Find the arc length, area and perimeter of the sector with
radius 21 cm and central angle 60°
Given:
radius (r) = 21 cm
central angle 60°
Arc length = ?, area of sector = ? and perimeter of sector = ?
As we know,
(a) Length of arc =
⇒ Length of arc = 22cm
(b) Area of sector =
⇒ area of sector = 231cm2
(c) perimeter of sector = r + r + length of arc
= 2(r) + length of arc
= 2(21) + 22
= 42 + 22 = 64cm
Find the arc length, area and perimeter of the sector with
radius 4.9 cm and central angle 30°
Given:
radius (r) = 4.9 cm
central angle 30°
Arc length = ?, area of sector = ? and perimeter of sector = ?
As we know,
(a) Length of arc =
⇒ Length of arc = 2.57cm
(b) Area of sector =
⇒ area of sector = 6.3cm2
( c) perimeter of sector = r + r + length of arc
= 2 ( r) + length of arc
= 2(4.9) + 2.57
= 9.8 + 2.57 = 12.37cm
Find the arc length, area and perimeter of the sector with
radius 14 cm and sector angle 45°
Given:
radius (r) = 14 cm
central angle 45°
Arc length = ?, area of sector = ? and perimeter of sector = ?
As we know,
(a) Length of arc =
⇒ Length of arc = 11cm
(b) Area of sector =
⇒ area of sector = 77cm2
(c) perimeter of sector = r + r + length of arc
= 2 ( r) + length of arc
= 2(14) + 11
= 28 + 11 = 39cm
Find the arc length, area and perimeter of the sector with
radius 15 cm and sector angle 63°
Given:
radius (r) = 15 cm
central angle 63°
Arc length = ?
area of sector = ?
perimeter of sector = ?
As we know,
(a) Length of arc =
⇒ Length of arc = 16.5cm
(b) Area of sector =
⇒ area of sector = 123.75cm2
(c) perimeter of sector = r + r + length of arc
= 2 ( r) + length of arc
= 2(15) + 16.5
= 30 + 16.5 = 46.5cm
Find the arc length, area and perimeter of the sector with
radius 21 dm and sector angle 240°
Given:
radius (r) = 21 dm
central angle 240°
Arc length = ?
area of sector = ?
perimeter of sector = ?
As we know,
(a) Length of arc =
⇒ Length of arc = 88dm
(b) Area of sector =
⇒ area of sector = 924 dm2
(c) perimeter of sector = r + r + length of arc
= 2 ( r) + length of arc
= 2(21) + 88
= 42 + 88 = 130dm
Find the angle subtended by an arc 88 cm long at the centre of a circle of radius 42 cm.
Given:
radius (r) = 42 cm
Arc length = 88 cm
central angle = ?
As we know,
Length of arc =
⇒ central angle = 12°
The arc length of a sector of a circle of radius 14 cm is 22 cm. Find its central angle.
Given:
radius (r) = 14 cm
Arc length = 22 cm
central angle = θ
As we know,
Length of arc =
⇒ central angle = 90°
Find the radius of a sector of a circle having a central angle 70°and an arc length of 44 cm.
Given:
radius (r) = ? cm
Arc length = 44 cm
central angle = 70°
As we know,
Length of arc =
⇒ radius = 36cm
Find the area and perimeter of the sector with
radius 10 cm and arc length 33 cm.
Given:
radius (r) = 10 cm
Arc length (a) = 33cm
area of sector = ?
perimeter of sector = ?
As we know,
(a) Area of sector =
⇒ area of sector = 165 cm2
(b) perimeter of sector = r + r + length of arc
= 2 ( r) + length of arc
= 2(10) + 33
= 20 + 33 = 53cm
Find the area and perimeter of the sector with
radius 55 cm and arc length 80 cm.
Given:
radius (r) = 55 cm
Arc length (a) = 80cm
area of sector = ?
perimeter of sector = ?
As we know,
(a) Area of sector =
⇒ area of sector = 2200 cm2
(b) perimeter of sector = r + r + length of arc
= 2 ( r) + length of arc
= 2(55) + 80
= 110 + 80 = 190cm
Find the area and perimeter of the sector with
radius 12 cm and arc length 15.25 cm.
Given:
radius (r) = 12 cm
Arc length (a) = 15.25cm
area of sector = ?
perimeter of sector = ?
As we know,
(a) Area of sector =
⇒ area of sector = 91.5 cm2
(b) perimeter of sector = r + r + length of arc
= 2 ( r) + length of arc
= 2(12) + 15.25
= 24 + 15.25 = 39.25cm
Find the area and perimeter of the sector with
radius 20 cm and arc length 25 cm.
Given: radius (r) = 20 cm
Arc length (a) = 25cm
area of sector = ?
perimeter of sector = ?
As we know,
(a) Area of sector =
⇒ area of sector = 250 cm2
(b) perimeter of sector = r + r + length of arc
= 2 ( r) + length of arc
= 2(20) + 25
= 40 + 25 = 65cm
Find the arc length of the sector of radius 14 cm and area 70 cm2
Given:
radius (r) = 14 cm
Arc length (a) = ?
area of sector = 70cm2
As we know,
Area of sector =
⇒ arc length = 10 cm
Find the radius of the sector of area 225 cm2 and having an arc length of 15 cm
Given:
radius (r) = ?
Arc length (a) = 15cm
area of sector = 225cm2
As we know,
Area of sector =
⇒ radius = 30 cm
Find the radius of the sector whose central angle is 140°and area 44 cm2.
Given:
radius (r) = ?
central angle (a) = 140°
area of sector = 44cm2
Area of sector =
⇒ r2 = 36
⇒ r = 6 cm
The perimeter of a sector of a circle is 58 cm. Find the area if its diameter is 9 cm.
perimeter of a sector = 58 cm
diameter = 9 cm
Radius (r) = d/2 = 9/2 = 4.5 cm
Perimeter = r + r + length of arc
= 2 ( r) + length of arc
58 =arc length + 9
Arc length (a)= 58 – 9 = 49cm
Area of sector =
⇒ area of sector = 110.25 cm2
Find the area of a sector whose radius and perimeter are 20 cm and 110 cm respectively.
perimeter of a sector = 110 cm
radius = 20 cm
Perimeter = r + r + length of arc
= 2 ( r) + length of arc
110 =arc length + 2(20)
Arc length (a)= 110 – 40 = 70cm
Area of sector =
⇒ area of sector = 700 cm2
Find the central angle of a sector of a circle having
area 352 cm2 and radius 12 cm
Given:
radius (r) = 12 cm
central angle = ?
area of sector = 352 cm2
As we know,
Area of sector =
⇒ central angle = 280°
Find the central angle of a sector of a circle having
area 462 cm2 and radius 21 cm
Given:
radius (r) = 21 cm
central angle = ?
area of sector = 462 cm2
As we know,
Area of sector =
⇒ central angle = 120°
Calculate the perimeter and area of the semicircle whose radius is 14 cm.
Given:
radius (r)= 14cm
(a) perimeter of semi-circle = 1/2 (2πr) + r + r
= π r + 2r
perimeter of semi-circle = 44 + 28 = 72cm
(b) area of semi-circle =
area of semi-circle = 308 cm2
Calculate the perimeter and area of a quadrant circle of radius 7 cm.
Given:
radius (r)= 7cm
(a) perimeter of quadrant-circle = 1/4 (2 π r) + r + r
= 1/2 π r + 2r
perimeter of quadrant-circle = 11 + 14 = 25cm
(b) area of quadrant-circle =
area of quadrant-circle = 38.5 cm2
Calculate the arc length of a sector whose perimeter and radius are 35 cm and 8 cm respectively.
perimeter of a sector = 35 cm
radius = 8 cm
Perimeter = arc length + r + r
= arc length + 2r
35 =arc length + 2(8)
Arc length (a)= 35 – 16 = 19cm
Find the radius of a sector whose perimeter and arc length are 24 cm and 7 cm respectively.
perimeter of a sector = 24 cm
arc length = 7 cm
Perimeter = arc length + r + r
= arc length + 2r
24 = 7 + 2(r)
2r = 24 – 7
r = 17/2
r = 8.5 cm
Time spent by a student in a day is shown in the figure. Find how much time is spent in
(i) school (ii) play ground (iii) other activities
Given:
Total time available is 24hr
(i) time spent in school =
time spent in school = 7hr
(ii) time spent in playground =
time spent in playground = 2hr
(iii) time spent in other activities =
time spent in other activities = 15hr
Three coins each 2 cm in diameter are placed touching one another. Find the area enclosed by them.
Given:
diameter of each coin = 2cm
⇒ Radius(r) = 1cm
Area enclosed by one coin = π r2
=
Area enclosed by all the 3 coin = 3× 22/7
= 9.43cm2
Four horses are tethered with ropes measuring 7 m each to the four corners of a rectangular grass land 21 m x 24 m in dimension. Find
(i) the maximum area that can be grazed by the horses
(ii) the area that remains ungrazed.
Given:
(i) length of rope = radius of circle.
At one corner area grazed =
At one corner area grazed = 38.5m
the maximum area that can be grazed by the horses = 4 × area at one corner
= 4 × 38.5
= 154m2
(ii) Area that remain ungrazed = total area of rectangle – area grazed by horses
Area that remain ungrazed = (21 m x 24 m) – 154
= 504 – 154
= 350 m2
Find the area of card board wasted if a sector of maximum possible size is cut out from a square card board of size 24 cm. [π = 3.14].
Given:
The area of square = (side)2 = 24× 24 cm2 = 576 cm2
Now, the area of area has to be maximum. It means that the side of the square will be radius of the sector.
Then, area of the sector =
=
= 452.57 cm2
Then, the wasted area will be given as the remaining area
= Area of square – Area of the sector
= 576 cm2 – 452.57 cm2
123.84 cm2
Find the area of the shaded portion in the adjoining figure
Given:
θ = 30°
radius of small part is 21cm
radius of bigger part is 42cm
Area of small sector =
⇒ area of small sector = 115.5 cm2
Area of large sector =
⇒ area of large sector = 462 cm2
Area of shaded part = 462 – 115.5
= 346.5cm2
Find the radius, central angle and perimeter of a sector whose length of arc and area are 4.4 m and 9.24 m2 respectively.
Given:
length of arc (a) = 4.4m
area = 9.24m2
(a) Area of sector =
⇒ radius of sector = 4.2m
(b) Area of sector =
⇒ central angle = 60°
(c) perimeter of sector = 2 ( r) + length of arc
= 2(4.2) + 4.4
= 8.4 + 4.4
= 12.8cm
Find the Lateral Surface Area (LSA), Total Surface Area (TSA) and volume of the cubes having their sides as
5.6 cm
Given:
side of cube (a) = 5.6cm
(a) LSA of cube = 4a2
= 4(5.6)2
⇒ LSA of cube = 125.44cm2
(b) TSA of cube = 6a2
= 6(5.6)2
⇒ TSA of cube = 188.16cm2
(c) Volume of cube = a3
= (5.6)3
⇒ Volume of cube = 175.62cm3
Find the Lateral Surface Area (LSA), Total Surface Area (TSA) and volume of the cubes having their sides as
6 dm
Given:
side of cube (a) = 6dm
(a) LSA of cube = 4a2
= 4(6)2
⇒ LSA of cube = 144dm2
(b) TSA of cube = 6a2
= 6(6)2
⇒ TSA of cube = 216dm2
(c) Volume of cube = a3
= (6)3
⇒ Volume of cube = 216dm3
Find the Lateral Surface Area (LSA), Total Surface Area (TSA) and volume of the cubes having their sides as
2.5 m
Given:
side of cube (a) = 2.5m
(a) LSA of cube = 4a2
= 4(2.5)2
⇒ LSA of cube = 25m2
(b) TSA of cube = 6a2
= 6(2.5)2
⇒ TSA of cube = 37.5m2
(c) Volume of cube = a3
= (2.5)3
⇒ Volume of cube = 15.625m3
Find the Lateral Surface Area (LSA), Total Surface Area (TSA) and volume of the cubes having their sides as
24 cm
Given:
side of cube (a) = 24cm
(a) LSA of cube = 4a2
= 4(24)2
⇒ LSA of cube = 2304cm2
(b) TSA of cube = 6a2
= 6(24)2
⇒ TSA of cube = 3456cm2
(c) Volume of cube = a3
= (24)3
⇒ Volume of cube = 13824cm3
Find the Lateral Surface Area (LSA), Total Surface Area (TSA) and volume of the cubes having their sides as
31 cm
Given:
side of cube (a) = 31cm
(a) LSA of cube = 4a2
= 4(31)2
⇒ LSA of cube = 3844cm2
(b) TSA of cube = 6a2
= 6(31)2
⇒ TSA of cube = 5766cm2
(c) Volume of cube = a3
= (31)3
⇒ Volume of cube = 29791cm3
If the Lateral Surface Area of a cube is 900 cm2, find the length of its side.
Given:
LSA = 900cm2
LSA of cube = 4a2
900 = 4(a)2
⇒ (a)2 = 225
⇒ length of side of cube = 15cm
If the Total Surface Area of a cube is 1014 cm2, find the length of its side.
Given:
TSA = 1014cm2
TSA of cube = 6a2
1014 = 6(a)2
⇒ (a)2 = 169
⇒ length of side of cube = 13cm
The volume of the cube is 125 dm3. Find its side.
Given:
Volume = 125dm3
Volume of cube = a3
125 = (a)3
⇒ (a) = 5
⇒ length of side of cube = 5dm
A container is in the shape of a cube of side 20 cm. How much sugar can it hold?
Given:
side of container (a) = 20cm
Volume of cube = a3
= (20)3
= 20 × 20 × 20
⇒ Volume of container = 8000cm3
A cubical tank can hold 64,000 litres of water. Find the length of the side of the tank.
Given:
Volume of tank = 64000 litres = 64m3
Volume of cube = a3
64000 = (a)3
⇒ (a)3 = (4)3
⇒ a = 4
⇒ length of side of tank = 4m
Three metallic cubes of side 3 cm, 4 cm and 5 cm respectively are melted and are recast into a single cube. Find the total surface area of the new cube.
Given:
three cubes of side 3 cm, 4 cm and 5cm
Total volume of all cube = volume of new cube
(3)3 + (4)3 + (5)3 = (a)3
27 + 64 + 125 = (a)3
216 = (a)3
⇒ a = 6
TSA of cube = 6a2
= 6(6)2
⇒ TSA of cube = 216cm2
How many cubes of side 3 cm are required to build a cube of side 15 cm?
Given:
side of small cube = 3cm
side of large cube = 15cm
volume of small cube = (3)3
= 27
volume of large cube = (15)3
= 3375
Number of required = volume of large cube/ volume of small cube
N = 3375/27
N = 125
Number of cubes required = 125
Find the area of card board required to make an open cubical box of side 40 cm. Also find the volume of the box.
Given:
side of cube (a) = 40cm
Area of open cubical box = 5a2
= 5(40)2
⇒ area of cube = 8000cm2
Volume of cube = a3
= (40)3
⇒ Volume of cube = 64000cm3
What is the total cost of oil in a cubical container of side 2 m if it is measured and sold using a cubical vessel of height 10 cm and the cost is ₹50 per measure.
Given:
side of container = 2m
side of vessel = 10cm = 0.1m
Volume of container = (2)3
= 8000000cm3
Volume of container = (10)3
= 1000cm3
Rate = 50 per 1000 m cube
So, cost =
Cost = Rs. 400000
A container of side 3.5m is to be painted both inside and outside. Find the area to be painted and the total cost of painting it at the rate of ₹75 per square meter.
Given:
side of container (a) = 3.5cm
Area of cubical box to be painted both sides = 12a2
= 12(3.5)2
⇒ Area of cubical box to be painted both sides = 147m2
total cost of painting it at the rate of Rs75 per square meter
then total cost of painting painted box = 147 × 75 = Rs11025
Find the L.S.A, T.S.A and volume of the cuboids having the length, breadth and height
respectively as
5 cm, 2 cm , 11cm
Given:
l = 5cm, b = 2cm and h = 11cm
(a) LSA of cuboid = 2(lh+hb)
= 2(5(11) + 2(11))
= 2(55+22)
= 2(77) = 154cm2
⇒ LSA of cuboid = 154cm2
(b) TSA of cuboid = 2(lh+hb+lb)
= 2(5(11) + 2(11) + 5(2))
= 2(55+22+10)
⇒ TSA of cuboid = 174cm2
(c) Volume of cuboid = l× b× h
= 5× 2× 11
⇒ Volume of cuboid = 110cm3
Find the L.S.A, T.S.A and volume of the cuboids having the length, breadth and height
respectively as
15 dm, 10 dm, 8 dm
Given:
l = 15dm, b = 10dm and h = 8dm
(a) LSA of cuboid = 2(lh+hb)
= 2(15(8) + 8(10))
= 2(120+80)
= 2(200) = 400dm2
⇒ LSA of cuboid = 400dm2
(b) TSA of cuboid = 2(lh+hb+lb)
= 2(15(8) + 8(10) + 15(10))
= 2(120+80+150)
⇒ TSA of cuboid = 700dm2
(c) Volume of cuboid = l× b× h
= 15× 10× 8
⇒ Volume of cuboid = 1200dm3
Find the L.S.A, T.S.A and volume of the cuboids having the length, breadth and height
respectively as
2 m, 3 m, 7 m
Given:
l = 2m, b = 3m and h = 7m
(a) LSA of cuboid = 2(lh+hb)
= 2(2(7) + 7(3))
= 2(14+21)
= 2(35) = 70m2
⇒ LSA of cuboid = 66m2
(b) TSA of cuboid = 2(lh+hb+lb)
= 2(2(7) + 7(3) + 2(3))
= 2(14+21+6)
⇒ TSA of cuboid = 82m2
(c) Volume of cuboid = l× b× h
= 2× 3× 7
⇒ Volume of cuboid = 42m3
Find the L.S.A, T.S.A and volume of the cuboids having the length, breadth and height
respectively as
20 m, 12 m, 8 m
Given:
l = 20m, b = 12m and h = 8m
(a) LSA of cuboid = 2(lh+hb)
= 2(20(8) + 8(12))
= 2(160+96)
= 2(256) = 512m2
⇒ LSA of cuboid = 512m2
(b) TSA of cuboid = 2(lh+hb+lb)
= 2(20(8) + 8(12) + 20(12))
= 2(160+96+240)
⇒ TSA of cuboid = 992m2
(c) Volume of cuboid = l× b× h
= 20× 12× 8
⇒ Volume of cuboid = 1920m3
Find the height of the cuboid whose length, breadth and volume are 35 cm, 15 cm and 14175 cm3 respectively.
Given:
l = 35cm, b = 15cm and volume = 14175 cm3
Volume of cuboid = l× b× h
14175 = 35× 15× h
⇒ h = 14175/ 525
⇒ h = 27
⇒ Height of cuboid = 27cm
Two cubes each of volume 64 cm3 are joined to form a cuboid. Find the L.S.A and T.S.A of the resulting solid.
Given:
volume of given cube = 64cm3
Side = a cm
Volume = (a)3
64 = (a)3
a = 4cm
two cubes are joined together.
Length of resultant cube = 4+4 = 8cm
Breadth = 4cm
Height = 4cm
(a) LSA of cuboid = 2(lh+hb)
= 2(8(4) + 4(4))
= 2(32+16)
= 2(48) = 96cm2
⇒ LSA of cuboid = 66m2
(b)TSA of cuboid = 2(lh+hb+lb)
= 2(8(4) + 4(4) + 8(4))
= 2(32+16+32)
⇒ TSA of cuboid = 2(80) = 160m2
Raju planned to stitch a cover for his two speaker boxes whose length, breadth andheight are 35 cm, 30 cm and 55 cm respectively. Find the cost of the cloth he has to buy if its costs ₹75 per sq.m.
Given:
l = 35cm, b = 30cm and h = 55cm
Total area to cover two speaker =2{ 2(lh + bh) + lb)}
= 2{2(35(55) + 30(55)) + 35(30)}
=2{ 2(1925 + 1650 )+ 1050}
=2{ 2(210) + 1050 }
= 2{7150 + 1050} = 16400 cm2 = 1.64m2
⇒ area = 1.64m2
Cost of 1m = Rs 75
Cost of 1.64 m2 = 1.64× 75
= Rs 123
Mohan wanted to paint the walls and ceiling of a hall. The dimensions of the hall is 20m x15m x6m. Find the area of surface to be painted and the cost of painting it at ₹78 per sq. m.
Given:
l = 20m, b = 15m and h = 6m
area of walls and ceiling to be renovated = 2(lh + bh) + lb)
= 2(20(6) + 15(6)) + 20(15)
= 2(120 + 90 )+ 300
= 2(210) + 300
= 420 + 300 = 720 m2
⇒ area = 720m2
Cost of 1m = Rs 78
Cost of 720 m2 = 720× 48
= Rs 56160
How many hollow blocks of size 30cm x 15cm x20cm are needed to construct a wall 60m in length, 0.3m in breadth and 2m in height.
Given:
sides of small cuboid = 30cm x 15cm x20cm and sides of large cuboid = 60m x 0.3m x 2m
volume of small cube = 30cm x 15cm x20cm
= 9000cm3 = 0.009m3
volume of large cube = 60m x 0.3m x 2m
= 36m
Number of hollow blocks required = volume of large cuboid/ volume of small cuboid
N = 36/0.009
N = 4000
Number of hollow blocks required = 4000
Find the cost of renovating the walls and the floor of a hall that measures 10m x 45m x 6m if the cost is ₹48 per square meter.
Given:
l = 10m, b = 45m and h = 6m
area of cuboid to be renovated = 2(lh + bh)+ lb
= 2(10(6) + 45(6)) + 10(45)
= 2(60 + 270) +450
= 660 + 450 = 1110
⇒ area of cuboid to be renovated zz= 1110m3
Cost of 1m = Rs 48
Cost of 1110 m2 = 1110× 48
= Rs 53280
The length of the arc of a sector having central angle 90° and radius 7 cm is
A. 22 cm
B. 44 cm
C. 11 cm
D. 33 cm
Given:
length of arc = ?
radius = 7cm
Arc length =
⇒ Arc length = 11cm
Option (C) is correct.
If the radius and arc length of a sector are 17 cm and 27 cm respectively, then the perimeter is
A. 16 cm
B. 61 cm
C. 32 cm
D. 80 cm
Given:
Perimeter = ?
length of arc = 27cm
radius = 17cm
perimeter of sector = arc length + r + r
= arc length + 2r
= 2(17) + 27
= 34 + 27 = 61cm
Option (B) is correct.
If the angle subtended by the arc of a sector at the center is 90°, then the area of the sector in square units is
A. 2πr2
B. 4πr2
C.
D.
Given:
radius = r
central angle = 90°
Area of sector =
Option (C) is correct.
Area of a sector having radius 12 cm and arc length 21 cm is
A. 126 cm2
B. 252 cm2
C. 33 cm2
D. 45 cm2.
Given:
radius = 12cm
arc length = 21cm
Area of sector =
⇒ area of sector = 126cm
Option (A) is correct.
The area of a sector with radius 4 cm and central angle 60° is
A.
B.
C.
D.
Given:
radius = 4cm
central angle = 60°
Area of sector =
⇒ central angle = 8π /3 cm2
Option (C) is correct.
If the area and arc length of the sector of a circle are 60 cm2 and 20 cm respectively, then the diameter of the circle is
A. 6 cm
B. 12 cm
C. 24 cm
D. 36 cm
Given:
area of sector = 60cm2
arc length = 20cm
Area of sector =
⇒ Radius of circle = 6cm
Diameter of circle = 12cm
Option (B) is correct.
The perimeter of a sector of a circle is 37cm. If its radius is 7cm, then its arc length is
A. 23 cm
B. 5.29 cm
C. 32 cm
D. 259 cm
Given:
Perimeter = 37cm
radius = 7cm
perimeter of sector = arc length + r + r
= arc length + 2r
37 = 2(7) + a
a = 37 - 14 = 23cm
Option (a) is correct.
A solid having six equal square faces is called a
A. cube
B. cuboid
C. square
D. rectangle
A solid having six equal square faces is Cube.
Option (a) is correct.
The quantity of space occupied by a body is its
A. area
B. length
C. volume
D. T.S.A
Quantity of space occupied by a body is its volume.
Option (c) is correct.
The LSA of a cube of side 1dm is
A. 16 dm2
B. 4 dm2
C. 2 dm2
D. 1 dm2
Given:
side of cube (a) = 1dm
LSA of cube = 4a2
= 4(1)2
= 4(1)
⇒ LSA of cube = 4dm2
Option (B) is correct.