Question 1
The diameter of a circle is 28 cm.
Find its :
(i) Circumference
(ii) Area.
Let r be the radius of the circle.
(i) 2r = 28 cm
circumference = 2πr
= 28π cm
= 88 cm
(ii) area = πr2
= π
= 196π cm2
= 616 cm2
Question 2
The circumference of a circular field is 308 m. Find is:
(i) Radius
(ii) Area.
Let r be the radius of the circular field
(i) 2πr = 308
⇒ r =
⇒ r =
⇒ r = 49 m
(ii) area = πr2
=
= 7546 m2
Question 3
The sum of the circumference and diameter of a circle is 116 cm. Find its radius.
Sol:Let r be the radius of the circle.
2πr + 2r = 116
2r( π + 1) = 116
r =
= 14 cm
Question 4
The radii of two circles are 25 cm and 18 cm. Find the radius of the circle which has a circumference equal to the sum of circumferences of these two circles.
Sol:Circumference of the first circle
S1 = 2π x 25
= 50π cm
Circumference of the second circle
S2 = 2π x 18
= 36π cm
Let r be the radius of the resulting circle.
2πr = 50π + 36π
2πr = 86π
r =
= 43 cm
Question 5
The radii of two circles are 48 cm and 13 cm. Find the area of the circle which has its circumference equal to the difference of the circumferences of the given two circles.
Sol:Circumference of the first circle
s1 = 2π x 48
= 96πcm
Circumference of the first circle
s2 = 2π x 13
= 26πcm
Let r be the radius of the resulting circle.
2πr = 96π - 26π
2πr = 70π
r =
= 35cm
Hence area of the circle
A = πr2
= π x 352
= 3850cm2
Question 6
The diameters of two circles are 32 cm and 24 cm. Find the radius of the circle having its area equal to the sum of the areas of the two given circles.
Sol:Let the area of the resulting circle be r.
π x (16)2 + π x ( 12 )2 = π x r2
256 π + 144 π = π x r2
400π = π x r2
r2 = 400
r = 20 cm
Hence the radius of the resulting circle is 20cm.
Question 7
The radius of a circle is 5 m. Find the circumference of the circle whose area is 49 times the area of the given circle.
Sol:Area of the circle having radius 85m is
A = π x ( 85 )2
= 7225πm2
Let r be the radius of the circle whose area is 49times of the given circle.
πr2 = 49 x ( π x 52 )
r2 = ( 7 x 5 )2
r = 35
Hence circumference of the circle
S = 2πr
= 2π x 35
= 220 m
Question 8
A circle of the largest area is cut from a rectangular piece of cardboard with dimensions 55 cm and 42 cm. Find the ratio between the area of the circle cut and the area of the remaining card-board.
Sol:Area of the rectangle is given by
A = 55 x 42
= 2310 cm2
For the largest circle, the radius of the circle will be half of the sorter side of the rectangle.
Hence r = 21cm.
Area of the circle = π x ( 21 )2
= 1384.74 cm2
Area remaining = 2310 - 1384.74
= 925.26
Hence
the volume of the circle: area remaining
= 1384. 74: 915. 26
= 3.2
Question 9
The following figure shows a square cardboard ABCD of side 28 cm. Four identical circles of the largest possible sizes are cut from this card as shown below.
Find the area of the remaining card-board.
Area of the square is given by
A = 282
= 784 cm2
Since there are four identical circles inside the square.
Hence the radius of each circle is one-fourth of the side of the square.
Area of one circle = π x 72
= 154 cm2
Area of four circle = 4 x 154cm2
= 616 cm2
Area remaining = 784 - 616
= 168 cm2
The area remaining in the cardboard is = 168 cm2.
Question 10
The radii of two circles are in the ratio 3: 8. If the difference between their areas is 2695π cm2, find the area of the smaller circle.
Sol:Let the radius of the two circles be 3r and 8r respectively.
area of the circle having radius 3r = π(3r)2
= 9πr2
area of the circle having radius 8r = π(8r)2
= 64πr2
According to the question
64πr2 - 9πr2 = 2695π
55r2 = 2695
r2 = 49
r = 7 cm
Hence the radius of the smaller circle is 3 x 7 = 21 cm
Area of the smaller circle is given by
A = πr2 =
Question 11
The diameters of three circles are in the ratio 3: 5: 6. If the sum of the circumferences of these circles is 308 cm; find the difference between the areas of the largest and the smallest of these circles.
Sol:Let the diameter of the three circles be 3d, 5d, and 6d respectively.
Now
π x 3d + π x 5d + π x 6d = 308
14πd = 308
d = 7
radius of the smallest circle =
Area = π x (10 .5)2
= 346.5
radius of the largest circle =
Area = π x ( 21 )2
= 1386
difference = 1386 - 346.5
= 1039.5 cm2
Question 12
Find the area of a ring-shaped region enclosed between two concentric circles of radii 20 cm and 15 cm.
Sol:Area of the ring = π( 20 )2 - π( 15 )2
= 400π - 225π
= 175π
= 549.7 cm2
Question 13
The circumference of a given circular park is 55 m. It is surrounded by a path of uniform width of 3.5 m. Find the area of the path.
Sol:Let r be the radius of the circular park.
2πr = 55
r =
= 8.75 m
area of the park = π x ( 8.75 )2 = 240.625 m2
Radius of the outer circular region including the path is given by
R = 8.75 + 3.5
= 12.25 m
Area of that circular region is A = π x ( 12.25 )2 = 471.625 m2
Hence area of the path is given by
Area of the path = 471.625 - 240.625 = 231 m2.
Question 14
There are two circular gardens A and B. The circumference of garden A is 1.760 km and the area of garden B is 25 times the area of garden A. Find the circumference of garden B.
Sol:Let r be the radius of the circular garden A
Since the circumference of the garden A is 1.760 km = 1760 m, we have,
2πr = 1760 m
⇒ r =
Area of garden A = πr2 =
Let R be the radius of the circular garden B.
Since the area of the garden, B is 25 times the area of garden A, We have,
πR2 = 25 x πr2
⇒ πR2 = 25 x π x 2802
⇒ R2 = 1960000
⇒ R = 1400 m
Thus circumference of garden B = 2πR
=
Question 15
A wheel has a diameter of 84 cm. Find how many completer revolutions must it make to cover 3.168 km.
Sol:The diameter of the wheel = 84 cm
Thus, the radius of the wheel = 42 cm
Circumference of the wheel = 2 x
In 264 cm, the wheel is covering one revolution.
Thus, in 3.168 km = 3.168 x 100000 cm, number of revolutions
covered by the wheel =
Question 16
Each wheel of a car is of diameter 80 cm. How many completer revolutions does each wheel make in 10 minutes when the car is traveling at a speed of 66 km per hour?
Sol:The car travels in 10 minutes
=
Circumference of the wheel = distance covered by the wheel in one revolution
Thus, we have,
Circumference = 2 x
Thus, the number of revolutions covered by the wheel in 1100000 cm =
Question 17
An express train is running between two stations with a uniform speed. If the diameter of each wheel of the train is 42 cm and each wheel makes 1200 revolutions per minute, find the speed of the train.
Sol:radius of the wheel =
Circumference of the wheel = 2π x 21 = 132 cm
Distance travelled in one minute
= 132 x 1200
= 158400 cm
= 1.584 km
Hence the speed of the train
=
= 95.04 km/hr.
Question 18
The minute hand of a clock is 8 cm long. Find the area swept by the minute hand between 8.30 a.m. and 9.05 a.m.
Sol:The time interval is 9.05 - 8.30 = 35 minutes
Area covered in one 60 minutes = π x 82 = 201 cm2
Hence area swept in 35 minutes is given by
A =
Question 19
The shaded portion of the figure, given alongside, shows two concentric circles. If the circumference of the two circles is 396 cm and 374 cm, find the area of the shaded portion.
Let R and r be the radius of the big and small circles respectively.
Given that the circumference of the bigger circle is 396 cm
Thus, we have,
2πR = 396 cm
⇒ R =
⇒ R = 63 cm
Thus, the area of the bigger circle = πR2
=
= 12474 cm2
Also, given that the circumference of the smaller circle is 374 cm
⇒ 2πr = 374
⇒ r =
⇒ r = 59.5 cm
Thus, the area of the smaller circle = πr2
=
= 11126.5 cm2
Thus the area of the shaded portion = 12474 - 11126.5 = 1347.5 cm2
Question 20
In the given figure, the area of the shaded portion is 770 cm2. If the circumference of the outer circle is 132 cm, find the width of the shaded portion.
From the given data, we can calculate the area of the outer circle and then the area of the inner circle and hence the width of the shaded portion.
Given that the circumference of the outer circle is 132 cm
Thus, we have, 2πR = 132 cm
⇒ R =
⇒ R = 21 cm
Area of the bigger circle = πR2
=
= 1386 cm2
Also given the area of the shaded portion.
Thus the area of the inner circle = Area of the outer circle - Area of the shaded portion
= 1386 - 770
= 616 cm2
⇒ πr2 = 616
⇒ r2 =
⇒ r2 = 196
⇒ r = 14 cm
Thus, the width of the shaded portion = 21 - 14 = 7 cm.
Question 21
The cost of fencing a circular field at the rate of ₹ 240 per meter is ₹ 52,800. The field is to be ploughed at the rate of ₹ 12.50 per m2. Find the cost of pouching the field.
Sol:Let the radius of the field is r meter.
Therefore circumference of the field will be 2πr meter.
Now the cost of fencing the circular field is 52,800 at rate 240 per meter.
Therefore,
2πr. 240 = 52800
r =
r = 35
Thus the radius of the field is 35 meter.
Now the area of the field will be :
πr2 =
Thus the cost of ploughing the field will be:
3850 x 12.5 = 48,125 rupees.
Question 22
Two circles touch each other externally. The sum of their areas is 58π cm2 and the distance between their centers is 10 cm. Find the radii of the two circles.
Sol:Let r and R be the radius of the two circles.
r + R = 10 ....(1)
πr2 + πR2 = 58π .....(2)
Putting the value of r in (2)
r2 + R2 = 58
( 10 - R )2 + R2 = 58
100 - 20R + R2 + R2 = 58
2R2 - 20R + 42 = 0
R2 - 10R + 21 = 0
( R - 3 )( R - 7 ) = 0
R = 3, 7.
Hence the radius of the two circles is 3cm and 7cm respectively.
Question 23
The given figures show a rectangle ABCD inscribed in a circle as shown alongside.
If AB = 28 cm and BC = 21 cm, find the area of the shaded portion of the given figure.
From the figure,
AB = 28 cm
BC = 21 cm
AC =
=
= 35.
Hence diameter of the circle is 35 cm and hence
Area = π x
Area of the rectangle = 28 x 21 = 588 cm2
Hence area of the shaded portion is given by
A = 962 - 588 = 374.5 cm2
Question 24
A square is inscribed in a circle of radius 7 cm. Find the area of the square.
Sol:Since the diameter of the circle is the diagonal of the square inscribed in the circle.
Let a be the length of the sides of the square.
Hence,
a =
a2 = 98
Hence the area of the square is 98 sq.cm.
Question 25
A metal wire, when bent in the form of an equilateral triangle of largest area, encloses an area of 484
Let 'a' be the length of each side of an equilateral triangle formed.
Now, the area of an equilateral triangle formed = 484√3 cm2
⇒
⇒ a2 = 4 x 484
⇒ a = 2 x 22 = 44 cm
Then, perimeter of equilateral triangle = 3a = 3 x 44 = 132 cm
Now, length of wire = perimeter of equilateral triangle = circumference of circle
⇒ circumference of circle = 132 cm
⇒ 2πr = 132 .....( r is radius of circle )
⇒ r =
∴ Area of circle = πr2 =
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