¡Á
ð‚èõ£†®™ Þ¬í‚èŠð´‹ «ð£¶ A¬ì‚°‹ ¹Fò èù„ªêšõèˆF¡ ªñ£ˆîŠðóŠ¹
裇è.
Two identical cubes of side 7 cm are joined end to end. Find the total and lateral surface area
of the new resulting cuboid.

ê˜è£˜ ä.ã.âv Üè£ìI

A) 400 ªê.e2 B) 480 ªê.e2
C) 500 ªê.e2 D) 490 ªê.e2
èù„ê¶óˆF¡ ð‚è Ü÷¾ = 7 ªê.e

7 ªê.e
7 7 →

7 7 7 7 ← 14 →
¹Fò èù„ ªêšõèˆF¡ c÷‹ (l) = 7+ 7 = 14 ªê.e
Üèô‹ (b) = 7 ªê.e
àòó‹ (h) = 7 ªê.e
ªñ£ˆîŠðóŠ¹ = 2 (lb + bh + hl)
= 2 ((14x7) + (7x7) + (14x7))
= 2 (98 + 49 + 98)
= 2 x 245
= 490 ªê.e2
Exercise sums :
1.
å¼ Ü¬øJ¡ c÷‹, Üèô‹ ñŸÁ‹ àòó‹ º¬ø«ò 12 e†ì˜, 9 e†ì˜ ñŸÁ‹ 6
e†ì˜, 1.5 e†ì˜ c÷‹ ªè£‡ì âˆî¬ù èù„ ê¶óŠ ªð†®è÷£™ Þ‰î ܬø¬ò
º¿¬ñò£è GóŠðô£‹?
The length, breadth and height of a room are respectively 12 metres, 9 metres and 6 meters.
How many cubic boxes are needed to fill the room if the side of each box is 1.5 metres?
(A) 1072 (B) 648 (C) 324 (D) 192
ܬøJ¡ èùÜ÷¾
(Volume of room) = lbh
= 12 x 9 x 6
= 648 m3
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 TNPSC èí‚°

å¼ èù„ê¶ó ªð†®J¡ èùÜ÷¾

3
(Volume of box) =a
3
= (1.5)
= 3.375 m3
648
Number of Boxes =
3.375
= 192

2.216 cm3 èù Ü÷¾ ªè£‡ì Þ¼ èù ê¶óƒèœ Þó‡¬ì»‹ «ê˜ˆ¶ å¼ èù

ªêšõè‹ à¼õ£‚èŠð´Aø¶. ܉î èù ªêšõèˆF¡ ªñ£ˆî ð‚è ðóŠH¡ ñFŠ¹
(ê.ªê.e.)
Two cubes each of volume 216 cm3 are joined to form a cubiod. The total surface area of the
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

resulting cuboid is (in cm2)

A) 120 B) 360 C) 300 D) 240

3
Volume of cube (èùÜ÷¾) = a
a 3 = 216
a = 3 216
a = 6 cm
l = 6+6=12 cm ; b = 6 cm ; h = 6 cm
èùªêšõèˆF¡ ªñ£ˆî ð‚è ðóŠ¹
(Surface area of cuboid) = 2 (lb+bh+lh)
= 2 [(12 x 6)+(6 x 6)+(12 x 6)]
= 2 [72 + 36+72 ]
= 2 x 180
= 360 cm 2

3. å¼ èù ê¶óˆF¡ èù Ü÷¾ 125 èù ªê.e. âQ™ Üî¡ ¹øŠðóŠð÷¾ âšõ÷¾?

The volume of a cube is 125 cm3. The surface area of the cube is
(A) 625 cm2 (B) 125 cm2 (C) 150 cm2 (D) 100 cm2
èù„ê¶óˆF¡ èùÜ÷¾
3
(Volume of cube) = A
A = 3 125
A=5
¹øŠðóŠð÷¾ = 6 x A2
[Total surface area] = 6 x 5 x 5
2
= 150 cm

4. 1 e ð‚è Ü÷¾ àœ÷ èùê¶ó ªð†®J™ âˆî¬ù 10 ªê.e ð‚è Ü÷¾œ÷ èùê¶óƒè¬÷
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 TNPSC èí‚°

¬õ‚èô£‹?
The number of small cubes with edge 10 cm that can be accommodated in a cubical box of
edge 1m is
(A) 10 (B) 100 (C) 1000 (D) 10000
èù„ê¶ó ªð†®J¡ èùÜ÷¾ = A3
A = 1 x 1 x1
A = 1
Volume of cubical box = (100 )3 [1 m = 100 cm]
= 1000000
èù„ê¶óˆF¡ èùÜ÷¾ = 103
= 1000

Volume of cubical box = 1000 000

1000
= 1000

ê˜è£˜ ä.ã.âv Üè£ìI

5. å¼ èù ê¶óˆF¡ ªñ£ˆî ¹øŠðóŠ¹ 384 ªê.e2 âQ™ Üî¡ èù Ü÷¾
The total surface area of a cube is 384cm2, then its volume is
A) 521cm3 B) 512 cm3 C) 412 cm3 D) 421cm3
èù„ê¶óˆF¡ ªñ£ˆî ¹øŠðóŠ¹
(Total surface area of cube) = 384 sq.cm
6a 2 = 384
64
384
2
a =
6
1

a 2 = 64
a=8
3
èùÜ÷¾ (Volume) = a
= 83

= 512cm3

6. 3 ªê.e x 18 ªê.e x 108 ªê.e Ü÷¾œ÷ èù ªêšõèˆFL¼‰¶ 3 ªê.e MO‹¹

Ü÷¾ ªè£‡ì âˆî¬ù èù„ê¶óƒèœ ªõ†ìô£‹?
How many cubes of 3 cm edge can be cut out of a cuboid of 3 cm x 18cm x108cm?
A) 216 B) 326 C) 36 D) 45

èùÜ÷¾ V = l x b x h
= 3 x 18 x108
= 5832
ð‚è‹ a = 3 cm
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3
V = a
= ( 3) = 27
3

1944
Volume of cuboid 5832 216
= 216
Volume of cube 27

7. å¼ èù„ê¶óˆF¡ å¼ ð‚èˆF¬ìò ²Ÿø÷¾ 20 ªê.e âQ™ Üî¡ èù Ü÷¾

The perimeter of one face of a cube is 20 cm. Its volume must be
(A) 215 cm3 (B) 200 cm3 (C) 125 cm3 (D) 8000 cm3
èù„ê¶óˆF¡ å¼ ð‚èˆF¡ ²Ÿø÷¾ (Perimeter of one face of a cube)
= 20 cm
4a = 20
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

5
20
a=
4
=5
èù„ê¶óˆF¡ èùÜ÷¾ (Volume of cube)
3
=a
( 5)
3

=
= 125 cm3

8. 4 3 ªê.e. ͬôM†ì‹ ªè£‡ì èù„ ê¶óˆF¡ èù Ü÷¾

What is the volume of a cube whose diagonal measure is 4 3 cm ?
(A) 30 cm3 (B) 46 cm3 (C) 60 cm3 (D) 64 cm3
Diagonal of a cube = 4 3 cm
Length of an edge of the cube be x cm
x 3 =4 3
x = 4cm
Volume of cube = ( 4 ) cm3
3

= 64cm3
9. 5e x 4e x 2 e Ü÷¾œ÷ å¼ °N ñíô£™ GóŠðŠð´Aø¶. å¼ èù e†ì¼‚°
ñí™ GóŠð Ý°‹ ªêô¾ Ï. 270 âQ™ ªñ£ˆî ªêô¬õ‚ 致H®.
Find the cost of filling a pit of dimensions 5mx4mx2m with soil if the rate of filling is Rs. 270
per cubic metre
A) Rs. 10,800 B) Rs. 1,080 C) Rs. 10,080 D) Rs. 18,000
l = 5 m; b = 4 m; h = 2 m
èùªêšõèˆF¡ èùÜ÷¾ (Volume of cuboid) = 5 x 4 x2 = 40 m3
ªñ£ˆî ªêô¾ (Total cost) = 270 x 40 = Rs.10800

10. å¼ èù ê¶óˆF¡ ð‚è‹ 50% àò˜ˆîŠð†ì£™ Üî¡ ªñ£ˆîŠ ¹øŠðóŠH™ ãŸð†ì

ÜFèKŠH¡ êîMAî‹ â¡ù-?

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Edge of a cube is increased by 50% then find percentage increase in total surface area of the
cube.
A) 108% B) 125 % C) 90% D) 80%
Solution:
Let the edge of the cube = a
Increased edge = a + 0.5 a = 1.5 a

6(1.5a ) 2 − 6a 2
% increase in the area = x100
6a 2
13.5a 2 − 6a 2 7.5 a 2
x100 = x100 = 125%
6a 2 6 a2

11. 12 ªê.e. 13 ªê.e... 25 ªê.e ÝAòõŸ¬ø º¬ø«ò ð‚è Ü÷¾è÷£è‚ ªè£‡ì 14

èù„ ê¶óƒèO¡ ªñ£ˆî èù Ü÷¬õ‚ 裇è
Find the total volume of 14 cubes whose edges are 12cm, 13cm ------------ 25 cm respectively
A) 621019 cm3 B) 269101 cm3
C) 962101cm3 D) 101269 cm3

ê˜è£˜ ä.ã.âv Üè£ìI

2

Sum of cubes of n natural number =  ( ) 

 n n+1 
2
Total volume of 14 cubes = 123 + 133+....253 
= [13 + 23+ ..... 253] - [13 + 23 + ..... 113]
2 2
 25(25 + 1)  11(11 + 1) 
=   −  
 2 2
 2 2
25x 26  11x 12 
13 6
=    − 
 2   2 

= [325]2 - [66]2
= 105625 - 4356
= 101269cm3

12. A â¡ð¶ ‘x’ Ü÷¾ ð‚è‹ àœ÷ èùê¶ó‹. B â¡ð¶ ‘x’ â‹ Ýó‹ ªè£‡ì «è£÷‹
A&Þ¡ èù Ü÷¾‚°‹ B&Þ¡ èù Ü÷¾‚°‹ Þ¬ì«ò àœ÷ MAî‹
A is cube of side x, B is sphere of radius x. The ratio of volume of A to volume of B is
A) 3 : 4 π B) 3π : 4 C) 4 π : 3 D) 4 : 3 π

A) 3 : 4 π B) 3π : 4 C) 4 π : 3 D) 4 : 3 π
Solution:
Volume of A = x3
4 3
Volume of B = πx
3
x3 3
Ratio of Volume A to Volume B = =
4 4π
π x3
3
Answer: 3 : 4 π
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13. 6 ªê.e., 8 ªê.e., 10 ªê.e. c÷ºœ÷ MO‹¹è¬÷‚ ªè£‡ì Í¡Á èù„ê¶ó

õ®M™ àœ÷ Þ¼‹H¬ù ༂A å¼ èù„ê¶óˆ¬î à¼õ£‚Aù£™ Ü‰î ¹Fò èù„
ê¶óˆF¡ MO‹H¡ c÷‹
Three cubes of iron whose edges are 6 c.m, 8 c.m, 10c.m. respectively are melted and formed
into a single cube. The edge of the new cube is
A) 12c.m B) 14c.m C) 16c.m D) 18c.m
a1 = 6 a 2 =8 a 3 =10
3 3
V1 =a 1 V2 =a 2 V3 =a 33
V1 =216cm3
V2 =512cm3
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

V3 =1000cm3

Volume of new cube V = V1 +V2 +V3

= 216+512+1000
A 3 = 1728
A = 3 1728 = 12cm

14. 4 ªê.e ð‚è‹ ªè£‡ì å¼ èù ê¶óˆF¡ ܬùˆ¶ ð‚èƒèÀ‹ CõŠ¹ õ‡í‹

ÌCò H¡ 1 ªê.e ð‚è‹ ªè£‡ì CPò èù ê¶óƒè÷£è HK‚èŠð†ì£™ ÜõŸP™
Þó‡´ î÷ƒèœ õ‡í‹ ÌêŠð†ì CPò èù ê¶óƒèœ âˆî¬ù Þ¼‚°‹?
A cube of side 4 cm is painted red on all of its surface and then divided into various smaller
cubes of side 1 cm each. Find the number of smaller cubes with two surfaces painted red.
(A) 4 (B) 8 (C) 12 (D) 24
Side = 4 cm
Bigger cube = a 3 = 64cm3
Smaller cube = a 3 = 1cm3
ªñ£ˆî èùê¶óƒèœ = 64
ªñ£ˆî‹ 8 èùê¶óƒèœ 3 ð‚è‹ õ‡í‹ ÌêŠð†´œ÷¶.
8 x 3 = 24
15. 4 ªê.e à¬ìò èù ê¶óñ£ù¶ 1 ªê.e à¬ìò CPò èù ê¶óƒè÷£è ªõ†ìŠð´Aø¶
âQ™ CPò èù ê¶óƒèO¡ ªñ£ˆî ðóŠ¹ ò£¶?
A 4 cm. cube is cut into 1 cm cubes. The total surface area of all the small cube is
A. 96 cm2 B. 24 cm2 C. 384 cm2 D. 284 cm2
Volume of 4 cm cube = 43 = 64
Volume of 1 cm cube = 13 = 1
64
CPò èù„ê¶óƒèO¡ â‡E‚¬è = = 64
1

èù„ê¶óˆF¡ ªñ£ˆîðóŠ¹ = 6 (side)2 = 6(1)2 = 6 cm2

2
CPò èù„ê¶óƒèO¡ ªñ£ˆîðóŠ¹ = 64 x 6 = 384 cm

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16. èùê¶óˆF¡ èù Ü÷¾ = ............................... èù Üô°èœ

A. a2 B. 2 a2 C. 3a3 D. a3
Answer: a 3

17. 20 ªêe ð‚è Ü÷¾ ªè£‡ì Fì èù ê¶ó à¼õˆFL¼‰¶, âˆî¬ù 5 ªê.e ð‚è
Ü÷¾ ªè£‡ì CÁ Fì èù ê¶óƒè¬÷ à¼õ£‚èô£‹?
From the solid cube having side of 20 cm, how many number of small cubes having side 5 cm
can be formed ?
A. 46 B. 64 C. 56 D. 48
Number of a cube of edge
= a3
4 4 4
20 x 20 x 20
= = 64
5 x 5 x 5
1 1 1

18. 16m c÷º‹, 14m Üèôº‹, 7m àòóº‹ ªè£‡ì èùªêšõèˆF¡ èù Ü÷¾ ñŸÁ‹

ê˜è£˜ ä.ã.âv Üè£ìI

ðóŠ¹ è‡´H®
Find the volume and surface area of cuboid 16 m long , 14 m broad and 7m high
A) 1568 m3,868 cm2 B) 1560 m3,864 cm2
C) 1560 m3,860cm2 D) 1320 m3,850 cm2
èù„ªêšõèF¡ èùÜ÷¾
=lxbxh
= 16m x 14m x 7m
3
Volume = 1568 m
ðóŠ¹ = 2(lb + bh + hl)
= 2( 16 x 14 + 14 x 7 + 7 x 16)
= 2(224 + 98 + 112)
= 2(434)
Area = 868 m 2
19. 2 ªê.e î®ñ¡ ªè£‡ì àœ Ü÷¾ 6 ªê.e c÷‹ 2 ªê.e Üèô‹ ñŸÁ‹ 1 ªê.e
àòó‹ àœ÷ ñóŠªð†®J™ ñóˆF¡ èù Ü÷õ£ù¶
A wooden box of 2 cm thickness has inner measurements 6 cm long 2 cm breath and 1 cm
height. The volume of the wood in the box is
(A) 96 cm3 (B) 64 cm3 (C) 288 cm3 (D) 300 cm3
ñóˆF¡ àœ èùÜ÷¾
Volume = l x b x h
= 6 x 2 x 1 = 12
ñóˆF¡ ªõO èùÜ÷¾
l = 6 + 2 + 2 = 10 cm
b = 2 + 2 + 2 = 6 cm
h = 1 + 2 + 2 = 5 cm
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Volume = 10 x 6 x 5 = 300
ñóˆF¡ èùÜ÷¾ = 300 - 12 = 288 cm3

20. å¼ èù„ ê¶ó õ®õ c˜ˆ ªî£†®J¡ ªè£œ÷÷¾ 27,000 L†ì˜ âQ™ Üî¡ ð‚è
Ü÷¬õ‚ 裇 (e†ìK™)
A cubical tank can hold 27,000 litres of water. Find the dimension of its side (in metres)
(A) 27m (B) 9m (C) 3 m (D) 6m
c˜ˆªî£†®J¡ ªè£œ÷÷¾ = 27, 000 Liters
27 000
V = a3 = = 27m3 [1m3 = 1000 litres]
1000
a 3 = 27
a = 3 27 = 3m
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

21. å¼ èù ê¶óˆF¡ ªõOŠðóŠ¹ 96 ê.ªê.e âQ™ èù Ü÷¾ 裇?

The surface area of a cube is 96 cm2. Find its volume
(A) 16 cm3 (B) 48cm3 (C) 64 cm3 (D) 27 cm3
èùê¶óˆF¡ ªõOŠðóŠ¹
= 96 cm 2
96 = 6(sides) 2
96
= (sides) 2
6
(sides) 2 = 16
side = 4
èù„ê¶óˆF¡ èùÜ÷¾ = (Sides)3
= 43 = 64 cm3

22. 5 ªê.e ð‚è Ü÷¾ ªè£‡ì èù„ ê¶óˆF¡ èù Ü÷¾ 裇

Find the volume of a cube whose side 5 cm
(A) 100 cm3 (B) 110 cm3 (C) 125 cm3 (D) 105 cm3
a = 5,Volume of cube(èùê¶óˆF¡ èùÜ÷¾)

a 3 = 53 = 125cm3

23.
2 ≤ x ≤ 5,-1 ≤ y ≤ 3 Ý™ ܬìð´Aø ªêšõè ð°Fò£ù¬õ
Area of the rectangular region 2 ≤ x ≤ 5,-1 ≤ y ≤ 3 is
(A) 9 units (B) 12 units (C) 15 units (D) 20 units
The base of this rectangle is the change in X = 5 - 2 = 3
Height Y = 3-(-1) = 4
Area = 3 x 4 = 12 units

24. å¼ èù ê¶óˆF¡ èù Ü÷¾ 729 ªê.e3 âQ™ Üî¡ Í¬ôM†ì‹ âšõ÷¾

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The volume of the cube is 729 cm3. What is the length of the diagonal
(A) 9 2 cm (B) 9 3 cm (C) 18 cm (D) 18 3 cm
Volume of cube (èùê¶óˆF¡ èùÜ÷¾) = 729 cm3
a3 = 729
a = 9 cm
Diagonal of cube = a 3
ͬôM†ì‹(Diagonal) = 3x9
= 9 3 cm

25.
ð‚è Ü÷¾ 24 cm à¬ìò å¼ èù„ê¶óˆF¡ ¹øŠðóŠ¹ ñŸÁ‹ ªñ£ˆîŠ ðóŠ¹ â¡ð¶
º¬ø«ò
The Lateral Surface area and Total Surface area of the cube having side 24 cm are respectively
(A) 3204 cm2 ,4365 cm2 (B) 4302 cm2, 6534 cm2
(C) 2304 cm2, 3456 cm2 (D) 2340 cm2, 5634 cm2
èù„ê¶óˆF¡ ¹øŠðóŠ¹(LSA of a cube) = 4a2
= 4 x 24 x 24

ê˜è£˜ ä.ã.âv Üè£ìI

= 2304 cm2
èù„ê¶óˆF¡ ªñ£ˆîðóŠ¹ = 6a2

= 6 x 24 x 24
= 3456 cm2

27.
å¼ èù„ ê¶óˆF¡ ªñ£ˆî õ¬÷ðóŠ¹ 384 m2 âQ™ Üî¡ ð‚è‹ âšõ÷¾
The total surface area of cube is 384 m2. Find the side of the cube
(A) 3 m (B) 8 m (C) 4 m (D) 6 m
TSA of a cube(ªñ£ˆî õ¬÷ðóŠ¹) = 384 m 2
6a2 = 384
64
384
a = 2
6
a = 64

a=8m

28. 12 cm ð‚è Ü÷¾ ªè£‡ì èù„ê¶óˆF¡ ð‚èŠ ðóŠ¹

The lateral surface area of a cube of side 12 cm is
(A) 144 cm2 (B) 196 cm2 (C) 576 cm2 (D) 664 cm2
èù„ê¶óˆF¡ ð‚èŠðóŠ¹ (LSA of a cube)
= 4a 2
= 4 x 12 x 12
= 576 cm 2

29. Þó‡´ èù ê¶óƒèO¡ èù Ü÷M¡ MAîƒèœ º¬ø«ò 8-:1 âQ™ Üî¡ ð‚è

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 TNPSC èí‚°

Ü÷¾èO¡ MAîˆ¬î‚ ÃÁ.

The ratio of volumes of two cubes is 8:1 then the ratio of its sides is.
(A) 8:1 (B) 2 2 :1 (C) 2:1 (D) 64:1
Volume of cube V = a 3
V1 =a13 V2 =a 23
3
8  a1 
= 
1  a2 
3 3
2  a1 
  = 
1  a2 
2 : 1 = a1 : a 2
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

30. Asolid having six equal square faces is called a ______

(A) cube (B) cuboid (C) square (D) rectangle
ÝÁ êñê¶óƒè¬÷ ºèƒè÷£è‚ ªè£‡ì à¼õ‹ _____
(A) èùê¶ó‹ (B) èù„ªêšõè‹ (C) ê¶ó‹ (D) ªêšõè‹

31.
å¼ Ü¬øJ¡ c÷‹ 5 m, Üèô‹ 3 m, ÜõŸP¡  ²õ˜èO¡ ðóŠð÷¾ 88 m2 âQ™,
Üšõ¬øJ¡ àòó‹ 裇è
A room is 5 meters long and 3 metres wide. The area of its four walls is 88 sq. metres. The
height of the room is (ex)
(A) 6.5 (B) 7.5 (C) 5.5 (D) 8.5
Room is the shape of cuboid
LSA of Cuboid = 2 (l + b) h
4488= 2 (5 + 3) h
5.5
44
h =
8

h = 5.5 cm

32.
å¼ èù ªêšõèˆF¡ Üèô‹, àòó‹, èùÜ÷¾ º¬ø«ò 10 ªê.e, 11 ªê.e ñŸÁ‹
3080 ªê.e3 âQ™ Üî¡ c÷ˆ¬î‚ è‡ìPè.
The breadth, height and volume of a cuboid are 10 cm, 11cm and 3080 cm3 respectively.Find
the length of the cuboid.
A) 21 ªê.e/cm B) 28 ªê.e /cm C) 24 ªê.e/cm D) 30 ªê.e/cm
Volume of cuboid = l x b x h
3083080 28= l x 10 x 11
308
l =
11
l = 28 cm

33. èù ªêšõèŠ ªð†® å¡P¡ Ü®Šð‚è‹ ê¶óñ£è¾‹ Ü„ê¶óˆF¡ ð‚è‹ a ªê.e.

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 TNPSC èí‚°

Ý辋 ªð†®J¡ àòó‹ b ªê.e. Ý辋 Þ¼ŠH¡ Üî¡ èù Ü÷¾

A cuboid box has the square base of side a cm and height b cm then volume of the box is
a 2b 2 3
A) a b cm / ªê.e
2 2 3 3
B) ab cm / ªê.e C) a b cm / ªê.e D)
2 3 3 2 3 3
cm / ªê.e3
3
ªêšõ芪ð†®J¡ èùÜ÷¾ = (Ü®Šð‚è‹)2 x àòó‹
= a 2b cm 3

34. 3 e x 2 e x 1e Ü÷¾èœ ªè£‡ì èùªêšõè c˜ˆ ªî£†®J¡ èù Ü÷¾ L†ìK™?

Volume of cuboid tank measuring 3mx 2mx1m in litres is
(A) 6 lit. / L†. (B) 5lit. (C) 6000 lit. (D) 5000 lit.
(Volume of a cuboid)
èùªêšõèˆF¡ èùÜ÷¾ = l x b x h m3

=3x2x1
= 6 m3
1 m = 1000 litres
6 m = 6000 litres

ê˜è£˜ ä.ã.âv Üè£ìI

35. å¼ èù ªêšõèˆF¡ c÷‹, Üèô‹ ñŸÁ‹ Ýö‹ ÝAòõŸP¡ Ã´î™ 19 ªê.e.
«ñ½‹ Üî¡ Í¬ô M†ìˆF¡ c÷‹ 5 5 ªê.e. âQ™ Üî¡ ªñ£ˆîŠ ðóŠ¹ â¡ð¶
The sum of the length, breadth and depth of a cuboid is 19 cm and its diagonal is 5 5 cm. Its
total surface area is
(A) 361 cm2 (B) 125 cm2 (C) 236 cm2 (D) 486 cm2
l + b + h = 19
l 2 + b2 + h2 =
5 5

( )
2
l 2 + b2 + h2 =5 5
ªñ£ˆî ðóŠ¹ (TSA) = 2 (lb + bh + lh)
= (l + b+ h)2 - (l2 +b2+ h2)
= 192 - 125
= 361 - 125
= 236 cm2

36. å¼ èù ªêšõèˆF¡ Ü®ŠðóŠ¹ ñŸÁ‹ èù Ü÷¾èœ º¬ø«ò 180 ê.ªê.e ñŸÁ‹

900 è.ªê.e âQ™ Üî¡ àòó‹ â¡ù?
Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm2
(A) 5cm (B) 6 cm (C) 7 cm (D) 0.2 cm
Area of cuboid = 180 cm 2

Volume of cuboid = 900

length x breadth 180
=
length x breadth x height 900
1 18 0
=
height 90 0

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90 = 18 x height
90
18
h = 5 cm

37.11 e. 10 e, 7 e Ü÷¾œ÷ èù„ ªêšõèˆF¡ èù Ü÷¬õ‚ 裇è

Find the volume of a cuboid whose dimensions are given by 11 m, 10 m and 7 m
(A) 800 m3 / e3 (B) 770 m3 / e3 (C) 740 m3 / e3 (D) 710 m3 / e3
l = 11 m, b = 10 m, h = 7 m
= l x b x h
= 11 x 10 x 7 = 770 m3
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

38. 2 e, 2 e ñŸÁ‹ 1 e ÝAòõŸ¬ø º¬ø«ò c÷, Üèô, àòóƒè÷£è à¬ìò èù

ªêšõèˆF¡ ªõOŠðóŠ¹ (in m2)
If the length, breadth and height of a cuboid are 2 m, 2 m and 1 m respectively, then its surface
area is (in m2)
(A) 8 (B) 12 (C) 16 (D) 24
l=2m b=2m h=1m
Total surface area = (lb + bh + hl)
= 2 (2 x 2 + 2 x1 + 1 x 2) = 2 (4 + 2+ 2)
= 2 (8) = 16
= 16 m2

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༬÷ ñŸÁ‹ «è£÷‹
Cylinder and Sphere

1. The line joining the centres of the circle is called axis, When we revolve a Rectangle about
one side as the axis of revolution a Right Circular Cylinder

2. The section obtained on cutting a right circular cylinder by a plane, which contains two
elements and parallel to the axis of the cylinder in the rectangle.

3. If a plane cuts the right cylinder horizontally parallel to the bases, then it’s a circle

4. Curved surface area[CSA / TSA] =2πrh Square Units

(༬÷J¡ õ¬÷ðóŠ¹) ê¶ó Üô°èœ

5. Total surface area of Right Circular Cylinder (TSA) = 2πr ( r+h ) cubic Units
(༬÷J¡ èù Ü÷¾) èù Üô°èœ

ê˜è£˜ ä.ã.âv Üè£ìI

2
6. Volume of Right Circular Cylinder V = πr h Square Units
(༬÷J¡ ªñ£ˆî ðóŠð÷¾) ê¶ó Üô°èœ

7. If the segments joining the centres of the circular bases are perpendicular to the two palnes of
the circles, the cylinder is called a Right Circular Cylinder

8. If the segments joining the centres of the circules are not perpendicular to the two palnes of
the circles, the cylinder is called a Oblique Circular Cylinder

Cylinder and Sphere sums

1. Base area of cylinder is 45 cm2 and its height is 8 cm. Find volume ?
å¼ à¼¬÷J¡ Ü®Šð‚è ðóŠð÷¾ 45 cm2. Üî¡ àòó‹ 8 cm âQ™ èù Ü÷¾ ò£¶ ?
༬÷J¡ Ü®Šð‚è‹ π r 2 = 45cm 2
h=8
Volume (èùÜ÷¾) = π r h
2

= 45 (8)
= 360 cm3

2. The ratio of radii of two cylinders are 3 : 5, then find their ratio of volumes ?
Þ¼ ༬÷J¡ ÝóƒèO¡ MAî‹ 3:5 âQ™ ÜõŸP¡ èù Ü÷¾èO¡ MAî‹ â¡ù
? (r=h)
r1 : r2 = 3 : 5
༬÷J¡ èùÜ÷¾

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π r12h : π r22h
3x3x3:5x5x5
27 :125

3. The ratio of the volume of a cone, a sphere and cylinder if each has the same radius and same
height is
êññ£ù Ýó‹ ñŸÁ‹ àòó‹ à¬ìò ˹, «è£÷‹, ༬÷ ÝAòõŸP¡ èù Ü÷¾èO¡
MAî‹
Sol:
Cone : Sphere : Cylinder
h=r
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

1 2 4 3
πr h : πr : π r 2h
3 3
1 3 4 3
πr : πr : π r3
3 3
1 4 1x3
: :
3 3 3
Ans: 1 : 4 : 3

4. Curved surface area of the solid sphere is 36 cm2. If the sphere is divided into two hemisphere
then total surface area of one of its hemisphere is
å¼ Fì‚«è£÷ˆF¡ õ¬÷ðóŠ¹ 36 cm2 Üî¬ù Þ¼ ܬó‚«è£÷ƒè÷£è HKˆî£ô, å¼
ܬó‚«è£÷ˆF¡ ªñ£ˆî ðóŠ¹ â¡ù ?
«è£÷ˆF¡ õ¬÷ðóŠ¹ = 4π r
2

4π r 2 = 36 cm2
36
π r2 =
4
π r2 = 9
ܬó‚«è£÷ˆF¡ ªñ£ˆîŠðóŠ¹
= 3π r 2
= 3(9)
= 27cm 2

5. A hemi spherical bowl of radius 30 cm is filled with soap paste. If this paste is made into cylinderical
soap cakes each of diameter 10 cm and height 2 cm how many cakes do we get ?
30 cm Ýóºœ÷ ܬó‚«è£÷ õ®õ °ŠH «ê£Š¹‚ Ãö£™ GóŠðŠð†´œ÷¶. Þ‰î «ê£Š¹
ìö‚ªè£‡´ 10 cm M†ìº‹ 2 cm àòóº‹ àœ÷ ༬÷ õ®õ «ê£Š¹ âˆî¬ù -?
h = 2 cm, r= 30 cm. r = 5 cm
ܬó‚«è£÷ˆF¡ èùÜ÷¾ = ༬÷J¡ èùÜ÷¾
(Volume of Hemisphere) = (Volume of cylinder)
2 3
π r = nπ r 2h
3

90
 TNPSC èí‚°

2
x 30 x 30 x 30 = n x 5 x 5 x 2
3
6 6 10
2 30 x 30 x 30
n= x
3 5x5x2
n = 360

6. A rectangular sheet of metal fail with dimension 66 cm x 12 cm is rolled to form a cylinder of

height 12 cm. Find the volume of the cylinder.
66 cm x 12 cm â‹ Ü÷¾‚ªè£‡ì å¼ à«ô£èˆî膮¬ù 12 cm àòóºœ÷ å¼
༬÷ò£è ñ£ŸPù£™ A¬ì‚°‹ ༬÷J¡ èùÜ÷¾ 裇è.
l = 66 cm
b = 12 cm
height cylinder h = 12 cm
circumference of the cylinder = 66 cm
2π r = 66
7 1 21
r = 66 x x = cm
22 2 2

ê˜è£˜ ä.ã.âv Üè£ìI

Volume of the cylinder
22 21 21
= π r 2h = x x x 12
7 2 2
=33 x 21 x 6 = 4158cm3

7. A cone of height 24 cm is made up of modeling clay. A child reshapes it in the form of a cylinder
of same radius as cone. Find the height of a cylinder.
èOñ‡ ªè£‡´ ªêŒòŠð†ì 24 ªê.e. àòó‹ à¬ìò å¼ Ã‹¬ð å¼ °ö‰¬î Ü«î
Ýóºœ÷ æ˜ à¼¬÷ò£è ñ£ŸÁAø¶ âQ™ ༬÷J¡ àòó‹ 裇.
Height of cone (ËH¡ àòó‹) = 24 cm
radius of cylinder (༬÷J¡ Ýó‹) = 24 cm
ËH¡ èùÜ÷¾ = ༬÷J¡ èùÜ÷¾
1 2
π r h = π r 2h
3
1 2
r h = r 2h
3
1
x 24 x 24 x 24 = 24 x 24 x h
3
h = 8cm

9
8. If the volume of sphere is π then its radius is
9 2
π è.ªê.e. èùÜ÷¾ ªè£‡ì «è£÷ˆF¡ Ýó‹ â¡ð¶.
2 9
Volume of sphere («è£÷ˆF¡ èùÜ÷¾) = π
2

91
 TNPSC èí‚°

4 3 9
πr = π
3 2

9 x 3 27
r3 = =
2x4 8
3
r=
2

9. If radius of one sphere is half of radius of another sphere then their volumes are in the ratio.
å¼ «è£÷ˆF¡ Ýóñ£ù¶ ñŸªø£¼ «è£÷ˆF¡ ÝóˆF™ ð£F âQ™ ÜõŸP¡
èùÜ÷¾èO¡ MAîñ£ù¶.
r
=r1 = , r2 r
2
Volume of sphere («è£÷ˆF¡ èùÜ÷¾)
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

4 4
π ( r1 )3 : π r23
3 3 3
r 3
  :r
2
r3 3
:r
8
r1 : r2 = 1: 8

10. A hemispherical bowl of radius 60 cm is filled with soap paste. If this paste is made into cylindrical
soap cakes, each of radius 6cm & height 2 cm. How many cakes do we get?
60ªê.e. Ýó‹ ªè£‡ì ܬó‚«è£÷ õ®õ ð£ˆFóˆF™ G¬øò «ê£Š¹‚Û àœ÷¶.
Üî¬ù 6 ªê.e. Ýóº‹, 2 ªê.e. àòóº‹ ªè£‡ì âˆî¬ù ༬÷ õ®õ «ê£Š¹è÷£è
ñ£Ÿøô£‹.
ܬó‚«è£÷ˆF¡ èùÜ÷¾ (Volume of Hemisphere) r = 60 cm
༬÷J¡ èùÜ÷¾ (Volume of cylinder) h = 2
2 3
π r = nπ r 2h
3
2 20
x 60 x 60 x 60 = n x 6 x 6 x 2
3
14400 = n x 72
14400
n=
72
n = 2000

11. The total surface area of a solid right circular cylinder is 231 cm2. Its curved surface area is two
thirds of the total surface area. Find the radius and height of the cylinder.
å¼ F‡ñ «ï˜õ†ì ༬÷J¡ ªñ£ˆî ¹øŠðóŠ¹ 231 ê.ªê.e. Üî¡ õ¬÷ðóŠ¹
ªñ£ˆî ¹øŠðóŠH™ Í¡P™ Þó‡´ ðƒ° âQ™ Üî¡ Ýó‹ ñŸÁ‹ àóò‹ 裇è.
We know that the total surface area of a cylinder = 2π rh + 2π r 2
Curved surface area (õ¬÷ðóŠ¹) = 2π rh + 2π r 2
2
According to the eqn. (2π rh + 2π r 2 ) = 2π rh
3
2 
2π r  x( h + r )  =
2π rh
3 

92
 TNPSC èí‚°

2
x( h + r ) =
h
3
2h + 2 r =
3h
h = 2r
2π rh + 2π r 2 = 231
2π r ( h + r ) =
231
2π r x 3r=231
231
2π r 2 =
3
2 7 7 1
r = 77 x x
22 2 2
7
r=
2
r = 3.5cm
h = 7 cm
h = 2r
h = 2 (3.5) = 7 cm

ê˜è£˜ ä.ã.âv Üè£ìI

12 The ratio between the base radius and the height of a solid right circular cylinder is 2 : 5. If its
3960
curved surface area is sq.cm. Find the height and radius.
7
å¼ F‡ñ «ï˜õ†ì ༬÷J¡ Ýóº‹ àòóº‹ 2:5 â¡ø MAîˆF™ àœ÷ù. Üî¡
3960
õ¬÷ðóŠ¹ ê.ªê.e. âQ™ ༬÷J¡ àòó‹ ñŸÁ‹ Ýó‹ 裇è.
7
r:h=2:5
r 2
=
h 5
2
r= h
5
Now, the curved surface area (CSA) = 2π rh (õ¬÷ðóŠ¹)
22 2 3960
2x x xhxh=
7 5 7
3960 x 7 x 5
h2 =
2 x 22 x 2 x 7
h 2 = 225
h = 15cm
2
r= xh
5
2
r = x 15 = 6
5
r = 6cm

Exercise Sums:

93
 TNPSC èí‚°

1. B â¡ø «è£÷ˆF¡ Ýó‹ A&J¡ ÝóˆF™ 1 ðƒ° âQ™ A&J¡ èù Ü÷¾‚°‹

B&J¡ èù Ü÷¾‚°‹ àœ÷ MAî‹ 4
The radius of sphere B is one fourth of radius of sphere A. Then the ratio of volume between
A and B is
Sol:
(A) 1 : 16 (B) 16 : 1 (C) 1 : 64 (D) 64 : 1
Radius of a sphere A («è£÷ˆF¡ Ýó‹) = r1
Radius of a sphere B («è£÷ˆF¡ Ýó‹) = r2
r
r2 = 1 B
4
«è£÷ˆF¡ èùÜ÷¾ MAî‹ (Ratio of volume between A and B)
3 3
4π
4π rr113 : 4π 4π rr223
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

4π 3 :
3 3
4π3 3rr11 :: 4π 3rr22
4π3 3

3  r 3 3
3
3
3
rr133 ::  r11 33
 r4 
rr1133 ::  r411 
1

433
 r4
rr133 :: r113
1 rr113
64
rr1133 :: 64
64
64 :1 :1 64 64
64
64 :1 :1

2. å¼ «è£÷ˆF¡ õ¬÷ð󊹋 èùÜ÷¾‹ êñ‹ âQ™ Üî¡ Ýó‹?

If the volume and surface area of a sphere are numerically the same, then its radius is
A) 3 units / Üô°èœ B) 2 units / Üô°èœ
C) 1 units / Üô°èœ D) 4 units / Üô°èœ
Sol:
Volume of Sphere («è£÷ˆF¡ èùÜ÷¾) = 4 πr 3
3
Surface area of sphere («è£÷ˆF¡ õ¬÷ðóŠ¹) = 4πr 2

4 33 2
πr = 4πr 2
3
r
=1
3
rr = 3 units

3. å¼ àœkìŸø «è£÷ˆF¡ ªõO ñŸÁ‹ àœ Ýóƒèœ º¬ø«ò 12 ªê.e ñŸÁ‹ 10

ªê.e. âQ™ Ü‚«è£÷ˆF¡ èù Ü÷¬õ‚ 裇è.
The outer and the inner radius of a hollow sphere are 12 cm and 10 cm. find its volume.

2
(A) 3050 ªê.e3 (B) 3049 ªê.e3
3
2 2
(C) 3060 ªê.e3 (D) 3059 ªê.e3
3 3
Sol:

94
 TNPSC èí‚°

R=12cm r =10cm
àœkìŸø «è£÷ˆF¡ èùÜ÷¾
4
4 π ( R 33 - r 33 )
=
= 4 π ( R 3 - r3 )
= 3 π ( R 3 - r3 )
3
3
4
4 x 2222 (1233 -1033 )
=
= 4 x 22 (1233 -1033 )
= 3
3 x 7 7 (12 -10 )
3
4 7
22
= 4 x 22 (1728-1000 )
= 4 x 22 (1728-1000 )
= 3
3 x 7 7 (1728-1000 )
3
4 7
22 104
= 4 x 22 x 728 104
= 4 x 22 x 728
104
= 3
3 x 7
104
71 x 728
3 71
9152
9152 1= 3050 2
1
2 cm33
=
= 9152 = 3050 2 cm33
= 2 2 = 3050 3 3 cm
2
9152 23 3
= = 3050 cm
3 3

4. 1 ªê.e. Ýóº‹, 5 ªê.e. àòóº‹ ªè£‡ì «ï˜ ༬÷J™ Þ¼‰¶

ªõ†® â´‚èŠð†ì I芪ðKò «è£÷ˆF¡ èùÜ÷¾ 裇è.

ê˜è£˜ ä.ã.âv Üè£ìI

Find the volume of a largest sphere inscribed from the right circular cylindrical wood whose
radius is 1 c.m. and height is 5 c.m.
4 5
A) π cm3 (B) π cm3
3 16
(C) 5π cm3 (D) 8π cm3
Sol:
The radius of the Sphere = Radius of cylindrical log of wood

Volume of the Sphere

4
= π r3
3
4
= π x 13
3
4
= π cm 3
3

5. æ˜ «è£÷ñ£ù¶ «ï˜õ†ì àœkìŸø ༬÷J™ ðìˆF™ è‡ìõ£Á ¬õ‚èŠð´Aø¶.

Ü‚«è£÷ñ£ù¶ ༬÷J¡ «ñ™, Ü®, ²ŸÁŠð°Fè¬÷ˆ ªî£´‹ð®
ܬñ‚èŠð†´œ÷¶. «è£÷ˆF¡ Ýó‹ ‘r’ â¡ø£™ ༬÷J¡ èù Ü÷¾ â¡ù?

95
 TNPSC èí‚°

A sphere is placed inside a right circular hollow cylinder so as to touch the top, base and the
lateral surface of the cylinder as shown in figure. If the radius of the sphere is r, the volume of
the cylinder is
8π
(A) 4π r3 (B) r3 (C) 2π r3 (D) 8π r3
3
Sol:
The volume of the cylinder = π r 2 h
The given, cylinder radius height = 2r
Volume of the given cylinder = π r 2 (2r ) = 2π r 3
Ans : 2π r 3

6. 16 cm M†ìº‹ 2cm àòóº‹ ªè£‡ì ༬÷¬ò ༂A 12 êñ «è£÷ƒèœ

TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

ªêŒòŠð´A¡øù. 嚪õ£¼ «è£÷ˆF¡ M†ì‹ â¡ù?

12 spheres of the same size are made from melting a solid cylinder of 16 cm diameter and 2 cm
height then the diameter of each sphere is
(A) 3 cm (B) 2cm (C) 3cm (D) 4cm
Sol:
Volume of conservation
4 4
12 44
x 4 π r 333 = π r1222 h
12 x
12 x 3 π π rr =
= π
π rr11 hh
3
16r 333 = 8 x 8 x 2
16r = 88 x 8 x 2
64
88 x 2
r 333 = 64 x2
r = 16x 2 64
16
162
22
3
r 33 = 8
rr = = 88
r=2
r=2
M†ì‹
d = 2r d = 2r
= 2 x 2 = 4cm
= 2 x 2 = 4cm

7. å¼ «è£÷ˆF¡ M†ì‹ 6 ªê.e Üî¬ù ༂A, 2 I.e M†ì‹ ªè£‡ì è‹Hò£è

ñ£ŸPù£™, Ü‰î‚ è‹HJ¡ c÷‹ ò£¶?
The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2mm. The
length of the wire is
(A) 12 m (B) 18 m (C) 36 m (D) 66 m
Sol:
Wire: d = 0.2 cm r = 0.1 cm Sphere d = 6 cm, r = 3 cm
Volume of Sphere = Volume of wire
4
4 π r 33 = π r 22 h
3 πr = πr h
3
4
4 x 3 x 3 x 3 = 0.1 x 0.1 x h
3 x 3 x 3 x 3 = 0.1 x 0.1 x h
3
4 x 9 = 0.01 x h
4 xx 9
436 9= = 0.01
0.01 xh
= 0.01 x hx h
36 = 0.01
36 0.01 xxhh
36
h = 36 = 3600cm (or)
h = 0.01 = 3600cm (or) 96
0.01
 3
4
x 3 x 3 x 3 = 0.1 x 0.1 x h
3
4 x 9 = 0.01 x h TNPSC èí‚°
36 = 0.01 x h
36
h= = 3600cm
2600 cm(or)
(or) = 36 m
0.01
= 36m
8. å¼ «è£÷ˆF¡ ÝóˆF¡ â‹ñFŠ¹‚°, «è£÷ˆF¡ ªè£œ÷÷¾ â‡í÷M™ Üî¡
¹øŠðóŠHŸ° êñ‹
For what value of radius of a sphere, the volume of the sphere is numerically equal to the sur-
face area of the sphere
A) 1 B) 2 C) 3 D) 4
Sol:
Volume of sphere = Volume of surface area

4
π r 3 = 4 π r2
3
r
=1
3
r=3
Ans : 3

ê˜è£˜ ä.ã.âv Üè£ìI

9. 12 ªê,e, Ýóºœ÷ å¼ «è£÷ °‡ì£ù¶ ༂A Í¡Á CÁ «è£÷ƒè÷£è
ñ£ŸøŠð´Aø¶. ºî™ Þó‡´ «è£÷ƒèO¡ Ýóƒèœ 6 ªê,e. ñŸÁ‹ 8 ªê,e.
Í¡ø£õ¶ «è£÷ˆF¡ Ýó‹
The metallic sphere of radius 12 cm is melted into three smaller spheres. If the ratio of two
smaller spheres are 6 cm and 8 cm, the radius of the third sphere is
(A) 14 cm (B) 16 cm (C) 10 cm (D) 12 cm
Sol:
12cm 6cm
8cm

Bigger Sphere r = 12

S1 → r1 = 6cm
S2 → r2 = 8cm
3 4
Volume of Sphere = πr
3

4
4 4
4 4 3 4
3 + 4 πr 3 4
3 3 3
4
4 πr
πr
3
3 =
= 4
4 πr
πr 13 + 4
4 πr 23 +
+ 4
4 πr
πr
3
33
3
3 πr 3 = 3 πr1
3
πr = 3 13 + 3 πr2
πr1 + 3
3 2 3 + 3 πr3
3
πr2 + 3 33
πr33
3
3
3 3
3 1
3 3
3 2
3 3
3 3
3
4 4
4
4 π
π rr 3333 =
=
4
4
4 π
π (
( rr13333 +r
+r
3
3 +r
3
+r 3)
3
3)
3
4
3
3
3
π
π
π r
r
r 3 =
=
=
4
3
3
3
π
π
π ( r
r
r1
1
1
1
3 +r
+r
+r
23
2
2
2
2
3 +r
+r
+r 3 )
33
3
3
3
3 3 3
1728
1728 =
= 216+512+r
216+512+r 33
3
1728
1728 =
=
1728 = 216+512+r 216+512+r
216+512+r 3
3
3
3
3
3
1728
1728 = 728+r 3 3

1728 = = 728+r
33
1728
1728 =
= 728+r
728+r333333
728+r
1728-728
1728-728 = rr3333
1728-7283 =
1728-728
1728-728 =
= rrr33333
=
1000
1000 = r 3
1000 = = rrrr333333
3
1000
1000 =
=
3

rr3 = 10
rr333 =
3
= 10
= 10
10
97
 1728 = 216+512+r33
1728 = 728+r33
1728-728 = r33
TNPSC èí‚°
1000 = r33
r3 = 10
Í¡ø£õ¶ «è£÷ˆF¡ Ýó‹ = 10 cm

10. 6 ªê.e Ýó‹ ªè£‡ì «è£÷ õ®õ ðÖQ™ 裟Á ªê½ˆîŠð´‹ «ð£¶, Üî¡ Ýó‹
12 ªê.e Ýè ÜFèK‚Aø¶. Ýó‹ð G¬ôJ½‹, ÞÁF G¬ôJ½‹ ðÖQ¡ èù
Ü÷¾èO¡ MAî‹ ò£¶ ?
A) 1 : 8 B) 2 : 7 C) 8: 1 D) 2: 3
As air is pumped into a spherical balloon the radius increases from 6 cm to 12 cm. The ratio
between volume of the balloon in the beginning and the end is
Sol:
Volume = 4 πr 3
3
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

4
4 πr 333 : 4 4 πr 333
4 πr 1 : 4 πr
4
4
3 πr 13 : 3 πr
: 2
3 πr
3 1 3 2
1
3 πr
2
23
34 34
4
4π rr1333 :: 4 4π rr2333
4 π
π r
3 π r1 : 3
1 : 4 π
π
3 r223
3
3 1 3 π r2
3

6 333 : 1233 3
6
633 :: 12 1233
6216: 12
216 ::: 1728
216 1728
1728
1216 : 8 : 1728
1
1 :: 8 8
1:8

11. å¼ F‡ñ ܬó «è£÷ˆF¡ M†ì‹ 2 ªê.e. âQ™ Üî¡ ªñ£ˆî ¹øŠðóŠ¹
The total surface area of a solid hemisphere of a diameter 2 cm is equal to
A) 12cm2 B) 12π cm2 C) 4π cm2 D) 3π cm2
Sol:
Diameter r = 2 cm
Ýó‹ = 1 cm
Total surface area
2
=
= 3πr
= 3πr 22
3πr
2
=
= 333 xxx π
= π xxx 11122
π
=3π
=3π sq.cm
=3π sq.cm
sq.cm

12. æ˜ Ã‹¹, æ˜ Ü¬ó «è£÷‹ ñŸÁ‹ æ˜ à¼¬÷ Í¡Á‹ å«ó Ü÷¬õ‚ ªè£‡ì
Ü®ð°F¬ò»‹, êññ£ù àòóˆ¬î»‹ à¬ìòùõ£Œ àœ÷ù. Þî¡ èù Ü÷¾èO¡
MAî‹ è£‡è.
A cone, a hemisphere and a cylinder stand on equal bases and have the same height. Find the
ratio of their volumes
A) 3 : 2 : 1 B) 1 : 2 : 3 C) 3 : 1: 2 D) 1 : 3: 2
Sol:
1
Volume of Cone VC = πr 2 h
3
2 2
Volume of Hemisphere VH = πr h
3
Volume of Cylinder VVcyC =πr 2 h
98
 TNPSC èí‚°

1 2
2
πr h : πr 2 h : πr 2 h
3 3
1 2
: :1
3 3
1:2:3

14. å¼ àœkìŸø «è£÷ˆFÂœ à†¹øñ£è å¼ ê˜‚èv ió˜ «ñ£†ì£˜ ¬ê‚AO™ ê£èê‹

ªêŒAø£˜. ܉î ê£èê ió˜ ê£èê‹ ªêŒ»‹ ðóŠð÷¾ 154 e2 âQ™ Ü‚«è£÷ˆF¡
à†¹ø Ýóˆ¬î 裇è.
A hollow sphere in which a circus motor cyclist performs his stunts, has an area of 154 m2
available to him for riding. Find the inner radius
A) 7 m B) 3.5m C) 4 m D) 6m
Sol:
2 2
4πr
4πr 2 =
Ares =of154m
2
154m
Sphere
2 2
= 4πr 2
2 22
2 22 154m 2 2
4πr
4
4x
4πr x 7= = 154mx
x rr 22 =
= 154
2
154
7
22
4 227777 x r 22 = 154
42 x x 14 x r =7 7154 1
1

ê˜è£˜ ä.ã.âv Üè£ìI

rr 22 = 714
14
= 7154
154 x
x 22
x
x 4
14
7
7 22
7 4
1
rr 2 = 2 14
154 x 7
2
2
2 x 1
= 154 x 1 x
1
1
1

2 49
49 22
22 4
4
rr 22 = = 4
2
21
1
4
49
rr 22 = 49
= 4949
rr = = 4 4
4
4
49
rr = 7 49
rr = =
= 24
7
4
2
7
rr = 7
= 3.5m
3.5m
2
2
rr = 3.5m
= 3.5m
15. å¼ Ü¬ó«è£÷ˆF¡ õ¬÷ðóŠ¹ 2772 ªê.e2 âQ™ ܬó«è£÷ˆF¡ ªñ£ˆî ¹øŠðóŠ¹
�
Surface Area of a hemisphere is 2772 cm2 then the total surface area of hemisphere is
(A) 4158 cm2 (B) 3172 cm2 (C) 3882 cm2 (D) 4258 cm2
Sol:
ܬó‚«è£÷ˆF¡ õ¬÷ðóŠ¹ (CSA of hemisphere) = 2πr 2
2
2πr
2πr 2222 == 2772
2772
2πr
2πr =
= 2772
2772 1 7
r 2222 = 2772 x 11 x 77
rr 2 = 2772 xx 12 xx 7
22
r = = 2772
2772 x 22 x 22
22
63
63
693
63 2 22
693
1386
63
693
1 7
x 11 x 77
1386
63
r 2222 = 2772
693
1386
693
rr 2 = 2772
1386
x x
2772 xx 1221 xx 227
1386
r = = 2772 22
2211 22
11
22
11
11
1
1 11
11
2 1 11
rr 222 =
= 63
63 x
x 7
7
1
1

rr 2 = 63 x
= 63 x 77
r= 441
r=
r= 441
441
r=
r=21441
r=21
r=21
r=21
99
 1386 1 7
r 2 = 2772 x x
2 22
1 11
1

2
r = èí‚°
TNPSC
63 x 7
r= 441
r=21
ܬó‚«è£÷ˆF¡ ªñ£ˆî ¹øŠðóŠ¹ (TSA of hemisphere) = 3πr 2
22 3
3πr 2 = 3 x x 21 x 21
7
= 4158 cm2
=4158 cm 2

16. 6 ªê.e Ýóºœ÷ «è£÷õ®õ à«ô£è‚ °‡´ ༂èŠð†´ 6 I.e M†ìºœ÷

CPò «è£÷õ®õ °‡´è÷£è õ£˜‚èŠð†ì£™ âˆî¬ù CPò «è£÷õ®õ °‡´èœ
A¬ì‚°‹?
Spherical metal ball or radius 6 cm is melted and casted into small spherical balls having diameter
6 mm.How many small balls can be casted
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

(A) 8000 (B) 1000 (C) 6000 (D) 2000

Sol:
4
π R 3 → Big
3
=
4 3
π r → Small
3
60 x 60 x 60
=
3x3x3
== 800
8000smaller
Smallerballs
balls

17. å«ó Ýóº¬ìò F‡ñ «è£÷ˆF¡ ªñ£ˆîŠðóŠ¹ (õ¬÷ðóŠ¹), F‡ñ ܬó‚«è£÷ˆF¡

ªñ£ˆî ¹øŠðóŠ¹ ñŸÁ‹ F‡ñ ܬó‚«è£÷ˆF¡ õ¬÷ðóŠ¹èÀ‚° àœ÷ MAî‹
The radius of solid sphere and solid hemisphere is same. Then the ratio of curved surface area
of solid sphere, total surface area of the solid hemisphere and curved surface area of solid hem-
isphere is
(A) 4 : 3 : 2 (B) 4 : 2 : 1 (C) 3 : 2 : 1 (D) 2 : 1 : 1
Sol:
CSA of Sphere : TSA of hemisphere : CSA of hemisphere
4π r 2 : 3π r 2 : 2π r 2
Ans : 4 : 3: 2

18. å¼ «è£Š¬ðò£ù¶ Ü¬ó‚ «è£÷ˆF¡ e¶ ༬÷ Þ¬í‰î õ®M™ àœ÷¶.

༬÷Š ð°FJ¡ àòó‹ 9 ªê.e ñŸÁ‹ «è£Š¬ðJ¡ ªñ£ˆî àòó‹ 12.5 e âQ™,
༬÷Š ð°FJ¡ Ýó‹
A cup is in the form of a hemisphere surmounted by a cylinder. The height of the cylindrical
portion is 9 cm and the total height of the cup is 12.5 cm. Then the radius of the cylindrical
portion is
(A) 3.5 cm (B) 2.5 cm (C) 21.5 cm (D) 9 cm
Radius = Total Height - 9
= 12.5 - 9
r = 3.5 cm

100
 TNPSC èí‚°

19. Þó‡´ «è£÷ƒèO¡ õ¬÷ðóŠ¹èO¡ MAî‹ 9 : 25 âQ™ ÜõŸP¡ èùÜ÷¾èO¡

MAî‹
The surface areas of two spheres are in the ratio of 9: 25. Then their volumes are in the ratio is
A) 81 : 625 B) 27 : 125 C) 729 : 15625 D) 27 : 75
Sol:
9
Þó‡´ «è£÷ƒèO¡ õ¬÷ðóŠ¹èO¡ MAî‹
25
Surface area = 4πr 2
2
4π
4π rr11122 = 99
4π rr2 222 = 25
4π 25
22
2
rr11122 = 99
rr2 222 = 25 25
22
rr1 3
11 = 3
rr2 = 55
22
3
4π r1 4π r1 3
èùÜ÷¾
= =3 3
r2 4π r2
4π

ê˜è£˜ ä.ã.âv Üè£ìI

93 9327 27
= 3 = 3 =
=
5 5125 125
27 : 125
27 : 125
27:125

20. å¼ «è£÷õ®õ F‡ñ à«ô£è âP °‡®¡ èù Ü÷¾ 310.464 âQ™ Üî¡

Ýóñ£ù¶
The volume of a sphere-shaped shot put is 310.464 cu. cm, then the radius is
A) 4.2 B) 4.8 C) 6.4 D) 8.4
Sol:
Volume = 310.464
4 44 πr 333 = 310.464
4 πr 3 = = 310.464
33 πr 310.464
33 πr = 310.464
13 7
3528
3
3528
77.616
xx 333 xxx 111 x
3528 3.528 77.616
rr 3333 == 310.464
77.616
3528
310.464
77.616
rr 3 = 310.464
77.616
r = = 310.464
310.464 xx 444 xx 22 224
22
22
74.088 4 22
rr 3333 = 74.088
74.088
rr = = 74.088 xxx 1000 1000
= 1000 1000 x 1000 1000
1000
1000
74088
74088
rr 3333 = 74088
74088
rr = =
= 10001000
1000
1000 33
 42  33 = 42
rr 33 = 33
=  42
 42
42   =
42
42
42
rr = 

=  10 
10  = 10
 = 10
 10
10   10
10
rr = 4.2 cm
rr = = 4.2
4.2
= 4.2 cm
cm
cm
21. å¼ õ†ìº‹ «è£÷º‹ êñ Ýó‹ àœ÷¬õ. õ†ìˆF¡ ð󊹂°‹ «è£÷ˆF¡
¹øŠð󊹂°‹ àœ÷ MAî‹ â¡ù?
A circle and a sphere are of same radius. What is the ratio of the area of the circle to the surface
area of the sphere?
(A) 1 : 1 (B) 1 : 2 (C) 1 : 4 (D) 1 :3
Sol:
101
 TNPSC èí‚°

Area of circle = surface area of sphere

πr 2 = 4 πr 2
1=4
Surface area of Sphere = 4
Area of circle = 1
Ratio = 1: 4

22. å¼ Ü¬ó‚«è£÷º‹, ˹‹ å«ó Ü®Šð‚èˆ¬î ªè£‡´œ÷ù. ÜõŸP¡ àòóº‹

êñ‹ âQ™ ÜõŸP¡ õ¬÷ðóŠH¡ MAî‹ â¡ù?
A hemisphere and a cone have equal bases. If their If their heights are also equal, than the ratio
of their curved surfaces will be
A) 1:2 B) 2 : 1 C) 1 : 2 D) 2 :1
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

Sol:
ܬó‚«è£÷ˆF¡ õ¬÷ðóŠ¹ : ËH¡ õ¬÷ðóŠ¹
2 2 2
2
2π π rr :: π
2
π rr rr +h
+h2 2

2 2
hhh2 =:: rrr2
2 2

2 2 2
2r
2rπ2 r::2 rr: rrπ2r+r
+rr2 2 +h 2
2
2
2
2 2r
r 2 :
2 : r:2 2 r 2
2 r 22
h
2r : r 2r
2. 2 : 2
2r22 .: r2xr:2 +r222 x r
2r2
2 r2 : 2 r2
2. 2 : 2

Ratio = 2 :1

23. A, B ñŸÁ‹ C ÝAòù ªõš«õÁ à«ô£èƒè÷£ô£ù åˆî Í¡Á «è£÷ƒè÷£°‹. B, C

¬ò‚ 裆®½‹ ÜFèñ£ù ñŸÁ‹ A¬ò‚ 裆®½‹ °¬øõ£ù â¬ì¬ò‚ ªè£‡ì¶
âQ™, Ü예F¬òŠ ªð£¼ˆî÷M™,
(A) B, Iè ÜFè ñFŠ¹¬ìò¶ (B) A, Iè ÜFè ñFŠ¹¬ìò¶
(C) B, Iè CPò ñFŠ¹¬ìò¶ (D) A, Iè CPò ñFŠ¹¬ìò¶
A, B and C are three identical spheres made of different metals. If B weighs more than C but
less than A, then the value of density of
(A) B is the greatest (B) A is the greatest
(C) B is the least (D) A is the least
Ans : A Iè ÜFè ñFŠ¹¬ìò¶ / A is the greatest (A>B>C)

24. å¼ «è£÷ˆF¡ M†ì‹ Þ¼ ñìƒè£Aø¶ âQ™ Üî¡ ¹øŠðóŠ¹ âˆî¬ù êîiî‹

ÜFèK‚°‹?
If the diameter of sphere is doubled, the increase in its surface area will be
(A) 100% (B) 200% (C) 300% (D) 400%
Sol:
Let radius be r

102
 TNPSC èí‚°

surface area = 4π r 2
new radius = 2r
New surface area = 4π (2r ) 2
= 16π r 2
16π r 2 − 4π r 2
Increase % = x100
4π r 2
4π r 2 (4 − 1)
x100
4π r 2
= 300%

Ans : 300%

25. ܬó‚«è£÷ˆF¡ èù Ü÷¾ 19404 Cu cm âQ™ Üî¡ Ýó‹ â¡ù?

Volume of a hemisphere is 19404 Cu cm. It’s radius is
(A) 10.5 cm (B) 17.5 cm (C) 21 cm (D) 42 cm
ܬó‚«è£÷ˆF¡ èùÜ÷¾ = 19404

ê˜è£˜ ä.ã.âv Üè£ìI

2
2
2
2 πr
πr
3
3 = 19404
3 = 19404
3
2
3 πr
πr 3 = 19404
= 19404
3
3 πr 3 =4851 19404
441
3
441
441
3 7
4851
9702
441
3 7
4851
9702
rr 3333 = 19404
4851
9702
x 3
3 x 7
7
rr = 19404 x x
441
9702
=
= 19404
4851
19404
9702 x
x 2
3
2 x
x 22
7
22
3
r = 19404 x 2
2111 x 22
22
11
11
1
11
21 22
1
11
1
rr 3333 = 44
44 x
x33x
x7 1
1
11
rr = =
= 44
44 x
x33x
x7
7
7
1

3
rr 33 =
3 9261
= 44 x3x7
9261
rr = 9261
= 9261
rr 3=
==21 cm
9261
21 cm
rr =
= 21
21 cm
cm
r = 21 cm
26. å¼ F‡ñ ܬó‚«è£÷ˆF¡ ªñ£ˆîŠ ¹øŠðóŠ¹ 675 π ê.ªêe âQ™, Üî¡ õ¬÷ðóŠ¹
â¡ð¶
If the total surface area of a solid semi sphere is 675π sq.cm, then its curved surface area is
(A) 243 π sq.cm (B) 340 π sq.cm (C) 450 π sq.cm (D) 240 π sq.cm
Sol:
ܬó‚«è£÷ˆF¡ ªñ£ˆî ¹øŠðóŠ¹
= 3πr 2
3πr 2 = 675π sq.cm
r 2 = 225

225
675 π
r2 =
3π
õ¬÷ŠðóŠ¹
2
=
= 2πr
2πr 2
=
= 2π
2π x
x 225
225
=450 sq.cm
=450 sq.cm
450π sq.cm
103
 TNPSC èí‚°

27. å¼ Ã‹H¡ Ü®Šð‚èˆF¡ Ýó‹ 2.1 ªê.e. àòó‹ 8.4 ªê.e. ܶ ༂èŠð†´ å¼
«è£÷ñ£è õ£˜‚èŠð†ì£™ «è£÷ˆF¡ Ýó‹ â¡ù?
A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere.
Find the radius of the sphere
(A) 2.4 cm (B) 2.1 cm (C) 2.2 cm (D) 2.3 cm
Sol:
Volume of Cone = 1 πrc 2 h
3
Volume of Sphere
4 3
= πrs
3
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

1 4
1 π rc 22 h = 4 π rs 33
1
13 π r 2 h = 4
3 π
4 rs 3
13 π
π rrcc 2 h = 3
4 π rs 3
h =
c h 3= 3 π rπ
r3
2
3
2
2 hπ = r 4r 3 3 rss 3
r3c 2 h = 4rs 3 3
c c s
r 2h =2 4rs 3
((rrccc2.1
2 h )=2 4r
2.1)=22 4r
h x ss8.4
3 = 4r 3
x 8.4 = 4rss 33
((r 2.13 )2 x
3 =) 9.261 x 8.4
8.4 =
= 4r
4rss 33
(rss2.1
2.1 =) x
9.261 8.4 = 4rs
rrs 33 === 9.261
2.1
9.261 cm
3
rss ==2.1 9.261 cm
rrs = = 2.1
2.1 cm
cm
rss = 2.1 cm

9
28. π è. ªêe èù Ü÷¾ ªè£‡ì «è£÷ˆF¡ Ýó‹ â¡ð¶
16
9
If the volume of a sphere is π then its radius is
16
4 3 3 2
(A) cm (B) cm (C) cm (D) cm
3 4 2 3
Sol:
«è£÷ˆF¡ èùÜ÷¾
4 3
= πr
3
44 3 99
4π rr 3 =
= 9 π
33 π π r 3 = 16
16 π
π
3 99 16 33
rr 333 =
= 9 xx 43
r = 16 16 x 4
16
27 4
rr 333 = 27
= 27
r = 64 64
3364
rr = = 3
r = 44 cm
4

29. Þ¼ «è£÷ƒèO¡ ðóŠ¹ MAî‹ 4:25 âQ™, Üî¡ èù Ü÷¾ MAî‹

If the surfacc areas of two spheres are in the ratio of 4:25 then the ratio of their volume is
(A) 4:25 (B) 25:4 (C) 125:8 (D) 8:125
Sol:
104
 TNPSC èí‚°

Surface area of Sphere

2 2
4πr
4πr 2 = = 4πR
4πR 2
4
4 :: 25
25
rr 2
:: 2
R
R 5 5
Ratio of the volume of two Sphere is
4 3
4
V 4 πr
πr 3
V 4
3 πr 33
V111 =
= 3
3 πr
V = 4
V 1
V222 = 43 πR
4
πR
3
3
V2 4
3 πR 33
3
3 πR
rr 333 3
=
= r33
= R
Rr3
= R 33 3
Rrr 33
=  rr 3
=  R
= 
=  R R 3
 R 33
=  2 2
2 3
=  52 
=
=  5 5 
 85 

ê˜è£˜ ä.ã.âv Üè£ìI

= 8
8
=
= 125 8
= 125 125
8
8 :125 125
8 :: 125 125
88 :: 125 125

30. å¼ «è£÷ˆF¡ ¹øŠðóŠ¹ = 144 π cm2 âQ™, Üî¡ èù Ü÷¾ (in cm3) ™ â¡ù?
The total surface area of a sphere is 144 π cm2 the volume of this sphere (in cm3) is
(A) 144 π (B) 72 π (C) 288 π (D) 576 π
Sol:
«è£÷ˆF¡ ¹øŠðóŠ¹
= 144π cm 2

4πr 22 =144π cm 22
4πr =144π cm
r 22 = 36
r = 36
r=6
r=6
èùÜ÷¾
44 33
= 43 πr
=
= πr
πr 3
33
44
= 43 π
=
= π xxx 666 xxx 666 xxx 666
π
33
=
= 288πm333 3
= 288πm
288πmcm

32. 9 ªê.e Ýóºœ÷ å¼ F‡ñ‚ «è£÷ˆ¬î‚ ªè£‡´ Ü«î Ýóºœ÷ ˹ ªêŒòŠð†ì£™
܂ËH¡ àòó‹ ò£¶?
A solid sphere of radius 9 cm is melted and cast into a shape of a solid cone of same radius,
find the height of the cone
(A) 9 cm (B) 81 cm (C) 27 cm (D) 36 cm
105
 TNPSC èí‚°

Sol:
ËH¡ èùÜ÷¾ = «è£÷ˆF¡ èùÜ÷¾
1
1 2 4
4 3
1 π π rr 222 h
h =
= 4 π
π rr 333
3
3 πr h = 3 3 πr
3 2
922 x h = 933 x 4
2 3 3
3
9
9 x xh h= =9 9 xx44
9 3333 x 4
h
h = 99 x x22 44
h= = 9 9 2
2
h = 36cm 9
h = 36cm
h = 36cm

33. r Ýó‹ ªè£‡ì å¼ «è£÷ˆF¡ èù Ü÷¾ Üî¡ ðóŠð÷¾ì¡ âî¬ù ªð¼‚°õîù£™

A¬ì‚Aø¶
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

The volume of a sphere of radius r is obtained by multiplying its surface area by

4 r 4r
(A) (B) (C) (D) 3r
3 3 3
Sol:
Surface area = 4πr 2
4 3
Volume = πr
3
2 4
4 r 33
4
4ππ rr 2 xx =
= 3 ππr
3
rr
xx =
= 3
3

11352
34. å¼ àœkìŸø «è£÷ˆF¡ èù Ü÷¾ cm3 ñŸÁ‹ Üî¡ ªõO Ýó‹ 8 ªêe.
7
âQ™ Ü‚«è£÷ˆF¡ àœ Ýóˆ¬î‚ 裇è

11352
Volume of a hollow sphere is cm3 . If the outer radius is 8 cm, find the inner radius of the
7
sphere
(A) 6 cm (B) 5 cm (C) 7 cm (D) 8 cm
Sol:
Volume of hollow Sphere
44
π (R -- rr 33 )
33
=
= π R
33
516
516 11
516 1
1032
1032 22
11352
11352 44 22 33 33
22
77
=
= xx
3 77
(88 -r-r )
44
= ( 512512 -- rr 3 )
3
516
516 =
33
129 3
516 xx 3 =
129
516 512 -- rr 33
= 512
41
1 1
387 -- 512
387 512 == - rr 33
-125 =
-125 = -- rr 33
rr =
= 55
r = 5 cm
106
 TNPSC èí‚°

35. 3 ªê.e Ýó‹ ªè£‡ì «è£÷ õ®õ ðÖQ™ 裟Á ªê½ˆîŠð´‹ «ð£¶ Üî¡ Ýó‹
9 ªê.e Ýè, ÜFèKˆî£™, ÜšM¼ G¬ôèO™ ðÖQ¡ èù Ü÷¾èO¡ MAî‚
裇è.
The radius of a spherical balloon increases from 3 cm to 9 cm as air is being pumped into it.
Find the ratio of volumes of the balloon in the two cases.
(A) 1:3 (B) 1:9 (C) 1:27 (D) 1:8
Sol:
Volume of the Sphere
4
4 πr 333
πr
3
3
4
4 π r 333
3 π r111 3
3 = 3xx33xx33
4 = 9 x 9 x 9
4 π r 333 9x9x9
3 π r222
3
1
1
=
= 27
27

ê˜è£˜ ä.ã.âv Üè£ìI

1
1 : 27
: 27
1:27

36. Þ¼ «è£÷ƒèO¡ ðóŠð÷M¡ MAî‹ 1 : 4 âQ™ ÜõŸP¡ èù Ü÷M¡ MAî‹

If the area of two spheres are in the ratio 1: 4 then the ratio of their volumes is
(A) 1:2 (B) 1:4 (C) 1:8 (D) 1:6
Sol:
4πr 2 1
4πr 222 = 14
4πR
4πr 2 = 1
4πR =
r 2 2 1 44
4πR 2 =
Rrr 22 = 114
Rr 22 =14
R= 4
Rrr = 112
R=2
R 2
Volume
4 33
πr 3
3 3
3
3 r 1
= =  = 
4 3 R 2
πR 3
3
1
=
8
= 1:8
1:8

37. «è£÷ õ®M™ àœ÷ ðÖ¡ èù Ü÷¾ V, ªõO¹ø ðóŠð÷¾ S à¬ìò¶ ðÖ¬ù
«è£œ õ®õ‹ ñ£ø£ñ™ 8 Vèù Ü÷¾‚° áFù£™ ¹¶ õ®M™ ªõO¹ø ðóŠð÷¾
â¡ù?
A balloon in the shape of a sphere has volume V and surface area S. If the balloon is blown up

107
 TNPSC èí‚°

so as to have a volume of 8 V without changing the spherical shape. What is the new surface
area?
(A) 8S (B) 4S (C) 2S (D) S
Sol:
4 3
Volume of sphere is V = πr
3
Surface area of sphere is S = 4π r 2
Now given that
Initial Volume = V
Surface area = S
Final Volume V ꞌ = 8v
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

4
V ' = 8x π r 3
4 3 4
π r '3 = 8x π r 3
3 3
3 3
r ' 8r

r ' = 2r
Final Surface
S' = 4π r'2

S ' = 4π (2r ) 2

S ' = 4π (4r 2 )

S = 4π r 2

S ' = 4S
Ans : 4S

38. ܬó‚«è£÷ õ®õ A‡íˆF¡ î®ñ¡ 0.25 ªê.e. Üî¡ à†¹ø Ýó‹ 5 ªê.e. âQ™

22
Ü‚A‡íˆF¡ ªõOŠ¹ø õ¬÷ðóŠ¬ð‚ 裇è. ( π = ).
7
The thickness of a hemispherical bowl is 0.25 cm. The inner radius of the bowl is 5 cm, Find

22
the outer curved surface area of the bowl. (Take π = ).
7

(A) 153.25 sq.cm. (B) 173.25 sq.cm. (C) 145.25 sq.cm. (D) 163.25 sq.cm.
Sol:
r = 5 cm
Thickness = 0.25 cm
R = 5 cm + 0.25 cm = 5.25 cm
Curved Surface area = 2πR 2
11 105 105
22 525 525
= 2 x x x
7 100 108
100
10 10
 TNPSC èí‚°
= 2πR 2
= 2πR 2
11 105 105
22 525 105525 105
11
= 2 x x 22 x525 525
= 72 x 100x 100x
710 10010 100
10 10
2
= 173.25 cm 2
= 173.25 cm

1
39. å¼ F‡ñ «è£÷ˆF¡ èù Ü÷¾ 7241 , è. ªê.e âQ™ Üî¡ Ýóˆ¬î 裇è.
7
1
If the volume of the solid sphere 7241 cubic cm., find its radius.
7
(A) 13cm (B) 10 cm (C) 11 cm (D) 12 cm
Sol:
4
Volume of Sphere = πr 3
3
4 3
4 1
1
7241
7241 +
+ 4
4 πr
πr1
13
3 =
=
7241
7241 +
+ 3 πr
πr7
73 =
=
3
3 7
12672
12672
1 3
1
7
50688
12672
4
1

ê˜è£˜ ä.ã.âv Üè£ìI

50688
12672
50688 4
1
4 πr 3
3=
50688
7 4
3 πr
πr 3
3=
=
576 7
7 3
3 πr =
7
576
1152
576
576
3 1
1152 1
12672
1152
576
12672
1152 7
7
1
1
12672
12672 x
x 3
3 x
x 7
7 =
= rr 333
7 x 3 x
x 3 x 22 =
= rr 3
7 22
77 3
22
2
22
2
2
1728 =
1728 = rr 3
3
2

1728
1728 =
=r r 3
rr = 12cm
rr =
= 12cm
12cm
= 12cm

40. å¼ «è£÷ˆF¡ èù Ü÷¾ ñŸÁ‹ ¹øŠðóŠ¹ ÝAòõŸP¡ MAî‹ â¡ù?

What is the ratio of volume to surface area of a sphere?
1 1
(A) r : 3 (B) 3 : r (C) : 1 (D) : r
Sol: 3 3
44 3
π
π rr 3
Volume
Volume = 33
=
Surface area
Surface area π rr 22
44 π
rr
=
=
33
rrr =:: 333

41. «è£÷ƒèO¡ ÝóƒèO¡ MAî‹ 4:7 âQ™ èù Ü÷¾èO¡ MAî‹

The ratio of radii of two spheres are 4: 7 then the ratio of their volumes
(A) 4:7 (B) 64:49 (C) 16:49 (D) 64:343
Sol:
Radius = 4 : 7
4
Volume = πr 3
3

109
 TNPSC èí‚°

4 3
π r1
3 43 64
= = 3 =
4 3 7 343
π r2
3
64 : 343
64:343

43. å¼ F‡ñ ܬó‚«è£÷ˆF¡ èùÜ÷¾ 29106 è.ªê.e. Í¡P™ Þó‡´ ðƒ° èù

Ü÷¾œ÷ ñŸ«ø£˜ ܬó‚«è£÷‹ ÞFL¼‰¶ ªê¶‚èŠð´ñ£ù£™ ¹Fò ܬó‚«è£÷ˆF¡
Ýó‹ â¡ù?
A volume of a solid hemisphere is 29106 cm3. Another hemisphere whose volume is two-third
of the above is carved out. Find the radius of the new hemisphere.
(A) 21.5 cm (B) 12 cm (C) 21 cm (D) 23 cm
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

2 9702
9702
= x 29106
Sol:
3
Volume2 29106
2 33 19404
9702
=3 πr11 x=29106
2 9702
New Sphere = x 29106
3
441 3
441
2 3 9702 4851
219404
4851

πr1 29106
=9702 πr13x=3 19404
x7
3
r 33
= 3
11
19404
4412 x 322x4417
r13 = 9702 4851
11 11
11 4851

2
29106 x 22
x
119702
3 x7 x3x7
3 29106
rr111 == 9261 r123 =x 22
11 2 x 22
r11 = 21cm1 1 1 11
1

r1 = 9261
r1 = 9261
44. A hemispherical
r = tank
r1 = 21cm 21cmof radius. 1.75 m is full of water. It is connected with a pipe which emp-
1
ties the tank at the rate of 7 litres per second. How much time will it take to empty the tank
completely?
Ýó‹ 1.75 e àœ÷ æ˜ Ü¬ó«è£÷ õ®õˆ ªî£†® ºŸP½‹ có£™ GóŠðŠð†´œ÷¶. å¼
°ö£J¡ Íô‹ M‚° 7 L†ì˜ iî‹ ªî£†®JL¼‰¶ c˜ ªõO«òŸøŠð´ñ£ù£™,
ªî£†®¬ò âšõ÷¾ «ïóˆF™ º¿õ¶ñ£è è£L ªêŒòô£‹?
(A) 27 minutes (B) 26 minutes (C) 72 minutes (D) 62 minutes
Sol:
2
Volume of tank = πr 3
3

2
2 2222
=
= 3 x
x 7 x
x 1.75
1.75 x
x 1.75
1.75 x
x 1.75
1.75
3 7
2
2 2222 175
175 175175 175
175
=
= 3 x
x 7 x
x 100 xx 100 x
x 100
3 7 100 100 100
11
11
2
2 22
22 7
7 7
7 7 7
=
= 3 xx 7 x x 4 xx 4 x
x 4
3 7 42 4 4
2
11x49 11x49
= 11x49 = 11x49 m33 110
 2 22
= x
= 1.75 x
x 1.75 x 1.75
1.75 x
x 1.75
1.75
3 7
2 22 175 175 175 TNPSC èí‚°
= x
= x x
x x x
x
3 7 100 100 100
11
11
2 22 7 7 7
= 2 x 22 x 7 x 7 x 7
3 7 4 4 4
2 2
11x49 11x49
= 11x49 = 11x49 m33
3x4x4 48
48
11x49
11x49 x 1000 litres
=
= x 1000 litres
48
= 11229.17
= 11229.17 litres
litres
1604.17
1604.17
Time taken
11229.17 11229.17
=
= 11229.17
1604.17
1604.17

= 11229.17 7
= 7
77= 1604.17
= 1604.17 secondsseconds
= 1604.17
= 1604.17 seconds
seconds
= 26.74=minutes
26.74 minutes
= 26.74 minutes
= 26.74=minutes
27 minutes
= 27 minutes
= 27 minutes
= 27 minutes

45. å˜ Ü¬ó«è£÷ˆF¡ èùÜ÷¾ 1152 π cm3 âQ™ Üî¡ õ¬÷‰î ªõOðóŠH¡

ðóŠð÷¾

ê˜è£˜ ä.ã.âv Üè£ìI

The volume of the hemi sphere is 1152 π cm3 then the curved surface area of the hemi sphere
is
(A) 242 π cm2 (B) 208 π cm2 (C) 288 π cm2 (D) 338 π cm2
Sol:
2
Volume of Hemisphere = πr 3
576
576 22 3
1152
1152
576 π
π =
= 2 π
π rr 333
1152 π = 3 r
3 π
33
576
576 xx 33 =
= rr 33
576 x 3 = r
rr 333 = 1728
r = = 1728
1728
rr = 12
r= = 12
12

Area of hemisphere
= 2πr 22
= 2πr 2
= 2πr
= 2π (12 )2
2

= 2π (12 )2
= 2π (12 ) 2
= 288π cm 2
= 288π cm 2
= 288π cm

46. å¼ Fì‚«è£÷ˆF¡ õ¬÷ŠðóŠ¹ 36 ªê.e.2 Üî¬ù Þ¼ Ü¬ó‚ «è£÷ƒè÷£è

HKˆî£™, å¼ Ü¬ó‚«è£÷ˆF¡ ªñ£ˆîðóŠ¹ â¡ù?
Curved surface area of a solid sphere is 36 cm2. If the sphere is divided into two hemisphere
then total surface area of one of its hemisphere is
(A) 9 cm2 (B) 12 cm2 (C) 18 cm2 (D) 27 cm2
Sol:
Area of Sphere = 4πr 2

111
 TNPSC èí‚°

9
9 1
1 22
22 x r 22
36
36 = = 4
4 xx xr
7
7
7
2
rr 2 =
= 9
9x x 7
22
22

= 3πr 22 area of hemisphere = 3πr 2
Surface
= 3πr
22 7 22 7
= 3 x 22 x 9 x 7 =3x x9x
=3x 7 x9x 22
7 22 7 22
= 27 cm 22 = 27 cm 2
= 27 cm

47. A cylindrical tank of diameter 28 cm is full of water. If 11 litres of water is drawn off, the water
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

level in the tank will drop by

M†ì‹ 28 cm Ýè‚ ªè£‡ì å¼ à¼¬÷ õ®õ‹ àœ÷ ªî£†®J™ î‡a˜ Gó‹H-
J¼‚Aø¶. ÞF™ 11 L†ì˜ î‡a˜ ªõO«òŸøŠð†ì£™ ªî£†®J™ î‡aK¡ Ü÷¾
âˆî¬ù àòó‹ °¬ø‰F¼‚°‹?
6 3
A) 17 cm B) 11 cm D) 10 ½ cm D) 14 cm
7 7
Sol:
Diameter of the tank = 28 cm
(༬÷ õ®õˆ ªî£†®J¡ M†ì‹)

Radius of the tank = 14 cm

(༬÷ õ®õˆ ªî£†®J¡ Ýó‹)

= πr h
2
Volume of Cylinder
(༬÷J¡ èùÜ÷¾)

V = 11 litres = 11 x 103 cm3

V = π r 2h
r = 14 cm
22
x (14 ) x h
2
11 x 103 =
7
7 1
h = 11 x 103 x x
22 14 x14
2 2

103 1000
h= =
56 56
125 6
= = 17 cm
7 7

48. Find the volume of a solid cylinder whose radius is 14 cm and height 30 cm
å¼ F‡ñ ༬÷J¡ Ýó‹ 14 ªê.e ñŸÁ‹ Üî¡ àòó‹ 30 ªê.e âQ™ Üš¾¼¬÷J¡
èù Ü÷¬õ‚ 裇è.
(A) 18380 cm3 (B) 18480 cm3 (C) 18580 cm3 (D) 18680 cm3
Sol:
Volume of the solid cylinder = ?
112
 TNPSC èí‚°

Radius of the tank r = 14 cm

(F‡ñ ༬÷J¡ Ýó‹)

Height of the Cylinder h = 30 cm

(F‡ñ ༬÷J¡ àòó‹)

V = π r 2h

22 2
= x 14 x14 x 30
7
=18480 cm3

49. In a cylinder, if radius is doubled and height is halved then what happens to the curved surface
area ?
(A) Halved (B) Doubled
(C) Does not change (D) Four times

ê˜è£˜ ä.ã.âv Üè£ìI

å¼ à¼¬÷J™ Ýó‹ Þ¼ ñìƒè£‚èŠð†´, àòó‹ ð£Fò£è °¬ø‚èŠð†ì£™ Üî¡
¹øŠðóŠ¹ â¡ùõ£°‹ ?
(A) ð£Fò£°‹ (B) Þ¼ ñìƒè£°‹
(C) ñ£ø£¶ Þ¼‚°‹ (C) ° ñìƒè£°‹
Sol:
h
r=2r = 2r h=
2
Curved surface area of the Cylinder =2πrh
(༬÷J¡ ¹øŠðóŠ¹)
h
=2π x ( 2 r ) x
2
=2πrh
Curved surface of the cylinder does not change
(༬÷J¡ ¹øŠðóŠ¹ ñ£ø£¶ Þ¼‚°‹)

50. If the capacity of a cylindrical tank is 1848 m3 and the diameter of its base is 14m, then find the
depth of the tank?
å¼ à¼¬÷ õ®õˆ ªî£†®J¡ ªè£œ÷÷õ£ù¶ 1848 e3 ñŸ-Á‹ Üî¬ìò M†ìñ£ù¶
14 e âQ™, Üî¬ìò Ýö‹ ò£¶? -
(A) 12 m (B) 14 m (C) 15 m (D) 18 m
Sol:
3
The capacity of a cylindrical tank = 1848 m
(༬÷ õ®õˆ ªî£†®J¡ ªè£œ÷÷¾)
Diameter = 14 m
(M†ì‹)
àòó‹(height) = ?
V = πr 2 h

113
 TNPSC èí‚°

22
= x 7 x7 x h
7
1848 1848 h = 12 m
h= =
22 x 7 154

51. A hollow cylindrical iron pipe is of length 35 cm. Its outer and inner diameters are 10 cm and 8
cm respectively. Find the weight of the pipe if 1 cu. cm of iron weight 7 gm.
å¼ àœkìŸø Þ¼‹¹ °ö£J¡ c÷‹ 35 ªê.e. Üî¡ ªõO ñŸÁ‹ àœ M†ìƒèœ
º¬ø«ò 10 ªê.e ñŸÁ‹ 8. ªê.e âQ™, Þ¼‹¹‚ °ö£J¡ â¬ì¬ò 裇è. (1. è.ªê.e
Þ¼‹H¡ â¬ì 7 Aó£‹)
(A) 6.93 kg (B) 9.90 kg (C) 7.53 kg (D) 7.93 kg
Sol:
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

A hollow cylindrical iron pipe is of length (h) = 35 cm

(å¼ àœkìŸø Þ¼‹¹ °ö£J¡ c÷‹)
Outer Diameter ( D1 ) = 10 cm
(ªõO M†ì‹)
r1
= 5 cm
Inner Diameter ( D 2 ) = 8 cm
(àœ M†ì‹)
r2
= 4 cm
Volume of Pipe = π ( r1 +r2 )( r1 -r2 ) x h

22
(5 + 4)(5 − 4)x35
7
22 5
= x 9 x 1 x 35
7
=990 cu.cm
1 cu. cm = 7g
990 cu. cm = ? x
x = 990 x 7 = 6930g (1 kg = 1000 g)
= 6.93 kg

52. The ratio of the radiiof two cylinders is 2 : 3 and the ratio of their heights is 5 : 3. The ratio of their
volume will be
Þ¼ ༬÷J¡ ÝóƒèO¡ MAî‹ 2 : 3 ñŸÁ‹ ÜõŸP¡ àòóƒèO¡ MAî‹ 5 : 3
âQ¡ Üî¡ èùÜ÷¾èO¡ MAî‹
A) 4 : 9 B) 9 : 4 C) 20 : 27 D) 27 : 20
Sol:
Volume of the cylinder
2
πr h πr 2 h
π x ( 2r ) x ( 5h ) π x ( 3r ) x ( 3h )
2 2
:

4 r 2 x5 h : 9 r 2 x3 h
20 : 27

114
 TNPSC èí‚°

53. The radii of two cylinder’s are in the ratio 3 : 5 and their heights are in the ratio 2: 3. Find the ratio
of their curved surface area
Þ¼ ༬÷J¡ Ýóƒèœ º¬ø«ò 3 : 5 â¡ø MAîˆF½‹ ÜõŸP¡ àòóƒèœ º¬ø«ò
2 : 3 â¡ø MAîƒèOL¼ŠH¡ ÜõŸP¡ õ¬÷ŠðóŠ¹èO¡ MAî‹ â¡ù?
A) 2 : 5 B) 5 : 2 C) 2 : 3 D) 3 : 2
Sol:
Curved surface of the cylinder
2πrh 2πrh
( ) ( )
πx 3r x 2h : ( ) ( )
πx 5r x 3h
6 : 15
2 : 5

54. The ratios of the respective heights and the respective radii of two cylinders are 1 : 2 and 2 : 1
respectively. Then their respective volumes are in the ratio
Þó‡´ à¼¬÷èO¡ àòóƒèœ º¬ø«ò 1 : 2 ñŸÁ‹ ÜõŸP¡ Ýóƒèœ º¬ø«ò
2 : 1 â¡ø MAîƒèO™ Þ¼ŠH¡ ÜõŸP¡ èù Ü÷¾èO¡ MAî‹
A) 4 : 1 B) 1 : 4 C) 2 :1 D) 1 : 2

ê˜è£˜ ä.ã.âv Üè£ìI

Sol:
Volumes are in the ratio of the two cylinders
(Þó‡´ à¼¬÷èO¡ èùÜ÷¾èO¡ MAî‹)
π r 2h π r 2h
( 2r ) x (1) h (1r ) x ( 2 ) h
2 2
:

4 r2 x 1 h : r 2 x2 h
4 : 2
2 : 1

55. The radius and height of cylinder and cone are equal. If the volume of cylinder is 120 cm3 then
the volume of cone is
å¼ Ã‹¹ ñŸÁ‹ ༬÷J¡ Ýóº‹ àòóº‹ º¬ø«ò êñ‹. ༬÷J¡ èù Ü÷¾
120 ªê.e3. âQ™ ËH¡ èù Ü÷¾
A) 90 cm3 B) 40 cm3 C) 30 cm3 D) 100 cm3
Sol:
3
Volume of the cylinder = 120 cm
(༬÷J¡ èùÜ÷¾)
Volume of the Cone = ?
(ËH¡ èùÜ÷¾)
1
Volume of the cone = πr 2 h
3
πr 2 h = 120

1 40
= x 120
3

115
 TNPSC èí‚°

3
= 40 cm

56. 50 circular plates each of radius 7 cm and thickness ½ cm are placed one above another to form
a solid right circular cylinder. What is the total surface area of the cylinder so formed?
7 ªêe Ýóº‹ 1/2 ªêe èùº‹ ªè£‡ì 50 õ†ìõ®õ î†´èœ å¡ø¡ «ñ™ å¡ø£è
Ü´‚èŠð†´ å¼ Fì «ï˜õ†ì ༬÷ à¼õ£A»œ÷¶. Þ‰î ༬÷J¡ ªñ£ˆî
¹øðóŠð÷¾ ò£¶?
A) 1230 cm2 B) 1332 cm2 C) 1408 cm2 D) 1560 cm2
Sol:
The radius of the base of the cylinder = 7 cm
(༬÷J¡ Ü®ð‚èˆF¡ Ýó‹)
Height of the cylinder is same as the thickness of 50 circular plates
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

(༬÷J¡ àòóñù¶ 50 õ†ì õ®õ èO¡ Ü옾‚° êñ‹)

1
Thickness of 1 plate = cm
2
1
Thickness of 50 plates = 50 x cm
(Ü예F) 2
=25
Total Surface area =2π r ( r + h )
(ªñ£ˆî ¹øŠðóŠð÷¾)
22
=2x x 7 ( 7 + 25)
7
= 44 x 32 = 1408 cm 2

57. The cylinder whose base is not in circular form is called

(A) Circular cylinder (B) Right circular cylinder
(C) Oblique cylinder (D) Irregular cylinder
æ˜ à¼¬÷J¡ Ü®Šð£è‹ õ†ì õ®M™ Þ™¬ô âQ™ ܶ âšõ£Á ܬö‚èŠð´‹?
(A) õ†ì ༬÷ (B) «ï˜ õ†ì ༬÷
(C) ꣌¾ ༬÷ (D) 心èŸø ༬÷
The Cylinder whose base is not in circular form is called Oblique Cylinder
æ˜ à¼¬÷J¡ Ü®ð£è‹ õ†ì õ®M™ Þ™¬ôªòQ™ ܶ ꣌¾ ༬÷ âù
ܬö‚èŠð´‹

58. If radii of two cylinders are in the ratio 5 : 3 and their heights are in the ratio 3 : 5 then ratio of
their volumes is
Þó‡´ à¼¬÷èO¡ ÝóƒèO¡ MAî‹ 5 : 3 âù¾‹ ÜõŸP¡ àòóƒèO¡ MAî‹
3 : 5 âù¾‹ Þ¼ŠH¡ ÜõŸP¡ èù Ü÷¾èO¡ MAî‹ â¡ù?
(A) 5 : 5 (B) 3 : 3 (C) 9 : 25 (D) 5 : 3
Sol:
Volume of the Cylinder = πr 2 h
(༬÷J¡ èùÜ÷¾)
π r 2h π r 2h

116
 TNPSC èí‚°

( 5r ) x ( 3) h ( 3r ) x ( 5) h
2 2
:
5 3
25 r 2 x 3 h : 9 r2 x 5 h
Ratio 5 : 3

59. The radii of two right circular cylinders are in the ratio 4:3 and their heights are in the ratio 7:4
then the ratio of their curved surface areas is in the ratio
Þó‡´ «ï˜õ†ì ༬÷èO¡ ÝóƒèO¡ MAî‹ 4 : 3 «ñ½‹ ÜõŸP¡ àòóƒèO¡
MAî‹ 7 : 4 âQ™ ÜõŸP¡ õ¬÷ðóŠ¹èO¡ MAî‹
(A) 3 : 5 (B) 5 : 3 (C) 3 : 7 (D) 7 : 3
Sol:
Curved surface area =
2π rh 2π rh
( 4r ) x ( 7 ) h : ( 3r ) x ( 4 ) h
( 4 r ) x (7) h : (3 r ) x ( 4 ) h
Ratio 7 : 3

ê˜è£˜ ä.ã.âv Üè£ìI

(MAî‹)

60. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is
filled with fruit juice to a height of 4 cm, then find the quantity of fruit juice to be prepared daily
in the hospital to serve 250 patients.
å¼ ñ¼ˆ¶õñ¬ùJ™ àœ÷ «ï£ò£O å¼õ¼‚°, Fùº‹ 7 ªê.e M†ìºœ÷
༬÷ õ®õ A‡íˆF™ õ®„ê£Á õöƒèŠð´Aø¶. ÜŠð£ˆFóˆF™ 4 ªê.e
àòóˆFŸ° ðö„ê£Á å¼ «ï£ò£O‚° õöƒèŠð†ì£™, 250 «ï£ò£OèÀ‚° õöƒèˆ
«î¬õò£ù ðö„꣟P¡ èù Ü÷¾ 裇è.
(A) 37.5 litres (B) 38.5 litres (C) 39.5 litres (D) 40.5 litres
Sol:
Diameter = 7
(M†ì‹)
7
r=
2
h = 4 cm

V= πr 2 h
22 7 7 2
= x x x 4
7 2 2
= 154 ml
For 250 patients,
250 x 154 = 38,500 ml
1000 ml = 1 l
38,500 ml = 38.5l
Ans: 38.5 litres

61. The radius and height of a cylinder are in the ratio 5:7. If its volume is 4400 cu.cm, find the radius
117
 TNPSC èí‚°

of the cylinder
å¼ à¼¬÷J¡ Ýó‹ ñŸÁ‹ àòóƒèO¡ MAî‹ 5:7 «ñ½‹ Üî¡ èù Ü÷¾ 4400
è.ªê.e âQ™ ܚ༬÷J¡ Ýóˆ¬î‚ 裇è.
(A) 6 cm (B) 8 cm (C) 10 cm (D) 12 cm
Sol:
Radius (Ýó‹) = 5 x

Height (àòó‹) = 7 x

Volume (èùÜ÷¾) = 4400

Volume= πr 2 h
4400 = π ( 5x ) x 7x
2
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

= π x 25x 2 x 7x
22
= x 25 x 7 x x 3
7
= 550 x 3
40 8
3 440 0 40
x = = =8
55 0 5
5

x3 = 8
x=2
Radius = 5 x 2 = 10 cm

62. Radius of a cylinder is increased by 10% then percentage increase in its volume is
æ˜ à¼¬÷J¡ Ýó‹ 10% àò˜ˆîŠð´‹ «ð£¶ Üî¡ èùÜ÷M™ ãŸð´‹ àò˜M¡
êîiî‹ ò£¶?
(A) 20% (B) 15% (C) 10% (D) 21%
Sol:
Radius (Ýó‹) = r
Increased % of r = 10%
(ÜFèKˆî ÝóˆF¡ êîiî‹)

Increased r = r + 10% of r
= (r + 10%) r

 10 
= r 1+ 
 100 
= r (1+0.1)
New Radius = 1.1 r
(¹Fò Ýó‹)
Original Volume = πr 2 h
(à‡¬ñò£ù èùÜ÷¾)
New Volume = π (1.1r ) x h
2

(¹Fò èùÜ÷¾)
118
 TNPSC èí‚°

( ) ( ) = π x 1.1 x 1.1 r 2 x h
2
= 1.21π r h
2 2
Percentage increase in voulume = 1.21π r h − πr h x 100%
(ÜFèKˆî èùÜ÷M¡ êîiî‹) πr 2 h

π r 2 h (1.21 − 1)
= x 100%
πr 2 h
= 0.21 x 100%
= 21%

63. The radii of two right circular cylinders are in the ratio of 3 :2 and their heights are in the ratio 5
: 3. Find the ratio of their curved surface areas.
Þó‡´ «ï˜õ†ì ༬÷J¡ ÝóƒèO¡ MAî‹ 3 : 2 â¡è. «ñ½‹ ÜõŸP¡
àòóƒèO¡ MAî‹ 5 : 3 âQ™ ÜõŸP¡ õ¬÷ðóŠ¹èO¡ MAî‹ è£‡è.
(A) 5 : 2 (B) 2 : 5 (C) 3 : 2 (D) 5 : 3
Sol:

ê˜è£˜ ä.ã.âv Üè£ìI

Curved Surface area = 2πrh
(õ¬÷ðóŠ¹)
2π rh 2π rh

( 3 r ) x ( 5) h : ( 2 r ) x ( 3) h
5 2
15 : 6
5 : 2
Ratio = 5 : 2

64. A solid right circular cylinder has radius 14 cm and height 30 cm. Its curved surface area is
å¼ F‡ñ «ï˜õ†ì ༬÷J¡ Ýó‹ 14 ªê.e. ñŸÁ‹ àòó‹ 30 ªê.e âQ™ Üî¡
õ¬÷ðóŠ¹
A) 2240 cm2 B) 2260 cm2 C) 2460 cm2 D) 2640 cm2
Sol:
Curved Surface area = 2πrh
(õ¬÷ðóŠ¹)
(Ýó‹) r = 14 cm
(àòó‹) h = 30 cm
22 2
Area = 2 x x 14 x 30
7
= 44 x 60
= 2640 cm 2
Curved surface area = 2640 cm 2

65. What is the difference between the total surface area and the lateral surface area of a solid cylinder
119
 TNPSC èí‚°

of radius “ r ” and height “ h ”.

å«ó àòóº‹ ( h ), Ýóº‹ ( r ) ªè£‡ì å¼ ªè†® ༬÷J¡ ªñ£ˆî ð󊹂°‹
¹øŠð󊹂°‹ àœ÷ «õÁ𣴠â¡ù?

1
(A) 2π r 2 (B) π r 2 (C) 2π r 2 h (D) 2π rh
2
Sol:
Total Surface area = 2πr ( h+r )
(ªñ£ˆî ðóŠð÷¾)
Lateral surface area = 2πrh
Difference between TSA and LSA
⇒ 2πr ( h+r ) - 2πrh
⇒ 2πrh +2πr 2 - 2πrh
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

⇒ 2πr 2

66. The volume of a cylinder is 81cm3. What is the volume of the cone that has the same radius and
height as the cylinder?
å¼ à¼¬÷J¡ èù Ü÷¾ 81 ªê.e3 ܉î ༬÷¬òŠ «ð£¡«ø Ýóº‹ àòóº‹
ªè£‡ì ËH¡ èù Ü÷¾ â¡ù?
(A) 27 cm3 (B) 243 cm3 (C) 81 cm3 (D) 162 cm3
Sol:
Volume of the cylinder
(༬÷J¡ èùÜ÷¾) V = πr 2 h = 81
1
Volume of the cone = πr 2 h
3
(ËH¡ èùÜ÷¾)
1 27
= x 81
3
3
= 27cm

67. Find the volume of a cyclinder whose height is 3.5 m and diameter of base is 4 meters
æ˜ à¼¬÷J¡ àòó‹ 3.5 m, Ü®M†ì‹ 4 m âQ™ ༬÷J¡ èù Ü÷¾
裇è
(A) 33 (B) 22 (C) 44 (D) 55
Sol:
Height (àòó‹) = 3.5
Diameter (M†ì‹) = 4 meter
Radius(Ýó‹) = 2 meter
Volume(èùÜ÷¾) = πr 2 h

22 0.5
= x 2 x 2 x 3.5
7
= 22 x 4 x 0.5
= 22 x 2.0
V= 44
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 TNPSC èí‚°

Base area of right circular cylinder is 80 cm2. If the height is 5 cm then the volume is
68.
å¼ «ï˜õ†ì ༬÷J¡ Ü®ð‚è ðóŠ¹ 80 ê.ªê.e. Üî¡ àòó‹ 5 ªê.e âQ™ Üî¡
èù Ü÷¾
400
A. 400 cm3 B. 16 cm3 C. 200 cm3 D. cm3
3
Sol:
Volume of the Cylinder = base x height
(ËH¡ èùÜ÷¾ = Ü®Šð‚è‹ x àòó‹)
= 80 x 5
= 400 cm3

69. If two cylinders of equal volumes have their heights in the ratio 2:3, then the ratio of their radii
is
å«ó èù Ü÷¬õ‚ ªè£‡ì Þ¼ ༬÷èO¡ àòóƒèO¡ MAî‹ 2:3 âQ™
ÜõŸP¡ ÝóƒèO¡ MAîñ£ù¶
(A) 6 : 3 (B) 5 : 3 (C) 2:3 (D) 3 : 2
Sol:

ê˜è£˜ ä.ã.âv Üè£ìI

Ratio of the heights = 2: 3 h1 = 2, h2 = 3
(àòóƒèO¡ MAî‹)
Volumes are equal
(èù Ü÷¾èœ êñ‹)
Volume of the cylinder = πr 2 h
(༬÷J¡ èùÜ÷¾)
π r12 h1 : π r2 2 h2
π x r12 x (2) = π x r2 2 x (3)
r12 3
=
r2 2 2
r1 3
=
r2 2

3: 2

70. The radius of a circular cylinder is the same as that of a sphere. There volumes are equal. The
height of the cylinder is
4 2
(A) times its radius (B) times its radius
3 3
(C) equal to its radius (D) equal to its diameter
å¼ «ï˜õ†ì ༬÷J¡ Ýó‹, Ýó‹ «è£÷ˆF¡ ÝóˆFŸ° êñ‹, «ñ½‹ ܬõèO¡
èù Ü÷¾èœ êñ‹ âQ™ «ï˜õ†ì ༬÷J¡ àòó‹ ò£¶?
4 2
(A) ñ샰 Üî¡ Ýó‹ (B) ñ샰 Üî¡ Ýó‹
3 3
(C) Üî¬ìò ÝóˆFŸ° êñ‹ (D) Üî¡ M†ìˆFŸ° êñ‹
Sol:
Radius of the circular cylinder = r
(«ï˜ õ†ì ༬÷J¡ Ýó‹)
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 TNPSC èí‚°

Radius of the Sphere = r

(«è£÷ˆF¡ Ýó‹)

Volumes are equal

(èù Ü÷¾èœ êñ‹)

4
π r 3 = π r2 h
3
4 times of its radius
3 ñ샰 Üî¡ Ýó‹
71. The ratio of heights of two cylinders of equal volumes is 1:3 then the ratio of their radii will be :
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

Þó‡´ à¼¬÷èO¡ êñ èù Ü÷¾ àœ÷ àòóƒèO¡ MAî‹ 1:3 âQ™ Üî¡

ÝóƒèO¡ MAî‹ â¡ù?
(A) 4 : 3 (B) 3: 2 3 (C) 2 : 3 (D) 3 : 3
Sol:
Height of the two cylinders be h1 and h 2
(༬÷J¡ àòó‹) = 1:3
Radius of the two cylinders be r1 and r2
(༬÷J¡ Ýó‹)
h1 1
=
h2 3
Volume of the Cylinder
(༬÷J¡ èùÜ÷¾)
V= πr 2 h
Volume of two cylinders
π r12 h1 = π r2 2 h 2

r12 h2
=
r2 2 h1
r12 3
=
r2 2 1
r1 3
=
r2 1
r1 3x 3
=
r2 1x 3
r1 3
=
r2 3
r1 : r2 = 3 : 3

72. A cylindrical shaped well of depth 20 m and diameter 14 m is dug. The dug out soil is evenly
spread to form a cuboid-platform with base dimension 20 m x 14 m. Find the height of the plat-
form.

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 TNPSC èí‚°

14 e M†ì‹ ñŸÁ‹ 20 e Ýöºœ÷ å¼ AíÁ ༬÷ õ®M™ ªõ†ìŠð´Aø¶.

Üšõ£Á ªõ†´‹ «ð£¶ «î£‡®ªò´‚èŠð†ì ñ‡ Yó£è ðóŠðŠð†´ 20e x
14eÜ÷¾èO™ Ü®Šð‚èñ£è‚ ªè£‡ì å¼ «ñ¬ìò£è ܬñ‚èŠð†ì£™,
Ü‹«ñ¬ìJ¡ àòó‹ 裇è
(A) 11 m (B) 22 m (C) 77 m (D) 33 m
Sol:
Height (àòó‹) h = 20 m

Diameter (M†ì‹) d = 14 m
2r = 14
Radius (Ýó‹) = 7 m
Volume of the cylinder = V = π r 2 h
(༬÷J¡ èùÜ÷¾)
22
= x7 2 x 20
7
3
= 22 x 7 x 20 m → (i )
length of the cuboid (èù ê¶óˆF¡ c÷‹) l = 20 m

ê˜è£˜ ä.ã.âv Üè£ìI

breadth (Üèô‹) b = 14 m
height (àòó‹) h = x
Volume of the Cuboid = l x b x g = = 20 x 14 x x m3 → (ii )
(èù ê¶óˆF¡ èùÜ÷¾)
(i ) = (ii )
22 x 7 x 20 m3 = 20 x 14 x x m3
11
22 x 7 x 20
x=
20 x 14
2

x = 11 m

73. A cylindrical iron pillar 49 cms high and 6 cms in radius is surmounted by a cone 14 cms high.
The volume of pillar is
å¼ Þ¼‹ð£ô£ù ༬÷ õ®õ ÉE¡ «ñŸ¹ø‹ ˹ ެ퉶œ÷¶. ༬÷J¡
àòó‹ 49 ªê.e, Ýó‹ 6 ªê.e, «ñ½‹ ËH¡ àòó‹ 14 ªê e âQ™ ÉE¡
èù Ü÷¾ ò£¶?
(A) 5981 cm2 (B) 6072 cm2 (C) 7012 cm2 (D) 8154 cm2
Sol:
2
Volume of cylindrical iron = πr h
(Þ¼‹ð£ô£ù ༬÷J¡ èùÜ÷¾)

r = 6 h = 49
22 7
= x 6 x 6 x 49
7
= 22 x 6 x 6 x 7
= 5544 cm3
1
Volume of cone = πr 2 h
3
123
 TNPSC èí‚°

(ËH¡ èùÜ÷¾)

r = 6 h = 14

1 22 2 2
= x x 6 x 6 x 14
3 7
= 22 x 6 x 2 x 2
= 528 cm3
Volume of Piller = Volume of cylindrical iron + Volume of cone
(ËH¡ èùÜ÷¾) (Þ¼‹ð£ô£ù ༬÷J¡ èùÜ÷¾) + (ËH¡ èùÜ÷¾)
= 5544 + 528
= 6072 cm3
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

74. The total surface area and curved surface area of the right circular cylinder of radius 7 cm and
height 20 cm are
å¼ F‡ñ «ï˜õ†ì ༬÷J¡ Ýó‹ 7 ªêe ñŸÁ‹ àòó‹ 20 ªêe âQ™ Üî¡
ªñ£ˆî ðóŠð÷¾ ñŸÁ‹ õ¬÷ðóŠð÷õ£ù¶
(A) Total surface area = 1188 cm2, Curved surface area = 880 cm2
(B) Total surface area = 1818 cm2, Curved surface area = 830 cm2
(C) Total surface area = 1188 cm2, Curved surface area = 830 cm2
(D) Total surface area = 1881 cm2, Curved surface area = 880 cm2
Sol:
Radius (Ýó‹) r = 7 m
Height (àòó‹) h = 20 m
Total surface area of cylinder = = 2πr ( r+h )
(༬÷J¡ ªñ£ˆî ðóŠð÷¾)

22
=2x x 7 x ( 7 + 20 )
7
= 2 x 22 x 27
= 1188 cm 2
Curved surface area = = 2πrh
(õ¬÷ðóŠ¹)

22
=2x x 7 x 20
7
= 2 x 22 x 20
= 880 cm 2
Total surface area = 1188 cm 2
Curved Surface area = 880 cm 2

75. Base area of the right circular cylinder is 30 sq.cm and its height is 6 cm then the volume of the
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 TNPSC èí‚°

cylinder is
å¼ «ï˜ õ†ì ༬÷J¡ Ü®ŠðóŠ¹ 30 ªê.e ñŸÁ‹ Üî¡ àòó‹ 6 ªê.e âQ™
Üî¡ èù Ü÷¾ ò£¶?
(A) 60cu.cm (B) 90cu.cm (C) 120 cu.cm (D) 180 cu.cm
Sol:
Volume of the cylinder = base x height
(༬÷J¡ èùÜ÷¾) = (Ü®Šð‚è‹ x àòó‹)
= 30 x 6
= 180 Cu. cm

76. The curved surface area of a right circular cylinder whose radius is a units and height b units is
a Üô°èœ Ýóº‹, b Üô°èœ àòóº‹ ªè£‡ì å¼ «ï˜õ†ì ༬÷J¡ õ¬÷ðóŠ¹
(A) π a2b ê. Üô°èœ (B) 2π ab ê. Üô°èœ
(C) 2π ê. Üô°èœ D) 2 ê. Üô°èœ
Sol:
Curved surface area of cylinder = = 2πrh

ê˜è£˜ ä.ã.âv Üè£ìI

(«ï˜ õ†ì ༬÷J¡ õ¬÷ðóŠ¹)
r = a h=b
= 2πab Sq. units

77. Right circular cylinder is a solid obtained by revolving ................... about its sides.
(A) Square (B) Rectangle
(C) Parellelogram (D) Rhombus
å¼ ................... Üî¡ ð‚èƒè¬÷ ¬ñò Ü„ê£è‚ ªè£‡´ ²öŸÁõî¡ Íô‹ å¼
«ï˜õ†ì ༬÷¬òŠ ªðøô£‹.
(A) ê¶ó‹ (B) ªêšõè‹
(C) Þ¬íèó‹ (D) ꣌ê¶ó‹
Sol:
Right circular cylinder is a solid obtained by revolving Rectangle about its sides
(å¼ ªêšõè‹ Üî¡ ð‚èƒè¬÷ ¬ñò Ü„ê£è‚ ªè£¡´ ²öŸÁõî¡ Íô‹
å¼ «ï˜õ†ì ༬÷¬òŠ ªðøô£‹.)

78. A rectangular sheet of metal foil with dimension 66 cm x 12 cm is rolled to form a cylinder of
height 12 cm. Find the volume of the cylinder.
(A) 2772 cm3 (B) 5148 cm3 (C) 4185 cm3 (D) 4158 cm3
66 ªê.e. x 12 ªê.e. â‹ Ü÷¾‚ ªè£‡ì å¼ à«ô£èˆ î膮¬ù 12 ªê.e. àòóºœ÷
å¼ à¼¬÷ò£è ñ£ŸPù£™ A¬ì‚°‹ ༬÷J¡ èù Ü÷¾ 裇è
Sol:
height (àòó‹) = 12 cm
Curcumference of the base of the cylinder = 66 cm
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 TNPSC èí‚°

(༬÷®¡ Ü®Šð‚è ²Ÿø÷¾)

2πr = 66
22
2x r = 66
7

3
6
66 7
r= x
2 22 2

21
cm r=
2
Volume of the cylinder = πr 2 h
(༬÷J¡ èù Ü÷¾)
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

11 3
22 21 21 6
= x x x 12
7 2 2

= 11 x 3 x 21 x 6

= 4158 cm 2

126
 TNPSC èí‚°
õ†ì‹, Þ¬íèó‹, èó‹, ꣌ê¶ó‹ ñŸÁ‹ êKõè‹
Circle, Parallelogram, Quadrilateral, Rhombus
& Trapezium

Sums:

1. 10 cm
10 cm 90°
cut

π r2
Area = xθ
360

22
x 10 x 10
= 7 x 90
360

ê˜è£˜ ä.ã.âv Üè£ìI

= 25π cm 2

2. The ratio of two angles is 4:9 and the ratio of supplementary angles is 8:3
Two angle (Þó‡´ «è£íƒèO¡ MAî‹) 4 : 9
Supplementary angle (I¬è G󊹂 «è£íƒèO¡ MAî‹) 8 : 3
(A) 60, 135 (B) 30,80 (C) 40, 90 (D) 80, 30

5 60,135
312, 27

(A) 4, 9 «è£íƒèO¡ MAî‹

I¬è G󊹂 «è£í‹

180 - 40 = 120
180 - 135 = 45
120 : 45
8:3

3.

49 m
56m

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 TNPSC èí‚°

Area of Path (ð£¬îJ¡ ðóŠð÷¾)

(R = 56, r = 49)
= π R 2 -π r 2
=π ( R 2 − r 2 )
22
= (562 − 492 )
7
22
= (3136 − 2401)
7
22 105
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

= (735)
7

= 2310 m 2

4. 
å¼ ê£Œê¶óˆF¡ å¼ ð‚è Ü÷¾ 8ªê.e, °ˆ¶òó‹ 12 ªê.e, ꣌ ê¶óˆF¡ ðóŠð÷¾
裇è.
Find the area of the rhombus whose side is 8 cm and height is 12 cm
A) 96cm2 B) 88cm2 C) 76cm2 D) 56cm2
Area of a rhombus (꣌ê¶óˆF¡ ðóŠð÷¾) = bh
= 8 x 12
= 96cm2

5. 
꣌ê¶ó‹ å¡P¡ ðóŠð÷¾ 4000 ê.e Üî¡ å¼ Í¬ôM†ì‹ 100e ñŸªø£¼ ͬô
M†ìˆF¡ Ü÷¾ 裇è.
The area of a rhombus is 4000 m2 and one of its diagonal 100 m find the other diagonal?
A) 80m B) 70m C) 60m D) 50m
Area of a rhombus (꣌ê¶óˆF¡ ðóŠð÷¾) = 12 (d1 x d2)
= 12 x d1d2 = 4000

= 12 x 100 x d2 = 4000

4000 x 2
d2 = 100

d2 = 80 m

6.  å¼ õò™ ꣌ê¶ó õ®M™ àœ÷¶. Üî¡ Í¬ôM†ì Ü÷¾èœ 70e, 80e ܉î
õò¬ô„ êñ¡ ªêŒò ê¶ó e†ì¼‚° Ï.3 iî‹ Ý°‹ ªêô¬õ‚ 裇è.
A field is in the form of a rhombus the diagonals of the field are 70 m and 80 m. Find the cost
of leveling it at the rate of 3 per square m
A) Ï.8200 B) Ï.8300 C) Ï.8400 D) Ï.2800
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 TNPSC èí‚°

Area of a rhombus (꣌ê¶óˆF¡ ðóŠð÷¾) = 12 d1d2

= 12 x 70 x 80
= 70 x 40
= 2800 ê.e
1 ê e†ì¼‚° Ý°‹ ªêô¾
(Cost of 1 m2) = Ï.3
2800 ê. e†ì¼‚° Ý°‹ ªêô¾
(Cost of 2800 m2) = 2800 x 3
= Ï.8400

7.  ðóŠð÷¾ 324 ê.ªê.e, Ü®Šð‚è‹ 27 ªê.e ªè£‡ì Þ¬íèóˆF¡ °ˆ¶òó‹ 裇è.

Find the height of a parallogram whose area is 324 cm2 and base is 27 cm
A) 10 ªê.e B) 11 ªê.e C) 12 ªê.e D) 13 ªê.e
Þ¬íèóˆF¡ ðóŠð÷¾ (Area of a Parallogram) = bh
bh = 324
27 x h = 324

ê˜è£˜ ä.ã.âv Üè£ìI

324
h = 27 = 12 ªê.e

8. 
å¼ Þ¬íèóˆF¡ Ü®Šð‚躋, ÜèŸø °ˆ¶òóº‹ º¬ø«ò 14 ªê.e, 8ªê.e
âQ™ Þ¬íèóˆF¡ ðóŠð÷¾ ò£¶?
Find the area of parallogram whose base and height are 14 cm and 8 cm
A) 112 ªê.e2 B) 115 ªê.e2 C) 118 ªê.e2 D) 120 ªê.e2
b = 14 ªê.e, h = 8 ªê.e
Þ¬íèóˆF¡ ðóŠð÷¾ (Area of a Parallogram) = bh
= 14 x 8
= 112 ªê.e2

9. 
å¼ i†´ ñ¬ùò£ù¶ èó õ®M™ àœ÷¶, Üî¡ å¼ Í¬ôM†ìˆF¡ c÷‹
250e ͬôM†ìˆF¡ âFªóF˜ ð‚èƒèO™ àœ÷ º¬ùèœ Þó‡´‹ ͬôM†ìˆ-
FL¼‰¶ 70e ñŸÁ‹ 80e ªî£¬ôM™ Þ¼ŠH¡, ñ¬ùJ¡ ðóŠ¹ ò£¶?
A plot of land is in the form of quadrilateral where one of its diagonals is 250 m long. The two
vertices on either side of this diagonal are 70 m and 80 m away what is the area of plot of land
A) 18750e2 B) 18780e2 C) 17850e2 D) 19850e2
èóˆF¡ ðóŠ¹ (Area of a Quadrilateral) = 12 (h1+h2) x d
= 12 (70 + 80) x 250
= 12 x 150 x 250
= 75 x 250
= 18750e2

10. 
å¼ êKõèˆF™ Þ¬íŠð‚èƒèO¡ Ã´î™ 18 ªê.e, °ˆ¶òó‹ 15 ªê.e âQ™,
Üî¡ ðóŠð÷¾.
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 TNPSC èí‚°

Find the area of a trapezium whose sum of the parallel sides 18 cm and height is 15 cm
A) 110 ªê.e2 B) 120 ªê.e2 C) 130ªê.e2 D) 135 ªê.e2
êKõèˆF¡ ðóŠð÷¾ (Area of a Trapezium) = 12 (a + b) x h
= 12 x 18 x 15
= 9 x 15
= 135 cm2

11. Þ¬íŠð‚èƒèO¡ Ã´î™ 20 ªê.e Ýè‚ ªè£‡ì êKõèˆF¡ ðóŠð÷¾ 80 ªê.e2

âQ™ °ˆ¶òó‹ _______
Find the height of the trapezium whose sum of the parallel sides 20 cm and area is 80 cm2
A) 2 ªê.e B) 4 ªê.e C) 6 ªê.e D) 8 ªê.e
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

1
êKõèˆF¡ ðóŠð÷¾ (Area of a Trapezium) = 2 (a + b) x h
1
2
x 20 x h = 80
4
80 x 2
h = 20
h = 4x2
h = 8 ªê.e

12. 
êKõèˆF¡ ðóŠð÷¾ 88 ªê.e2, ªêƒ°ˆ¶ ªî£¬ô¾ (àòó‹) 8ªê.e. êKõèˆF¡
Þ¬íŠð‚èƒèO™ å¼ ð‚èˆF¡ c÷‹ 10ªê.e âQ™ ñŸªø£¼ ð‚èˆF¡ c÷ˆ¬î‚
裇è.
The area of a trapezium is 88 cm2 and its height is 8 cm. If one of the parallel side is 10 cm
then find the other side
A) 11 ªê.e B) 12 ªê.e C) 10 ªê.e D) 15 ªê.e
A = 88 ªê.e, a = 10 ªê.e, h = 8 ªê.e, bc = ?
êKõèˆF¡ ðóŠð÷¾ (Area of a Trapezium) = 12 (a + b) x h
1
2
(10 + 6) x 8= 88
(10 + 6) = 88 x 2
8
= 11 x 2
10 + b = 22
b = 22 - 10
b = 12 ªê.e

13. 
å¼ «î£†ìñ£ù¶ êKõè‹ õ®M™ àœ÷¶. Üî¡ Þ¬íŠð‚èƒèœ 40e, 30e,
Þ¬íŠð‚èƒèÀ‚° Þ¬ì«ò àœ÷ ªî£¬ô¾ 25e, ܈«î£†ìˆF¡ ðóŠð÷¾
裇è.
Find the area of field in the shape of a trapezium with parallel sides of length 40 m and 30 m
and the distance between the parallel sides is 25 m.
A) 825e B) 850e2 C) 900e2 D) 875e2
a = 40 e, b = 30e, h = 25e
êKõèˆF¡ ðóŠð÷¾ (Area of a Trapezium) = 12 h (a + b)
= 12 x 25 (40+30)
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 TNPSC èí‚°

= 12 x 25 x 70

= 25 x 35 = 875e2

14. ¹™ªõOJ™ àœ÷ å¼ è¬ìJ™ Ý´ å¡Á 7‹ c÷ºœ÷ èJø£™ è†ìŠð´Aø¶.

ܶ «ñ»‹ ÜFèð†ê ð°FJ¡ ðóŠð÷¬õ‚ 裇è.
A goat is tied to a peg on a grass land with a rope 7 m find the maximum area of the land it can
graze
A) 77e2 B) 154e2 C) 188e2 D) 214e2

Ý´ «ñ»‹ ÜFèð†ê ðóŠ¹ (Maximum area grazed by goat) = õ†ìˆF¡ ðóŠ¹ (Area
of a Circle)
= πr2
= 227
x7x7
= 22x7 = 154e2

15. å¼ õ‡® ê‚èó‹ 1000 ²ŸÁèœ ²ö™õ 88A.e Éó‹ ïè˜Aø¶ âQ™ Ü„ê‚èóˆF¡

ê˜è£˜ ä.ã.âv Üè£ìI

Ýó‹ 裇è
A wheel makes 1000 revolution is covering distance of 88 km the radius of wheel will be?
A) 7e B) 14e C) 21e D) 28e
1000 ²ŸÁèÀ‚° ªî£¬ô¾ = 88A.e = 88000e
1 ²ŸÁèœ ªî£¬ô¾ = 88000 1000
= 88
2πr = 88
2x x r 22 = 88
7
42
88x7
r = 22x2 = 14e

16. å¼ õ†ìˆF¡ M†ì‹ 84ªê.e âQ™, Üî¡ è£™õ†ìˆF¡ ²Ÿø÷¾ _______ Ý°‹.
Find the perimeter of quadrant of a circle whose diameter is 84 cm.
A) 120ªê.e B) 150ªê.e C) 160ªê.e2 D) 180ªê.e
D = 84
84
r = 2 = 42 ªê.e
25r
裙õ†ìˆF¡ ²Ÿø÷¾ (Perimeter of a quadrant of acircle) = 7
6
25x42

= 7
= 150 ªê.e

17. Ýó‹ 63e âQ™ ܬóõ†ìˆF¡ ²Ÿø÷¾

Find the perimeter of semicircle whose radius is 63 m
A) 364e B) 36.4e C) 324e D) 32.4e
r = 63e
36r
ܬóõ†ìˆF¡ ²Ÿø÷¾ = 7
131
 TNPSC èí‚°

36x63
= 7
36x63
= 70
36x9
= 10
= 324
10
= 32.4e

18. Ýó‹ 35ªê.e ªè£‡ì ܬóõ†ìˆF¡ ðóŠð÷¾, ²Ÿø÷¾ 裇è.

The radius of a semicircle is 35 cm. Find its area and perimeter
A) 1900ªê.e2, 180ªê.e B) 1925ªê.e2, 160ªê.e
C) 1925ªê.e2, 180ªê.e D) 1900ªê.e2, 160ªê.e
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

πr2
ܬóõ†ìˆF¡ ðóŠ¹ (Area of a semi circle) = 2
5
1122
= 7
x 35x235

= 11x5x35
= 1925ªê.e2
36r
ܬóõ†ìˆF¡ ²Ÿø÷¾ (Perimeter of a semi circle) = 7
= 36x735
= 36x5
= 180ªê.e

19. ªð£¼ˆ¶è:
1. õ†ìˆF¡ ðóŠð÷¾ & πr2
πr2
2. ܬóõ†ìˆF¡ ðóŠð÷¾ & 2
25r
3. 裙õ†ìˆF¡ ²Ÿø÷¾ & 7
36r
4. ܬóõ†ìˆF¡ ²Ÿø÷¾ & 7

Match the following
1. Area of a circle - πr2
πr2
2. Area of a semicircle - 2
25r
3. Perimeter of quadrant of a circle - 7
36r
4. Perimeter of a semicircle - 7
A) 1234 B) 4321 C) 1243 D) 2134

20. 
ܬóõ†ì õ®Mô£ù ̃è£M¡ Ýó‹ 21e å¼ e†ì¼‚° Ï.5 iî‹ Ü„
²ŸÁ«õL ܬñ‚è Ý°‹ ªêô¬õ‚ 裇è.
A park is in the shape of a semicircle with radius 21 m. Find the cost of fencing it at the cost
of Rs. 5/m
A) Ï.500 B) Ï.520 C) Ï.530 D) Ï.540
²ŸÁ«õL ܬñ‚è
36r
ܬóõ†ìˆF¡ ²Ÿø÷¾ (Perimeter of semi circle) = 7

132
 TNPSC èí‚°

36x21
= 7
= 36x3
= 108e
1 e†ì¼‚° Ý°‹ ªêô¾ (Cost of 1 m) = Ï.5
108 e†ì¼‚° Ý°‹ ªêô¾ (Cost of 108 m) = 108x5 = Ï.540

21. 
2 ªê.e M†ìºœ÷ Í¡Á ï£íòƒèœ 塬ø å¡Á ªî£´ñ£Á ¬õ‚èŠð†ì£™
ÜõŸÁœ ܬìð´‹ ð°FJ¡ ðóŠ¬ð‚ 裇è.
Three coins each 2 cm in diameter are placed touching one another find the area enclosed by
them?
A) 0.151ªê.e2 B) 0.161ªê.e2
C) 0.171ªê.e2 D) 0.18ªê.e2
1 1
A B
1 1

1 1

ê˜è£˜ ä.ã.âv Üè£ìI

C

õ†ìˆF¡ Ýó‹(Radius of circle) = M†ì‹ = 2 = 1ªê.e

2 2
õ†ìƒè÷£™ ÅöŠð´‹ð°FJ¡ ðóŠ¹ (The area enclosed by circle)
= (êñð‚è ∆ABC - ¡ ðóŠ¹) (Area of a equilateral triangle ABC) -- 3 (õ†ì«è£ùŠ ð°FJ¡
ðóŠ¹) (Area of a sector)
= ( 23 a ) - 3 (366R xπr )
2 2

= ( 3 x2x2 ) -3 ( 360 x 7 x1x1 )

60 22
4
6
11
(
= 3 - 3 16 x 22
2 7
)
= 3 - 11
7
7 3 - 11
= 7
7x1.732 - 11
=
7
1.124
= 7
= 0.161 ªê.e2

22. 
õ†ì«è£íŠð°FJ¡ ðóŠð÷¾ 462 ê.ªê.e Ýó‹ 21ªê.e âQ™ ¬ñò‚«è£í‹
_______
The area of a sector of a circle of radius 21 cm is 462 cm2. Find its central angle
A) 60o B) 120o C) 180o D) 210o
õ†ì«è£íŠð°FJ¡ ðóŠ¹ (Area of the sector) = 462 ê.ªê.e

133
 TNPSC èí‚°

θ
360
xπr2 = 462
θ 22
360 x 7 x21x21 = 462
20
42 6 12060
θ 462x360x7
360 = 22x21x21
11 3 7
θ = 120o

23. ðóŠð÷¾ 225 ê.ªê.e ñŸÁ‹ M™L¡ c÷‹ 15 ªê.e ªè£‡ì õ†ì‚«è£íŠð°FJ¡
Ýó‹ 裇è.
Calculate the area of a sector whose area and arc length are 225 cm2 and 15 cm respectively
A) 15 ªê.e B) 18 ªê.e C) 21 ªê.e D) 30 ªê.e
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

õ†ì‚«è£íŠ ð°FJ¡ ðóŠ¹ (Area of the sector) = 225 ê.ªê.e

lr
2 = 225
lx15
2 = 225
225x2
M™L¡ c÷‹ (l)(length of arc) = 15
= 30 ªê.e

24. ªð£¼ˆ¶è:
1
1. ꣌ê¶óˆF¡ ðóŠ¹ & 2
(a+b)h
2. êñð‚è º‚«è£íˆF¡ ðóŠ¹ & bh
3. Þ¬íèóˆF¡ ðóŠ¹ & 3 a2
4
1
4. êKõèˆF¡ ðóŠ¹ & d xd
2 1 2
Match the following
1. Area of a Rhombus - 12 (a+b)h
2. Area of a equailateral triangle - bh
3. Area of a parallelogram - 3 a2
4
4. Area of a Trapezium - 1 d1xd2
2
A) 1234 B) 3214 C) 4321 D) 3421

Exercise Sums

1. å¼ õ†ìˆF¡ Ýó‹ 25% ÜFèK‚èŠð†ì£™ ðóŠ¹ ÜFèK‚°‹ êîiî‹

If the radius of a circle is increased by 25% then its area is increased by ........
(A) 50% (B) 25% (C) 56.25% (D) 46.25%
Sol. :
Ýó‹ ðóŠ¹
(Radius) (Area)
π (r2 ) : π (r2 )
π (10
Note:
2
) : π (12.5 )
2

Note25%
134
125
 π (r2 ) : π (r2 )
π (102 ) : π (12.52 ) TNPSC èí‚°

Note25%
125
10 x =100 : 156.25%
12.5
π (r ) : π (r2 )
2 10 0

π (102 ) : π (12.52 )
100 : 156.25%
Note25%
125
10 x
Increased 10056.25%
by : 156.25%
10 0

2. 7 e Ýóºœ÷ å¼ õ†ì õ®õ ¬ñî£ùˆ¬î ²ŸP ªõO¹ø‹ 7 e ÜèôˆF™ å¼

ð£¬î àœ÷¶ âQ™ ð£¬îJ¡ ðóŠð÷¾ -_____  π = 22 
 7 

A circular ground of radius 7 m has a path of width 7 m around it on its outside. The area of
the path is approximately equal to _____
(A) 154 ê.ªê.e. / sq.m. (B) 308 ê.ªê.e. / sq.m.
(C) 462 ê.ªê.e. / sq.m. (D) 616 ê.ªê.e. / sq.m.
Sol:
Dig:

ê˜è£˜ ä.ã.âv Üè£ìI

A2 A1

7m
7m

A1 = πr 2 Area of a big circle (ªðKò õ†ìˆF¡ ðóŠð÷¾)
22
= 22 x 14 x 14
= 7 x 14 x 14
7
= 616 m2
= 616
A 2 = πr 2 Area of a small circle (CPò õ†ìˆF¡ ðóŠð÷¾)
22
= 22 x 7 x 7
= 7 x 7 x7
7
= 154 m2
= 154
ð£¬îJ¡ ðóŠð÷¾ = 616 - 154
(The area of path) = 462 sq.m

3. 1.75 Ýó‹ ªè£‡ì å¼ ê‚èó‹ à¬ìò å¼ õ‡® 11 A.e. Éóˆ¬î èì‚è âˆî¬ù
²ŸÁèœ ²Ÿø «õ‡´‹ ?
The radius of a wheel is 1.75. How many revolutions will it make in travelling 11 km?
A) 10 B) 100 C) 1000 D) 10000
Sol. :

Circumference of a circle (õ†ìˆF¡ ²Ÿø÷¾) = 2πr

22 .25
135 =2x x 1.75
7
 TNPSC èí‚°
= 2πr
= 2πr
= 2πr 22 .25
= 2 x 22 x 1.75
.25
= 2 x 22 x 1.75
.25
=2x 7 7 x 1.75
= 44 x 70.25
= 44 x 0.25
=
= 44
11 xm0.25
= 11 m
= 11 m

11 A.e Éó‹ èì‚è ²ŸÁ‹ ²ŸÁèœ = 11 km

11 m
km converted to m
1000
11000
=
11 m [ 1 km = 1000m]
== 1000
1000 ²ŸÁèœ
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

4. å¼ õ†ìˆF¡ Ýó‹ 50% °¬ø‰î£™ Üî¡ ðóŠ¹ âˆî¬ù êîiî‹ °¬ø»‹?

If the radius of a circle decreased by 50%, find the % decrease in its area.
(A) 74% (B) 75% (C) 76% (D) 95%
Sol :
õ†ìˆF¡ Ýó‹ (radius) r = 10 cm â¡è
A1 = πr 22
A1 = πr
r 22 = 10 x 10 = 100
r = 10 x 10 = 100

50
Note: 10 5 x =5
100 2

õ†ìˆF¡ ÝóˆF™ 50% °¬ø‰î£™

r 2 = 5 x 5 = 25
MˆFò£ê‹ = 100 - 25 = 75
MˆFò£ê‹ % = 75%

5. 48 e Ýóñ£è‚ ªè£‡ì õ†ì õ®õŠ ̃è£M¡ ªõOŠ¹øˆF™ 4 e ÜèôˆF™

êñ„Yó£ù õ†ìŠð£¬î ܬñ‚èŠð´Aø¶. ÜŠð£¬îJ¡ ðóŠ¬ð 裇è.
A uniform circular path of width 4 m is laid out around a circular park of radius 48 m. Find
the area of the circular path.
(A) 1256 m2 / e2 (B) 1255 m2 / e2
(C) 400 m2 / e2 (D) 1254 m2 / e2
Sol:
A2 A1

48m
4m

A1 = πr 2 (ªðKò õ†ìˆF¡ ðóŠð÷¾)
22
= x 522
7
136
 TNPSC èí‚°

A 2 = πr 2 (CPò õ†ìˆF¡ ðóŠð÷¾)

22
= x 482
7
22
= 22 5222 - 48 22
2
( )
ð£¬îJ¡ ðóŠð÷¾ 7 52 -=482 522 - 482
= 22 ( )( )
7 522 - 4872
= 22
7 ( 52+48 )( 52-48 )
( )
= 22 22
= 22 = 7 ( 52+48 = )( 52-48 ) )( 52-48)
( 52+48
7
22 ( 52+48 )(
7 52-48 )
7 x 100 x 4
= 22
7 x 100=x22
= 22 4 x 100 x 4
= 7 x
= 3.14 100
400 4
x 7
7 x
= 3.14 x 400
2 = 3.14 x 400
= 1256 3.14 xm400
= 1256 m 22 = 1256 m 2
= 1256 m
Area of the circular path = 1256 m 2

6.
The length of a chain used as the boundary of a semicircular park is 72m. What is the area of
the park?
ܬó õ†ì õ®Mô£ù ̃è£M¡ «õLò£èŠ ðò¡ð´ˆîŠð†ì êƒALJ¡ c÷‹ 72 e

ê˜è£˜ ä.ã.âv Üè£ìI

âQ™ ̃è£M¡ ðóŠð÷¾ ò£¶ ?
(A) 77 e2 (B) 91 e2 (C) 126 e2 (D) 308 e2.
Sol:
Circumference of a semicircle
= π r(ܬóõ†ìˆF¡
+ 2r ²Ÿø÷¾)= π r + 2r
= π r + 2r π r + 2r = 72
π r + 2r = 72
π r + 2r = 72
r (π + 2 ) =72 r (π + 2 ) =72
r (π + 2 ) =72

 22 
r  22 + 2  = 72
r  22 7
22 + 2  = 72

rr  22 7 + 2 
 = 72
r  22 7 + 22  =
+14 = 72
72
 7 + 

 
r  7
22 + 14  = 72
r  22 7+ 14  = 72
rr  22
22 7+ + 14
14  = 72
r  36 77  = = 72
72

r  36 7 = 72 

r  36 7  = 72
rr  36
7  = 72
36
r  77 822  =
= 72
72
 7 82  7
7
r = 72 x
r = 72 82 x 36
82
77
rr = = 72
72
8 x 36
x 74
r = 72 x 36 36
4

r = 14 36 4
r = 14 4
4
rr = 14
r= = 14
14
π r2
Area if aπ semi
r 2 = circle
π r 2 (ܬóõ†ìˆF¡ ðóŠð÷¾) =
= 2
2 2 22 2

22 222 2 x 14 x 14
x 147 x x1414 x 14 = 7
= 7 = 2
2 2 = 22 x 14
= 22 x 14
= 22 x 14
2
= 308 m 2
= 308 m=2 308 m
137
 =
2
22 2
x 14 x 14
= 7
TNPSC èí‚° 2
= 22 x 14
2
= 308 m

7. If the radius of a circle is doubled, area is multiplied by

å¼ õ†ìˆF¡ Ýó‹ Þ¼ ñìƒè£ù£™, Üî¡ ðóŠð÷¬õ .............. Ý™ ªð¼‚è «õ‡´‹
A) 3 B) 2 C) 4 D) 8
Sol:
õ†ìˆF¡ ðóŠð÷¾ = π r 2
Ýó‹ Þ¼ ñìƒè£ù£™ = π ( 2r )
2

2
= π ( 2r )=2 π 4r 2
2
= π 4r = 4π r 2
2
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

= 4π r
2
2

ðóŠð÷¾ 4 Ý™ ªð¼‚è «õ‡´‹

8. If the area and arc length of the sector of a circle are 60 square metre and 20 metre respec-
tively. then find the diameter of the circle
å¼ õ†ì‚«è£íŠ ð°FJ¡ ðóŠ¹ 60 ê¶ó e†ì˜ ñŸÁ‹ M™L¡ c÷‹ 20 e†ì˜ âQ™
õ†ìˆF¡ M†ìˆ¬î‚ 裇è.
(A) 6 e†ì˜ (B) 12 e†ì˜ (C) 24 e†ì˜ (D) 36 e†ì˜
Sol:
õ†ì«è£íŠð°FJ¡ M™L¡ c÷‹ : õ†ì«è£íŠð°FJ¡ ðóŠ¹ = õ†ìˆF¡
²Ÿø÷¾ : õ†ìˆF¡ ðóŠð÷¾
20 : 60 = 2π r : π r 2
1 : 3 = 2:r
1 2
=
3 r
r =6
r=6
M†ì‹ = 2r
=2x6
= 12 m

9. Find the length of a chord which is at a distance of 15 cm from the centre of a circle of radius
25 cm.
å¼ õ†ìˆF¡ ¬ñòˆFL¼‰¶ 15 cm ÉóˆF™ å¼ ï£‡ ܬñ‰¶œ÷¶. õ†ìˆF¡
Ýó‹ 25 cm âQ™ ï£E¡ c÷‹ 裇è
(A) 45cm (B) 40 cm (C) 42cm (D) 50cm

r
cm o
25 15cm
A B C

138
 TNPSC èí‚°

By pythagoras theorem (Hî£èóv «îŸøˆF¡ð®)

2
OA
= 2
OB 22 + AB 22
OA
= 2
OB2
+ AB
25
= 2
15 + AB 2
25
= 1522 + AB 22
2522 − 152 = AB 2
25 − 15 = AB

AB 22
= 2522 − 1522
AB 2
= 252 − 152
AB
=
AB
= 2
= (2525
2522+− 15
−15)(25
1522 − 15)
AB 22= (25
= 252+−15)(25
152 − 15)
AB = (25
= 25 +−15)(25
15
AB = = (25
= (25
40 x10
40 x10 − 15)
+ 15)(25 − 15)
AB = + 15)(25 − 15)
AB = = (25 + 15)(25 − 15)
400xxx10
40 1010
AB
AB = = 40 400
40 x10
AB = 2040 x10
AB
AB = = 20400
400
AB = 400
AB
AB =
AC
AC = 20
2 400
20 x 20
2 x 20
AB
AC = 20
22 of
xx 20
AC == 20
AB
Length
Length of
the Chord = 40 cm
20
the Chord = 40 cm
AC
Length= 2 x 20
Length of the
AC = 2 ofx the Chord
20 Chord == 40
40 cm
cm
Length of the Chord = 40 cm
Length of the Chord = 40 cm

ê˜è£˜ ä.ã.âv Üè£ìI

10. The area of a circle is 220 cm2, then the area of the square inscribed in the circle is
å¼ õ†ìˆF¡ ðóŠð÷¾ 220 ê.ªê.e, âQ™ Üšõ†ìˆFŸ°œ ܬñ»‹ ê¶óˆF¡
ðóŠð÷¾
A) 120 ê.ªê.e B) 140ê.ªê.e C) 135 ê.ªê.e D) 250 ê.ªê.e
Sol:

Area of a circle (õ†ìˆF¡ ðóŠð÷¾) = π r 2
22 22
220 =
220 = 2222
22 rr 2
220
220 = = 22 7
7 rr 22
220 = 7 72 r
220
220 = rr7222
220 = rr 2
220
22 =
22
220 = r7 7
22
22 =r 22 7
7
10 22= r 22 7
10 = rr 2
10
10 = = r7
7
1022 = 7 7
rr 2 = 70
= 70
7
rr 2 =
2 70
= 70
rr === 70
70
70
rr =
= 7070
r = 70
ê¶óˆF¡ ͬôM†ì‹ = 2r
Diagonal of the Square = = 2 70

139
 TNPSC èí‚°

Diagonal (ͬôM†ì‹)
Side of a square (ê¶óˆF¡ ð‚è‹) =
2
2 70
35
= = 2 35
2
= a 22
Area of a square (ê¶óˆF¡ ðóŠð÷¾) = a 2
=a
(( ))
2

( )
2
= 2 35 2 = 2 35
= 2 35
= 4 x35 = 4 x35
== 44 xx3535
= 140cm 22 = 140cm 2
= 140cm
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

11. The area of a semi-circle of raidus 7 c.m. is

7 ªê.e Ýó‹ àœ÷ ܬó õ†ìˆF¡ ðóŠ¹
(A) 7 ªê.e.2 (B) 777 ªê.e.2 (C) 77 ªê.e.2 (D) 7777 ªê.e.2
Sol:
r=7
ܬó õ†ìˆF¡ ðóŠð÷¾ (Area of a semicircle)
π r2
=
2
11
22 7 x7
= x = 77cm 2
7 2

12. The length of the arc of a sector having central angle 90° and radius 7 cm is
õ†ì ¬ñòˆF™ °‹ «è£í‹ 90° Üî¡ Ýó‹ 7 ªêe âQ™, õ†ì M™L¡ c÷‹
(A) 22 ªêe (B) 44 ªêe (C) 11 ªêe (D) 33 ªêe
Sol:
r = 7 cm
²Ÿø ÷ ¾
Length of the arc (õ†ì M™L¡ c÷‹) =
4
2πr
=
4
11
22
2 x x 7
= 7
4
2

= 11 cm

13. Calculate the radius of a sector whose area and arc length are 60 cm2 and 20 cm respectively
å¼ õ†ì‚«è£íŠ ð°FJ¡ ðóŠ¹ 60 ªê.e.2 Üî¡ M™L¡ c÷‹ 20 ªê.e âQ™
Ýó‹ 裇è.
(A) 3 ªê.e (B) 6 ªê.e (C) 4 ªê.e (D) 5 ªê.e
Sol:
140
 TNPSC èí‚°

M™L¡ c÷‹ = 20 cm

2 π rA 10
2πrA = 20 cm
360 = 20 cm
360 10

A 10
= cm
360 πr
õ†ì‚«è£íŠð°FJ¡ ðóŠ¹ = 60 cm 2
πr 2 A
= 60 cm 2
360
10
πr2 x cm = 60 cm 2
πr
r x 10 = 60
60
r= = 6 cm
10

14. The circumference of a circular park is 264m. Find the area of the park

ê˜è£˜ ä.ã.âv Üè£ìI

õ†ìõ®õŠ ̃è£M¡ ²Ÿø÷¾ 264e. ̃è£M¡ ðóŠð÷¾ 裇è
A) 5544 ê.e B) 5444 ê.e C) 5554 ê.e D) 5534 ê.e
Sol:
õ†ìˆF¡ ²Ÿø÷¾ = 264 m
2
2 πr
πr =
= 264264
132
2 πr = 264 132
2
22 πr = 264
132
22 x r =132132
22
7 x r = 132
7 x
22 r = 132
7 x 12r66 = 132
7 126 7
7
rr == 132
132
12 x
x 7
12 x 22
6
r = 132 72
22
r = 132 x 22 2
rr = 42 22
2
= 42 2
r = 42
r = 42
õ†ìˆF¡ ðóŠð÷¾
= πr 2
22 6
= x 42 x 42
7
= 5544 cm 2

15. A Scooter wheel makes 50 revolutions to cover a distance of 88m then the radius of the
Wheel is
å¼ vÆìK¡ ê‚èóñ£ù¶ 50 º¿„ ²ŸÁ‚èO™ 88 e&¬ò èì‚Aø¶ âQ™ ܉î
vÆìK¡ ê‚èóˆF¡ Ýó‹
A) 24ªê.e B) 48 ªê.e C) 28ªê.e D) 12ªê.e
Sol:
Distance = ²ŸÁèO¡ â‡E‚¬è x ²Ÿø÷¾
Distance
Circuference =
²ŸÁèO¡ â‡E‚¬è
141
 TNPSC èí‚°

176
8800
2πr = (Convert to cm)
50
4
2 8
22 16
2 x x r = 176
7
r
=4
7
r = 28 cm
ê‚èóˆF¡ Ýó‹ = 28 cm

16. The diameter of a circle is 10cm. “P” is the point lying outside the circle. From that point
“P” two tangents are drawn to the circle. The length of each tangent is 12 cm. What is the
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

distance between “P” and the centre of the circle.

10 ªê.e M†ìºœ÷ õ†ìˆF¡ ªõO«ò “P” â¡ø ¹œO àœ÷¶. ¹œO “P”JL¼‰¶
õ†ìˆFŸ° Þó‡´ ªî£´«è£´èœ õ¬óòŠð†´œ÷¶. 嚪õ£¼ ªî£´«è£†®¡
c÷º‹ 12 ªê.e âQ™ õ†ìˆF¡ ¬ñòˆFŸ°‹ ¹œO “P”‚°‹ Þ¬ìŠð†ì ªî£¬ô¾
â¡ù?
(A) 12 ªê.e (B) 13 ªê.e (C) 15 ªê.e (D) 10 ªê.e
Sol:
P

12 cm

A 5 cm O 5 cm B

Hî£èóv «îŸøˆF¡ð®
P

12 cm

P 5 cm P

OP 2 = PB2 + OB2
= 122 + 52
= 144 + 25
OP 2 = 169
OP = 169
OP = 13cm

17. A sector of 1200, cut out from a circle, has an area of 9 3 sq.cm. Find the radius of the circle
7 3
å¼ õ†ìˆFL¼‰¶ ªõ†®ªò´‚èŠð†ì 1200 àœ÷ å¼ õ†ì‚ ÃÁ, 9 ê¶ó ªê.e
7
142
 TNPSC èí‚°

ðóŠð÷¾ à¬ìò¶. õ†ìˆF¡ Ýóˆ¬î‚ 裇è.

(A) 1 ªê.e (B) 2 ªê.e (C) 4 ªê.e (D) 3 ªê.e
Sol:
θ 3
θ o x πr 22 = 9 3
360 θ x πr =
θ o x πr 22 = 9 7733 9
360 θθ oo x πr 2 = 9 733
360
120 222 = 66
120oo xxx πr
360
360 πr
22 = 9
x r92277= 66
360 xx 22
360
120
120 227 xx rr 227= 66
7
66
360
120 x 227 x r2 = 7
66
77 xx rr 2 =
3
120
360 3 x 22 = 66
77
360
22 x =
360
22
360
3 2 7
7 66 77
3 x r2 =
3
33 xx rr 22 =
22
22 66
3 x r32 =
22 66
33 xx rr632 =
22 =
66
663
3
2 63 = 663
r32 = 66 x
r 2 = 66 6 x
3
63 33
22
rr 22 = = 66
6663 x 22
6 x 3322
rr 22 = = 66
66 xx 2222
r2 = 9 22 2
r2 = 9 22 2
2
rr 22== = 9
39cm 2
rr 2== 3 cm
= 99
rr = = 33 cm
cm
rr == 33 cm
cm

ê˜è£˜ ä.ã.âv Üè£ìI

18. Arc length of a circular sector is equal to the radius r of the sector. Area of that circular sector
is equal to
å¼ õ†ì‚ «è£íŠ ð°FJ¡ M™L¡ c÷ñ£ù¶ Üî¡ Ýó‹ r ‚° êññ£ù£™ Üšõ†ì‚
«è£íŠð°FJ¡ ðóŠð÷¾
π r2 r2
(A) r2 (B) (C) (D) 2r2
2
2 2
Answer: r
2
19. Area of circle is equal to the area of a rectangle having perimeter of 50 cm and the length is
more than its breadth by 3 cm. What is the diameter of the circle?
å¼ õ†ìˆF¡ ð󊹋, ªêšõèˆF¡ ²Ÿø÷¾‹ êññ£è àœ÷¶, ܬõ 50 ªê.e.
ªêšõèˆF¡ c÷‹ Üèôˆ¬î Mì 3 ªê.e. ÜFèñ£è àœ÷¶ âQ™ õ†ìˆF¡ M†ì‹
âšõ÷¾?
(A) 7 ªê.e (B) 21 ªê.e (C) 28 ªê.e. (D) 14 ªê.e.
Sol:
Üèô‹ = x â¡è
c÷‹ = x +3 â¡è
2 ( l + b) = 50
22 (( ll + + b)
b) = = 5050
2 ( lx ++3b)+=x50 ) = 50
22 (( lxx ++3 +3b)+ +=xx50)) =
= 50
50
22 (( l2x x++3 b)+=x=50
+3) ) = 50
50
222 ((( 2l2x xx++3 b)+=x=50
+3)
+3) 50
)50
= 50
224x(( +6 2xx x+3
=+3) + x=
50 =))50= 50
224x(( +6 2 x+3
=+3) +
50 x
= = 50
50
42x( +6 2 x =+3) 50= 50
42x( +6 =
=2x50+3) 50
- 6= 50
44xx +6 =
= 50 50= --5066
44xx +6 =
50 50
44xx +6 =
= = --5066
44
44
50
44xx = =11 44
50cm -6
44xxx==
= 44
50cm -6
x4xx=
==11 1144 cm
4xx==11 44cm
4xx==11 44cm
x = 11 cm
x = 11 cm
143
 TNPSC èí‚°

Üèô‹ = 11 cm
c÷‹ = x + 3 = 11 + 3 = 14â¡è

ªêšõèˆF¡ ðóŠð÷¾ = l x b
= 14 cm x 11 cm
= 154 cm 2
õ†ìˆF¡ ðóŠð÷¾ = 154
πr 222 = 154
πr = 154
222 = 154
πr
22 x r 222 = 154
7 x r 2 = 154
22
7 x r 7 = 154
7 14
7
7 7
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

r 222 = 154
14
14 x 7
r = 154 14 7 x 22
7

14 7 7
r22 = 154 x x22 2
2
r 222 = 49 2222 2
2

r = 49 2
rr = 2
7
==749
r=7
M†ì‹ = 2 x r
=2x7
= 14 cm

20. Calculate the area of a sector whose radius and are length arc 6cm and 20cm respectively
Ýó‹ 6 ªê.e M™L¡ c÷‹ 20 ªê.e, ªè£‡´œ÷ õ†ì‚ «è£íŠ ð°FJ¡ ðóŠ¬ð‚
裇è
A) 120 ªê.e2 B) 80 ªê.e2 C) 60 ªê.e2 D) 100 ªê.e2
Sol:
P Q
m
O 6c

20 cm

lr
«è£íŠð°FJ¡ ðóŠ¹ =
2
Area of arc 3

20 x 6
=
2
= 60cm 2

21.
The arc length of a sector is 66 cm and the central angle is 300 . Find the Radius
M™L¡ c÷‹ 66cm ñŸÁ‹ ¬ñò‚«è£í‹ 300 ªè£‡ì õ†ì‚ «è£íŠ ð°FJ¡ Ýó‹
裇
A) 166 ªê.e B) 140 ªê.e C) 122 ªê.e D) 126 ªê.e
õ†ì‚«è£íŠð°FJ¡ c÷‹
l = rθ

144
 TNPSC èí‚°
l = rθ
l = rθ
66 = r x 30oo
66 = r x 30
π
66 = r x π
66 = r x 6
6
2
22
6
66 = r x 7
6
18 2
36 = r x
7
18 x 7 = r
126 = r
r = 126cm

22. Find the area of a semi circle of radius 28 cm

Ýó‹ 28 ªê.e à¬ìò ܬó õ†ìˆF¡ ðóŠð÷¾ 裇
A. 144 ê.ªê.e B. 1220 ê.ªê.e C. 238 ê.ªê.e D. 1232 ê.ªê.e

ê˜è£˜ ä.ã.âv Üè£ìI

Sol: 2
πr
= πr 22 πr 2
= πr2 ðóŠð÷¾
ܬóõ†ìˆF¡ =
= 2 2
22
2 4
22 x 28 4 x 28 22 4

= 227 x 28 4
x 28 x 28 x 28
7 x 28 x 28 = 7
= 2
= 711 2 2
2
22
11 x 4 x 28 11
= 2211
x 4 x 28 22 x 4 x 28
= 22 x 24 x 28 =
=
= 1232 cm 2 2 2
2
= 1232 cm 22 = 1232 cm 2
= 1232 cm

23. Find the radius of a sector whose arc length and area are 27.5 cm and 618.75 cm 2 respective-
ly
M™L¡ c÷‹ 27.5 ªê.e. ðóŠð÷¾ 618.75 ê.ªê.e ªè£‡ì õ†ì‚ «è£íŠ ð°FJ¡
Ýó‹ 裇.
A) 35 ªê.e B) 25 ªê.e C) 45 ªê.e D) 55 ªê.e
Sol:
θ
o
x πr 2 = 618.75------(i)
360
θ
x 2πr=27.5------(ii)
360o
(i)%(ii)
(i) ÷ (ii)
θ
x πr2
360 o
618.75
=
θ 27.5
o
x2π r
36

145
 TNPSC èí‚°

r 618.75
=
2 27.5
618.75
r= x2
27.5
r = 45

24. Calculate the perimeter of a quadrant of a circle of radius 21 cm.

A) 65 cm B) 75 cm C) 44 cm D) 88 cm
21 ªê.e Ýóºœ÷ 裙õ†ìŠ ð°FJ¡ ²Ÿø÷¾ 裇
A) 65 ªê.e B) 75 ªê.e C) 44 ªê.e D) 88 ªê.e
Sol:
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

21

21

Perimeter of a Quadrant =  π +2  r
2 
 11 
22
= +2  x 21
7x 2 
 
 11+14  3
= = x 21 == 25 x 3 75
 7 

25. Three equal circles of radius 3cm touch one another in outside .find the area enclosed by
them
3 ªê.e Ýóºœ÷ Í¡Á õ†ìƒèœ 塬øªò£¡Á ªõO«ò ªî£´‹«ð£¶ ÜõŸø£™
ÅöŠð´‹ ð°FJ¡ ðóŠ¹
A) 10.88 ªêe2 B) 1.45 ªêe2 C) 6.11 ªêe2 D) 29.73 ªêe2
Sol:

A A

= Area of equilateral triangle ABC - 3 times area of the sector

3 2 θ
= a -3 o
x πr 2
4 360
11
3 3 3 60 22
= x 6 x 6 - 3 x x x3x3
4 360 7
2 6
2

146
 TNPSC èí‚°

99
=9 3 -
7
= 15359 - 14.14
= 1.45 cm 2

26. The perimeter and the area of a semicircle of radius 14 cm is

14 ªê.e Ýóºœ÷ ܬó õ†ìˆF¡ ²Ÿø÷¾ ñŸÁ‹ ðóŠð÷¾ º¬ø«ò
(A) 36; 308 (B) 72; 308 (C) 308; 36 (D) 308; 72
Sol:
14 cm

ܬóõ†ìˆF¡ ²Ÿø÷¾ = P = ( π+2 ) r

 22 
= + 2  x 14
 7 
36
= x 14 = 72 cm
7

ê˜è£˜ ä.ã.âv Üè£ìI

P = 72 cm
πr 2
ܬóõ†ìˆF¡ ðóŠð÷¾ =
2
11
22 2
x 14 x 14
= 7
2
= 22 x 14
= 308 cm 2

27.
The area of a circular field is 13.86 hectares. The cost of fencing it at the rate of 20 paise per
metre is
å¼ õ†ì õ®õ ªõOJ¡ ðóŠð÷¾ 13.86 ã‚è˜. âQ™, å¼ e†ì˜‚° 20 ¬ðê£ iî‹
Ü¬î ²ŸP «õL-«ð£ì Ý°‹ M¬ô
(A) Ï.277.20 (B) Ï. 264 (C) Ï. 324 (D) Ï. 198
Sol:
ðóŠð÷¾ = 13.86 ã‚è˜
Convert to m 2 [ 1 hectare = 1000 m2]
2
πr
πr 22 =
= 138600
138600
πr 2 = 138600
6300
πr2 2 = 138600
6300
12600 7
rrπr2 = =
12600
138600
138600
6300
x 7
2 = 138600 x 7
12600
22
6300
r 2 = 138600 12600
6300 x 72
22
r 2 = 138600 12600 x 22
72
rrr 22 = = 6300x7
138600 x
= 6300x7
22
2

r 22 = 6300x7 22
2
2
rr 2= = 6300x7
44100
rr = = 44100
6300x7
rr == 44100
210m
= 210m
44100
r = 210m
44100
r = 210m
r = 210m

147
 TNPSC èí‚°

²Ÿø÷¾
22 30
2πr = 2 x x 210
7
= 1320
å¼ «õL «ð£ì Ý°‹ M¬ô
20
= 132 0 x
10 0
= Rs. 264

28.
A wheel makes 20 revolutions to cover a distance of 66 m. Then the diameter of the wheel is
_______ m
å¼ ñA؉F¡ ê‚èó‹ 66 e ªî£¬ô¾ èì‚è 20 ²ŸÁèœ ²ŸPù£™ Ü„ê‚èóˆF¡
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

M†ì‹ ........... - e Ý°‹

(A) 1.05 (B) 1.04 (C) 1.03 (D) 1.02
Sol:
Distance
Circumference= --------------------------------
²ŸÁèO¡ â‡E‚¬è
33
66
2πr =
20
10

22 33
xr= 2x
7 10
44 33
xr=
7 10
3
33 7 21
r= x =
10 44 40
4

r = 0.525
r = 0.525 m
M†ì‹
d = 2 x 0.525
dd==1.05
1.05 m

29. The perimeter of semicircle with radius 28 cm is

Ýó‹ 28 ªê.e à¬ìò ܬóõ†ìˆF¡ ²Ÿø÷õ£ù¶
(A) 88 cm (B) 116 cm (C) 144 cm (D) 56 cm
Sol:
ܬóõ†ìˆF¡ ²Ÿø÷¾ P = ( π+2 ) r
 22 
= + 2  x 28
 7 
 22+14  4
=  x 28
 7 
= 36 x 4 = 144 cm
148
 TNPSC èí‚°

30. The radius of a circle is increased by 1% what is the increase percentage in its area?
å¼ õ†ìˆF¡ Ýó‹ 1% àò¼Aø¶ Üšõ†ìˆF¡ àò˜‰¶œ÷ ðóŠ¬ð êîiîˆF™
ÃÁ?
(A) 1% (B) 1.1% (C) 2% (D) 2.01%
Sol:
a=1
 a2 
Percentage increase in area =  2a + %
 100 
 12 
= 2 x 1 + 
 100 
= ( 2 + 0.01)
= 2.01%

31. The ratio of the circumference of two circles is 2:3. What is the ratio of their areas?

ê˜è£˜ ä.ã.âv Üè£ìI

Þ¼ õ†ìƒèO¡ ²Ÿø÷¾èO¡ MAî‹ 2:3 âQ™ Üî¡ ðóŠ¹èO¡ MAî‹ ò£¶?
(A) 2:3 (B) 4:9 (C) 3:2 (D) 9:4
Sol:
Þó‡´ õ†ìƒèO¡ Ýóƒèœ º¬ø«ò r1 and r2
õ†ìˆF¡ ²Ÿø÷¾ = 2πr

2πr1 2
=
2πr2 3
r12 4
2
=
r2 9
= 4 : 9

32. A silver wire when bent in the form of square encloses an area of 121 sq.cm. If the same wire
is bent in the form of a circle. Find the radius of Circle.
å¼ ªõœO è‹H¬ò õ¬÷ˆ¶ å¼ ê¶óñ£è ñ£ŸøŠð´Aø¶. ê¶óˆF¡ ðóŠ¹ 121 ê.ªêe.
Ü«î è‹H¬ò õ¬÷ˆ¶ å¼ õ†ìñ£è ñ£ŸPù£™ ܉î õ†ìˆF¡ Ýó‹ â¡ù?
A. 11 ªêe B. 7 ªêe C. 3.5 ªêe D. 14 ªêe
Sol:
ê¶óˆF¡ ð‚è‹ = area
= 121
= 11cm
ê¶óˆF¡ ²Ÿø÷¾ = 4a
= 4 x 11
= 44cm
õ†ìˆF¡ ²Ÿø÷¾ 2πr=44
2 2
22 4 149
2 x x r = 44
 TNPSC èí‚°
2πr=44
2πr=44
2 2
2
22 42
22 xx 22 xx rr =
4
= 44
44
77
rr
=1
77 = 1
rr = = 77
r = 7 cm

33. Calculate the area of a quadrant of a circle of radius 21 cm?

21 ªê.e Ýóºœ÷ 裙õ†ìŠ ð°FJ¡ ðóŠð÷¾ 裇.
(A) 346.5 ê.ªê.e (B) 322.7 ê.ªê.e
(C) 308.8 ê.ªê.e (D) 288.7 ê.ªê.e
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

Sol:
2
裙õ†ìˆF¡ ðóŠð÷¾ = πr
4
11
22 3
x 21 x 21
= 7
4
2

33 x 21
=
2
= 346.5cm 2

34. The area of a circle ie 154 sq.cm. What is the circumference of a circle?
õ†ìˆF¡ ðóŠ¹ 154 ê¶ó ªê.e âQ™ õ†ìˆF¡ ²ŸÁ õ†ìˆF¡ Ü÷¾ â¡ù?
(A) 28 ªê.e (B) 44 ªê.e (C) 88 ªê.e (D) 108 ªê.e
Sol:
õ†ìˆF¡ ðóŠð÷¾ πr 2 = 154
2 7
22 14
x r 2 = 154
7
2
r =7x7
r=7
²Ÿø÷¾ = 2πr
22
=2x x 7
7
= 44 cm

35. The radius and length of arc of a sector are 10 cm and 15 cm respectively. Find its perimeter
å¼ õ†ì‚«è£íŠ ð°FJ¡ M™L¡ c÷‹ 15 ªê.e ñŸÁ‹ Ýó‹ 10 ªê.e âQ™ Üî¡
²Ÿø÷¬õ‚ 裇è
(A) 15 ªê.e (B) 35 ªê.e (C) 25 ªê.e (D) 30 ªê.e
Sol:

150
 TNPSC èí‚°

10 cm

P Q
15 cm

²Ÿø÷¾ = l + 2r
= (15 + 2 x 10 )
= (15 + 20 )
= 35 cm

36.
The area of a sector of a circle of radius 21 cm is 231 cm. Find its central angle.
Ýó‹ 21 ªê.e ñŸÁ‹ ðóŠð÷¾ 231 ªê.e ªè£‡´œ÷ õ†ì«è£íŠ ð°FJ¡
¬ñò‚«è£íˆ¬î 裇è.
(A) 45° (B) 30°=  a  2 (C) 60° (D) 120°
 360  πr
Sol:  

ê˜è£˜ ä.ã.âv Üè£ìI

 2a  2
õ†ì «è£íŠð°FJ¡ ðóŠð÷¾ 21 a =  22  πr3
231 = x 360 x 21 x 21
360 a 7 22 2 2
21 231 = 120 21 x a x 32221 x 21 60 3
360231
120 60= 7 x x 21 x 21
1= a 360 7
120
60 60

a=60
o a
1=
60
a=60o
37. In a circular path, the radii of 2 concentric circles are 56 m and 49 m. Find the area of the
circular path
å¼ õ†ìŠð£¬îJ¡ Þ¼ ªð£¶¬ñò õ†ìƒèœ Ýó‹ 55 e ñŸÁ‹ 49 e âQ™ õ†ìŠ
ð£¬îJ¡ ð󊹂 裇è
(A) 3210 e2 (B) 2310 e2 (C) 3120 e2 (D) 2130 e2
Sol:

r
49 m

R
56 m

R = 56 r = 49
ðóŠð÷¾
= π ( R 2 -r 2 )
22
=
7
( 562 - 492 )

22
= x ( 56 + 49 ) x ( 56 - 49 )
7

151
 TNPSC èí‚°

22
= x 105 x 7
7
= 2310 m 2

38. The radius of a cart wheel is 35 cm. How many revolution does it make in travelling a distance
of 154 m
ñ£†´ õ‡®„ ê‚èóˆF¡ Ýó‹ 35 ªê.e. ܶ 154 e ªî£¬ô¾ èì‰î£™, Ü„ê‚èó‹
âˆî¬ù º¿„²ŸÁèœ ²ŸPJ¼‚°‹?
(A) 70 (B) 189 (C) 119 (D) 86
Sol:
õ†ìˆF¡ ²Ÿø÷¾ = 2πr
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

22 5
=2x x 35
7
= 220 cm
²öŸCèO¡ â‡E‚¬è = 154m (convert into cm)
220
70
140
1540 0
=
22 0
2

= 70

39. If the circle with diameter 4 cm is inside of the circle with radius 10 cm. Then the Maximum
possible distance between the centres is
4 cm M†ìˆ¬î ªè£‡ì õ†ì‹ 10 cm Ýóº¬ìò õ†ìˆFÂœ àœ÷ «ð£¶ Þ¼
¬ñòƒèÀ‚° Þ¬ì«ò Þ¼‚è Þò½‹ ÜFèð†ê Éó‹ â¡ù?
(A) 6 (B) 7 (C) 8 (D) 9
Sol:
cm
10

4 cm

ªõO õ†ìˆF¡ Ýó‹ = 10
àœ õ†ìˆF¡ Ýó‹ = 2

= 10 - 2 = 8

40. A silver wire when bent in the form of a square encloses an area of 484 sqm. If the same wire
22
is bent in the form of a circle, then find the diameter of the circle. (use π = )
7
å¼ ªõœO‚ è‹H õ¬÷‚èŠð†´ ê¶óñ£è ñ£ŸÁ‹ «ð£¶. Üîù£™ ܬìð´‹ ð°FJ¡
ðóŠ¹ 484 ê.e. Ü«î ªõœO‚ è‹H¬ò õ†ìñ£è õ¬÷‚èŠð´Aø¶ âQ™ õ†ìˆF¡
22
M†ì‹ â¡ù? ( π = âù‚ªè£œè)
7

152
 TNPSC èí‚°

(A) 14 e (B) 28 e (C) 24 e (D) 7 e

Sol:
ê¶óˆF¡ ð‚è‹
= Area
= 484
= 22 cm
ê¶óˆF¡ ²Ÿø÷¾
= 4a
= 4 x 22
= 88 cm
õ†ìˆF¡ ²Ÿø÷¾
2πr= 88
2
2 4
22 8
2 x x r = 88
7
r
=2
7

ê˜è£˜ ä.ã.âv Üè£ìI

r=2x7
r = 14
M†ì‹
d = 2r = 2 x 14
d = 28m

41. If the diameter of a bicycle wheel is 63 cm, then the distance covered by its wheel in 20 revo-
lutions is
å¼ IFõ‡®„ ê‚èóˆF¡ M†ì‹ 63 ªê.e âQ™, ܶ 20 ²ŸÁèœ ²ŸPù£™ èì‚°‹
ªî£¬ôM¡ Ü÷¾
(A) 50 e 28 ªê.e. (B) 43 e 34 ªê.e.
(C) 51 e 30 ªê.e. (D) 39 e 60 ªê.e.
Sol:
²Ÿø÷¾ = 2πr
9
22 63
= 2 x x
7 2
= 198
ªî£¬ô¾ = ²Ÿø÷¾ x ²ŸÁèO¡ â‡E‚¬è

= 198 x 20
= 3960 cm
= 39m 60cm
22
42. Area of semi circle whose diameter 0.14 cm is (Take π = )
7

153
 TNPSC èí‚°

22
M†ì‹ 0.14 ªê.e ªè£‡ì ܬó õ†ìˆF¡ ðóŠð÷¾ ( π = - âù‚ ªè£œè)
7
(A) 0.0077 cm 2
(B) 0.77 cm
2
(C) 0.077 cm2
(D) 0.7 cm2
Sol:
ܬóõ†ìˆF¡ ðóŠð÷¾ πr 2 d = 0.14, r = 0.07
=
2
11 0.01
22 0.07 x0.07
= x
7 2
= 0.007693
2
= 0.0077 cm
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

43. The inner circumference of a circular race track, 14 m wide, is 440 m. Find the radius of the
outer circle
14 e†ì˜ Üèôº¬ìò å¼ õ†ìñ£ù æ´ð£¬îJ¡ àœ ²Ÿø÷¾ 440 e†ì˜ Ý°‹.
ªõO õ†ìˆF¡ Ýóˆ¬î‚ 裇è
(A) 85 e (B) 84 e (C) 70 e (D) 80 e
Sol:

R

àœ õ†ìˆF¡ ²Ÿø÷¾
= 2πr
22
=2x xr
7
10 44
440 = xr
7
r = 70

ªõO õ†ìˆF¡ Ýó‹

= 70 + 14
= 84 m

44. Find the perimeter of a square whose area is equal to that of a circle with perimeter 2πr
2πr ²Ÿø÷¾ àœ÷ õ†ìˆF¡ ðóŠð÷¾‹ å¼ ê¶óˆF¡ ðóŠð÷¾‹ êñ‹ âQ™ ê¶óˆF¡
²Ÿø÷¾ - â¡ù?
(A) 2πr (B) π r (C) 4 π (D) 4 π r
Sol:
Given,
Perimeter of a circle = 2π r
It area = π r 2 radius = r

154
 TNPSC èí‚°

area of a square = a 2
a = π r
2 2

a = π r
Perimeter of square = 4 a
4 π r
4 π r

45. Find the perimeter of a sector whose radius and central angle are 18 cm and 210° respective-
ly.
Ýó‹ 18 ªê.e ñŸÁ‹ ¬ñò‚ «è£í‹ 210° âù‚ ªè£‡ì õ†ì «è£íŠ ð°FJ¡
²Ÿø÷¬õ‚ 裇è.
(A) 120 ªê.e (B) 110 ªê.e (C) 102 ªê.e (D) 108 ªê.e
Sol:
õ†ì «è£íŠð°FJ¡ ²Ÿø÷¾
θ
= x 2πr + 2r

ê˜è£˜ ä.ã.âv Üè£ìI

360o
3 11
210 22 6
= x 2 x x 18 + 2 x 18
360 7
12
6
3

= 66 + 36
=102 cm

46. In a circle of radius 10 cm, an arc subtends an angle of 90º at the centre. Find the area of
major sector.
10 ªê.e. Ýóºœ÷ å¼ õ†ìˆF™, Üî¡ ¬ñòˆF™ 90º «è£íˆ¬î å¼ õ†ìM™
à¼õ£‚°Aø¶ âQ™ IèŠ ªðKò õ†ì‚ «è£íŠð°FJ¡ ðóŠ¬ð‚ 裇è.
1650 1650
(A) 3 ê.ªê.e (B) 9 ê.ªê.e
1650 1650
(C) 11 ê.ªê.e (D) 7 ê.ªê.e

θ
õ†ì «è£íŠð°FJ¡ ðóŠ¹ = θo x πr 2 2
θ = 360 o x πr
= o
x πr 2
360 11
360
90 2211 5
= 90 x 11 22 x 105 x 10
90 22 5 = 360 x 7 x 10 x 10
= x x 10 x 10360
4 7
360 7 24
12
4 1
2
550
1
= 550
550 = 7
= 7
7
= õ†ìˆF¡ ðóŠð÷¾ & õ†ì «è£íŠð°FJ¡ ðóŠ¹
155
 TNPSC èí‚°

550
= πr 22 - 550
= πr 22 - 550
550
7
==πr
πr -- 7
22 77 x 10 - 550
= 22 x 10 550
= 22
722 x 10 x 10 - 550550
7
== 7 xx10 10xx1010-- 7
2200
77 - 550 77
= 2200 550
= 22007 - 550
2200 550
7
== 7 -- 7
165077 cm 277
= 1650 2
= 16507 cm 22
1650
== 7 cm cm
77
47. If the cost of fencing a circular path at the rate of Rs. 5 per metre is Rs. 1100, then the radius
of the park is
(A) 35 m (B) 7 m (C) 55 m (D) 11 m
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

å¼ õ†ì õ®õ ̃è£MŸ° «õL «ð£ì å¼ e†ì¼‚° Ï. 5 iî‹ Ï. 1100 Ý°‹

âQ™, ̃è£M¡ Ýó‹ â¡ð¶
(A) 35 e (B) 7 e (C) 55 e (D) 11 e
²Ÿø÷¾
220
2πr x 5 = 1100
5
22 10
2 x x r = 220
7
r
=5
7
r=5x7
r = 35 m

48. Calculate the area of a sector whose diameter and arc length are 30cm and 26cm respectively.
M†ì‹ 30 ªê.e, M™L¡ c÷‹ 26 ªê.e ªè£‡´œ÷ õ†ì «è£íŠ ð°FJ¡ ðóŠ¹
裇è,
(A) 195 ê.ªê.e (B) 175 ê.ªê.e
(C) 165 ê.ªê.e (D) 185 ê.ªê.e
Sol:
õ†ì «è£íŠð°FJ¡ c÷‹
θ
= x 2πr
360o
11
θ 22 3
26 = x 2 x x 15
360 7
90
18

θ=99.3630
θ
= x πr 2 θ
360
õ†ì o
«è£íŠð°FJ¡ ðóŠ¹ = x πr 2
11 1
360o
99.3630 22 3 5 11 1

= x x 15 x 15 99.3630 22 3 5
360 7 = x x 15 x 15
180 360 7
36
12
180
36
4 12
4
= 195 cm 2 2
= 195 cm156
 = x πr
360o
11 1
99.3630 22 3 5
= x x 15 x 15
360 7
180 TNPSC èí‚°
36
12
4

2
= 195 cm

49. The ratio of the area of a circle to the area of its semicircle is
å¼ õ†ìˆF¡ ðóŠð÷MŸ°‹ Üî¡ Ü¬ó õ†ìˆF¡ ðóŠð÷MŸ°‹ Þ¬ì«ò»œ÷
MAî‹
(A) 1:2 (B) 2:1 (C) 4:1 (D) 1:4
Sol:
The area of the circle = π r 2

1
area of semi circle = π r 2
2
The ratio of the area of a circle to the area of semicircle
π r2 2
=
1 2 1
πr
2
2:1

ê˜è£˜ ä.ã.âv Üè£ìI

50. å¼ Ì‰«î£†ì‹ ꣌ ê¶ó õ®M™ àœ÷¶. Üî¡ Í¬ô M†ìƒèœ 18 e, 25 e
̉«î£†ìˆF¡ ðóŠð÷¾ 裇è.
A flower garden is in the shape of a rhombus. The length of its diagonals are 18 m and 25 m.
find the area of the flower garden.
(Ü) 450 e2 / m2 (Ý) 225 e2 / m2
(Þ) 324 e2 / m2 (ß) 18 e2 / m2
Sol:
d1 = 18m, d 2 = 25m

Area of Rhombus = d1 x d 2
2
9
18 x 25
=
2
= 225 m 2

51. å¼ ê£Œê¶óˆF¡ ð‚èˆF¡ c÷‹ 5 e, Üî¡ å¼ Í¬ôM†ìˆF¡ c÷‹ 8 e âQ™

Üî¡ ñŸªø£¼ ͬôM†ìˆF¡ c÷‹ ò£¶?
The length of side of a rhombus is 5 m and one of its diagonal is 8 m. Then what is the length
of other diagonal ?
(A) 5m (B) 7 m (C) 6 m (D) 8 m
Sol:
c÷‹ = 5 e
ͬôM†ìˆF¡ c÷‹ = 8 e
Using Pythagoras therom

157
 TNPSC èí‚°

2 2
d  d 
a 22 =  d11  2 +  d 22  2
a =  2  +  2 
 2 2  2 
 44 2 2
8 d 
522 =  8  +  d 22  2
5 =  2  +  2 
 2   2 
 
2
 d2 
25 - 16 =  
 2 
d2
9=
2
d2
3=
2
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

d2 = 3 x 2 = 6m
Other Diagonal = 6 m

52. å¼ ê£Œ ê¶óˆF¡ ͬôM†ì Ü÷¾èœ 20 ªê.e, 12 ªê.e âQ™ Üî¬ìò

ðóŠð÷¾
The area of the rhombus whose diagonals are 20cm, 12cm is
(Ü) 120 ªê.e2 / cm2 (Ý) 200 ªê.e2 / cm2
(Þ) 400 ªê.e2 / cm2 (ß) 600 ªê.e2 / cm2
Sol:
d1 = 20 cm, d 2 = 12 cm

Area of Rhombus = d1 x d 2
2
10
20 x 12
=
2
= 120 cm 2

53. ꣌ê¶óˆF¡ ðóŠ¹ =

Area of Rhombus =
(A) ͬôM†ìƒèO¡ ªð¼‚è™ / Product of diagonal

1 1
(B) (ͬôM†ìƒèO¡ ôî™) / (sum of diagonals)
2 2

1 1
(C) × Í¬ôM†ìƒèO¡ ªð¼‚è™ / × product of diagonals
2 2
1
(D) ͬôM†ìƒèO¡ Ã´î™ / Sum of diagonals x Product of digonals
2

54. W›‚è‡ì ßÁèÀœ ªñŒò£ù ßÁ â¶/â¬õ?

(i) â™ô£ ê¶óƒèÀ‹ ꣌ê¶óƒèœ Ý°‹
(ii) â™ô£ Þ¬íèóƒèÀ‹ ªêšõèƒèœ Ý°‹
158
 TNPSC èí‚°

(iii) â™ô£ Þ¬íèóƒèÀ‹ èóƒèœ Ý°‹

(iv) â™ô£ êKõèƒèÀ‹ ꣌ê¶óƒèœ Ý°‹
Which of the following statement(s) is/are true statement(s)?
(i) All squares are rhombuses also
(ii) All parallelograms are rectangles also
(iii) All parallelograms are quadrilaterals also
(iv) All trapeziums are rhombuses also
(A) (i) only (B) (ii), (iii) (C) (i), (ii), (iii) (D) (i), (iii)

55. å¼ i†´ ñ¬ùò£ù¶ èó õ®M™ àœ÷¶. Üî¡ å¼ Í¬ô M†ìˆF¡ c÷‹
100 e. ͬôM†ìˆF¡ âFªóF˜ ð‚èƒèO™ àœ÷ º¬ùèœ Þó‡´‹ ͬô M†ìˆ-
FL¼‰¶ 50e ªî£¬ôM™ Þ¼ŠH¡. ñ¬ùJ¡ ðóŠ¹ ò£¶ ?
A plot of land is in the form of a quadrilateral where one of its diagonals is 100m long. If two
vertices on either side of this diagonals are 50 m away from the diagonal. Find the area of the
plot of land.
(A) 5000 m2 / e2 (B) 1000 m2 / e2

ê˜è£˜ ä.ã.âv Üè£ìI

(C) 10000 m2 / e2 (D) 500m2 / e2
Sol:
1
Area of Quadrilateral = d (h1 + h1 )
2
1
= x 100 (50 + 50)
2
= 50 x 100
= 5000 m 2

56. å¼ ï£ŸèóˆF¡ «è£íƒèœ ∠ A, ∠ B, ∠ C, ∠ D ÝAòù 2 : 3 : 5 : 8 â¡ø

MAîˆF™ ܬñ‰¶œ÷ù âQ™ ∠ D-¡ ñFŠ¹
If angles ∠A, ∠B, ∠C , ∠D of a quadrilateral are in the ratios 2: 3 :5: 8 then ∠D equals .
A) 800 B) 1000
C) 1600 D) 1800
Sol:
A = 2x B = 3x C = 5x D = 8x
2x + 3x + 5x + 8x = 360o

18x = 360o
20
360
x=
18
x = 20
∠A=40o ∠B=60o ∠C=100o ∠D=160o
Ans : 1600

159
 TNPSC èí‚°

57.
å¼ ê£Œ¾ èóˆF¡ ðóŠð÷¾ 1440e2 Þó‡´ Þ¬í‚ «è£´‚° Þ¬ì«ò àœ÷
ªêƒ°ˆ¶ «è£†®¡ àòó‹ 24e. «ñ½‹ Þó‡´ Þ¬í «è£´èO¡ MAî‹ 5 : 3
âQ™ ªðKò Þ¬í«è£†®¡ c÷ñ£ù¶
The area of field is the shape of trapezium measures 1440 m2 . The perpendicular distance be-
tween parallel sides is 24m. If the ratio of parallel sides is 5:3, then the length of longer parallel
side is
A) 45 m B) 60 m C) 75m D) 120 m
Sol:
꣌¾ èóˆF¡ ðóŠð÷¾ = 1440 m2
(Trapezium)
h = 24 m
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

ð‚è‹ 5x and 3x
èóˆF¡ ðóŠð÷¾ = ½(Sum of side) x height

1
1440 = ( 5x + 3x ) x24
2
1440
= 8x
12
120
x= = 15
8
Longer side = 5x = 5 x 15
x =x 75 = 75cm
m

58. å¼ ï£ŸèóˆF¡ c÷‹ 20 ªê.e, âF˜ ªêƒ°ˆ¶ àòó‹ 7 ªê.e- ñŸÁ‹ 10 ªê.e-
Þ¼‰î£™ ÜõŸP¡ ðóŠð÷¾ 裇è.
Calculate the area of the quadrilateral whose length is 20 cm and perpendicular distance to the
diagonal from opposite vertices be 7 cm and 10 cm
(A) 160 cm2 (B) 110 cm2 (C) 100 cm2 (D) 170 cm2
Sol:
1
Area of the quadrilateral = d (h1 + h1 )
2
1
= x 20 (7+10)
2
= 10 x 17
= 170 cm 2

59. ͬôM†ìƒèœ 6 ªê.e. ñŸÁ‹ 8 ªê.e. ªè£‡ì ꣌ ê¶óˆF¡ ðóŠ¹

Area of rhombus with diagonals as 6 cm and 8 cm is
(A) 12 cm2 (B) 18 cm2 (C) 24 cm2 (D) 36 cm2
Sol:
d1x d 2
Area of the Rhombus =
2

160
 TNPSC èí‚°

4
6x 8
= = 24cm 2
2

60. å¼ êKõèˆF¡ Þ¼ Þ¬í ð‚èƒèO¡ MˆFò£ê‹ 4 ªê.e. «ñ½‹ ÞšM¼

Þ¬í ð‚èƒèÀ‚A¬ì«ò àœ÷ ªêƒ°ˆ¶ Éó‹ 19 ªê.e. êKõèˆF¡
ðóŠð÷¾ 475 ªê.e 2
âQ™ Üî¡ Þ¼ Þ¬í ð‚èƒèO¡ Ü÷¾èœ
ò£¬õ -?
The difference between two parallel sides of a trapezium is 4 cm. The perpendicular distance
between them is 19 cm. If the area of the trapezium is 475 cm2, find the lengths of the parallel
sides
A) 20 cm and 16 cm B) 27 cm and 23 cm
C) 27 cm and 20 cm D) 25 cm and 23 cm
a − b = 4 → (i )
Area of Trapezium = ½ x (a + b) h
475 = ½ x (a + b) x 19
950 = (a +b) x 19

ê˜è£˜ ä.ã.âv Üè£ìI

50 = a + b .............. (ii)

Add (i) and (ii)

a −b = 4
a +b = 50
2a = 54
a = 27
(i ) ⇒
27 − b =4
27 − 4 =b
23 = b
Parallel sides 27cm and 23cm

61. å¼ «î£†ì‚è£ó˜ êKõè õ®M™ ²õ˜ å¡P¬ù ܬñ‚è F†ìI´Aø£˜. êKõèˆF¡

c‡ì ºî™ õK¬ê‚° 100 ªêƒèŸèœ «î¬õŠð´Aø¶. H¡¹ 嚪õ£¼
õK¬êJ¡ Þ¼¹øº‹ 1 ªêƒè™ °¬øõ£è ¬õ‚è «õ‡´‹. Üšõ®õ¬ñŠH™ 25
õK¬êèO¼ŠH¡ Üõ˜ õ£ƒè «õ‡®ò ªêƒèŸèO¡ â‡E‚¬è âˆî¬ù?
A gardener plans to construct a trapezoidal shaped structure in his garden. The longer side of
side of trapezoid needs to start with a row of 100 bricks. Each row must be decreased by 1
bricks on each end and the construction should stop at 25th row. How many bricks does he need
to buy?
(A) 1500 (B) 1700 (C) 1900 (D) 2000
Sol:
100,98,96.....
100 + 98 + 96 + .....
d = a2 − a1 = 98 − 100 161
 TNPSC èí‚°
100,98,96.....
100,98,96.....
100 + 98 + 96 + .....
100 + 98 + 96 + .....
d = a2 − a1 = 98 − 100
d = a2 − a1 = 98 − 100
d = −2
d = −2
a=100 d=-2 n=25
n
S=
n [ 2a + (n − 1)d ]
2
25
S 25
= [ 2x100 + (25 − 1)(−2)]
2
25
S=25 [ 200 + 24(−2)]
2
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

25
=SS 25 25 25 [[ 200 − 48 ]
=
= S 2525 22 [ 200
200 − 48]]
− 48
252 76
SS 25 = 25
25 x
76
152
76
= x 152
25 = 2
S 25 22 x 152
SS 25 =
= 1900
1900
25 = 1900
S 25

62. å¼ êKõèˆF¡ ðóŠ¹ 960 cm2. Üî¡ Þ¬í ð‚èƒèœ 40 cm ñŸÁ‹ 60 cm âQ™
Þ¬í ð‚èƒèÀ‚° Þ¬ìŠð†ì Éóˆ¬î 裇.
Area of trapezium is 960 cm2, The parallel sides are 40 cm and 60 cm.Find the distance be-
tween parallel sides
A)18.2 cm B) 19.2 cm C) 20.4 cm D)21.4 cm
Sol:
1
Area of Trapezium = (a + b) x h
2
1
960
= (40 + 60) x h
2
1 50
960 = x 100 x h
2

19.2
960
=h
50
h = 19.2cm

63. å¼ Þ¬íèóˆF¡ å¼ ð‚è‹ 18 ªê.e å¼ ð‚èˆFL¼‰¶ Ü´ˆî ð‚èˆFŸ° Ýù

ªî£¬ô¾ 8 ªê.e âQ™ ÜšM¬íèóˆF¡ ðóŠ¹
One side of a parallelogram is 18cm and its distance from the opposite side is 8 cm. The area
of the parallelogram is
A) 160 cm2 B) 230 cm2 C) 144 cm2 D) 140 cm2
Sol:
Area of parallelogram = b x h
= 18 x 8
= 144 cm 2 162
 TNPSC èí‚°
=bxh
=bxh
= 18 x 8
= 18 x 8
= 144 cm 22
= 144 cm

64.
å¼ Þ¬íèóˆF¡ ðóŠð÷¾ 49.92 ªê.e2 Üî¡ Ü®Šð‚è‹ 7.8 ªê.e âQ™ °ˆ¶òó‹
â¡ù?
Find the height of a parallelogram whose area is 49. 92 cm2 and base is 7.8 cm.
A) 5.6 cm B) 6.6 cm C) 5.4 cm D) 6.4 cm
Sol:
Area of parallelogram = b x h
49.92 = 7.8 x h
49.92
49.92 6.4
49.92
= h= h
7.8
7.8
h =h6.4 cm
= 6.4

65. æ˜ Þ¬íèóˆF¡ ªðKò «è£íñ£ù¶ Üî¡ CPò «è£íˆF¡ Þ¼ ñ샬è Mì

ê˜è£˜ ä.ã.âv Üè£ìI

30° °¬øõ£è Þ¼‚Aø¶ âQ™ CPò «è£í‹ ñŸÁ‹ ªðKò «è£í‹ º¬ø«ò
In a parallelogram, if the larger angle is 30° less than twice the smaller angle then the smaller
and larger angles are respectively.
(A) 20°, 50° (B) 30°, 80° (C) 50°, 70° (D) 70°, 110°
Sol:
180 ( Condition )
x+ y =
x 2 y − 30 ( Given )
=
2 y − 30 + y =
180
3 y − 30 =
180
3=y 180 + 30

70
210
y=
3
y = 70°
x 2 y − 30
=
=x 2 x 70 − 30
=x 140 − 30
x = 110
70 ,110

66. Which of the following statement is false in a Parallelogram?

A. The opposite sides are parallel
B. The opposite angles and sides are equal
C. The diagonals are equal
D. The diagonals bisect each other

163
 TNPSC èí‚°

æ˜ Þ¬íèóˆF™ ⶠîõø£ù ßÁ?

(Ü) âF˜Š ð‚èƒèœ Þ¬íò£°‹
(Ý) âFªóF˜ «è£íƒèœ ñŸÁ‹ ð‚èƒèœ êññ£°‹
(Þ) ͬô M†ìƒèO¡ c÷ƒèÀ‹ êññ£°‹
(ß) ͬô M†ìƒèœ 塬øªò£¡Á Þ¼ êñ‚ ÃP´‹

67.
Ü®Šð‚è‹ 9 ªê.e ñŸÁ‹ àòó‹ 5 ªê.e âù¾‹ à¬ìò Þ¬íèóˆF¡ ðóŠð÷¾
裇è
Find the area of a parallelogram whose base and height are 9 cm and 5 cm respectively
(A) 45 cm2 / 45 ê.ªê.e (B) 40 cm2 / 40 ê.ªê.e
(C) 14 cm2 / 14 ê.ªê.e (D) 22.5cm2 / 22.5 ê.ªê.e
TNPSC GROUP- I,II,IV & VAO │ (èEî‹ ¬è«ò´)

Sol:
b = 9, h = 5
Area of a parallelogram = b x h
= 9 x 5 = 45 cm2

164
 ê˜è£˜ ä.ã.âv Üè£ìI
èí‚°
TNPSC

165
(èEî‹)
TNPSC
TNPSC GROUP-
GROUP- I ││ (èEî‹ ¬è«ò´)
I,II,IV & VAO

Anna Nagar West, Chennai - 600012

Phone: 9962600037, 9962600038
www.sarkariasacademy.com