In the previous section we completed the discussion on Area of sectors. In this section we will see some solved examples related to this chapter in general.
Solved example 21.25
Perimeter of two circles are in the ratio 2:3
(i) What is the ratio of their radii?
(ii) What is the ratio of their areas?
Solution:
1. Let the radii of the two circles be r1 and r2. Then the perimeters are 2πr1 and 2πr2.
2. Given that 2πr1 : 2πr2 = 2:3
3. Dividing both sides by 2π we get r1 : r2 = 2:3 [This is the answer for part(i)]
4. Ratio of areas = πr12 : πr22
5. Dividing both sides of the ratio by π, we get:
Ratio of areas = r12 : r22
6. From (3) we have: r1⁄r2 = 2⁄3 ⇒ r1 = 2r2⁄3 . Substituting this value of r1 in (5), we get:
7. Ratio of areas = (2r2⁄3)2 : r22
8. Dividing both sides of the ratio by r22 we get:
Ratio of areas = 4⁄9 :1 ⇒ Ratio of areas = 4:9 [This is the answer for part(ii)]
Solved example 21.26
A wheel of radius 20 cm rolls along. How much would it move ahead after 10 rotations?
Solution:
Consider the circle in fig.21.43 below. The circle represents a wheel
1. Position (1) shows the position when the wheel is just about to roll to the right. At that moment, the point of contact of the wheel with the ground is marked as 'A'.
2. When the wheel rolls, point A moves along the circumference of the wheel. When A touches the ground again, one rotation is completed. This is shown as position (2)
3. So the distance travelled after one rotation is equal to the perimeter of the wheel
4. So the distance travelled after 10 rotations = 10 × perimeter = 10 × 2πr = 20πr
5. In this problem, r is given as 20 cm. So the distance travelled = 20 × 3.14 × 20 = 1256 cm
Solved example 21.27
Two semicircular pieces are cut off from a rectangular sheet as shown in the fig.21.44 below.
Find the area of the remaining part
Solution:
1. Total area = area of the rectangle = 50 × 20 = 1000 cm2.
2. Consider any one semicircular part. It reaches from the top edge to bottom edge of the rectangle.
3. So the diameter of the semicircle = height of the rectangle = 20 cm
4. So radius = 10 cm
5. Area of the semicircular part = half the area of a circle with 10 cm radius
= 1⁄2 × πr2 = 1⁄2 × π × 102 = 50π cm2
6. There are two semicircular parts. Their total area = 2 × 50π = 100π cm2.
7. So area of the remaining part = 1000 - 100π = 1000 - (100×3.14) = 1000 - 314 = 686 cm2
Solved example 21.28
The length of an arc of a circle is 4 cm. What is the length of an arc with the same central angle, in a circle of double the radius?
Solution:
1. We are given the length of an arc. No other details are given. So let us assume the required details:
Let the radius be r, and let the central angle be x
2. So length of arc will be equal to (πr⁄180 × x) cm
3. But the length of arc is given as 4 cm. So we can write:
(πr⁄180 × x) = 4
4. Now we are given another arc. It has the same central angle. But radius = 2r. We want the length of this new arc
5. We can write: length of new arc = (π(2r)⁄180 × x)
⇒ length of new arc = 2 × (πr⁄180 × x)
6. But from (3), the quantity inside brackets is 4 cm. So we get:
7. length of new arc = 2 × 4 = 8 cm
Solved example 21.29
In the fig.21.45 below, the sector OAB has a radius of 6 cm, and central angle 60o.
Find the area of the blue region
Solution:
1. If a sector has central angle 60o, we can separate out an equilateral triangle from it.
Solved example 21.30
The area of a circular table is 31400 cm2. What is it's radius? And it's perimeter?
Solution:
1. Area of a circle = πr2 = 3.14 × r2 = 31400
⇒ r2 = 31400⁄3.14 = 10000 ⇒ r = √10000 = 100 cm
2. Perimeter = 2πr = 2 × 3.14 × 100 = 628 cm
Solved example 21.31
In fig.21.46 below, OPQR is a sector of a circle with centre O and radius 8 cm.
Find the area of the yellow region.
Solution:
1. The central angle of the sector OPQR is given as 90o. So this sector will cover one fourth of the area of the full circle. We can write:
Area of sector OPQR = 1⁄4 × πr2 = 1⁄4 × π × 82 = 16π cm2.
2. Now, OPR is a triangle. It is a right triangle. We have it's base and height. So area of ΔOPR
= 1⁄2 × b × h = 1⁄2 × 8 × 8 = 32 cm2.
3. So area of the yellow region = (16π - 32) = 16(π - 2) cm2.
Solved example 21.32
The length of the pendulum of a clock is 12 cm. It makes an angle of 45o in one swing. What is the distance travelled by the tip of the pendulum in on swing?
Solution:
1. Fig.21.47 below shows one swing of the pendulum
2. In one swing, the pendulum travels from P to Q. It is an arc of radius 12 cm, and central angle 45o.
3. So we want the length of arc PQ. We can use theorem 21.1
According to the theorem, For every 1o central angle, the length of arc will be (πr⁄180) cm
4. So for 45o, the length of arc will be (π×12⁄180) × 45 = (π×12⁄4) = 3π cm
Solved example 21.33
In the fig.21.48 below, 'O' is the centre of the circle, and OPQR is a rectangle.
OP = 8 cm and OR = 15 cm. Find the area of the yellow region
Solution:
1. Consider a diagonal OQ of the rectangle. It's length can be obtained as follows:
OQ2 = OP2 + PQ2
⇒ OQ2 = (8)2 + (15)2 = 6•4 + 225 = 289 (∵ PQ = OR = 15 cm)
⇒ OQ = √289 = 17
2. But OQ is the radius of the circle. So we have all the details required to find the area of the yellow region:
• It's radius = 17 cm
• It's central angle = 90o
3. The yellow region is one fourth of the total area of the circle. So it's area
= 1⁄4 × πr2 = 1⁄4 × π × 172 = 72.25π cm2.
Solved example 21.34
In the fig.21.49 below, both sectors OAB and OPQ have the same centre O, and the same central angle 45o.
The sum of the radii is 18 cm. Area of the shaded portion ABQP is 18π. Find the radius of both the sectors.
Solution:
1. If we subtract the area of the inner sector OPQ from the area of the outer sector, we will get the area of the shaded portion ABQP
2. Area of the inner sector OPQ = (πr2⁄360) × x = (π×OP2⁄360) × 45 = (π⁄8)×OP2 cm2
3. Area of the outer sector OAB = (πr2⁄360) × x = (π×OA2⁄360) × 45 = (π⁄8)×OA2 cm2.
4. So area of ABPQ = [(π⁄8)×OA2 - (π⁄8)×OP2 ] = (π⁄8)×[OA2 - OP2 ] cm2.
5. But area of ABPQ is given as 18π. So we can equate the two:
(π⁄8)×[OA2 - OP2 ] = 18π.
⇒[OA2 - OP2 ] = 18 × 8 = 144
⇒(OA-OP)(OA+OP) = 144 [∵ (a2 - b2) = (a+b)(a-b)]
6. But (OA+OP) is given as 18 cm. So we can write:
(OA-OP) × 18 = 144 ⇒ (OA-OP) = 8 cm. Now we have two equations:
7. OA + OP = 18
8. OA - OP = 8
9. From (8) we get: OA = 8 + OP
10. Substituting this value of OA in (7), we get:
8 + OP + OP = 18 ⇒ 8 + 2OP = 18 ⇒ 2OP = 10 ⇒ OP = 5 cm
11. Substituting this value of OP on (8) we get: OA = 8 + 5 = 13 cm
Solved example 21.35
Find the area of the yellow portion in fig.21.50 below
Solution:
1. Area of the green sector = (πr2⁄360) × x = (π×62⁄360) × 60 = 6π cm2
2. Total area = πr2 = π × 62 = 36π cm2.
3. So area of the yellow portion = 36π - 6π = 30π cm2.
Another method:
1. Central angle of the yellow portion = 360 -60 = 300o.
2. So area = (πr2⁄360) × x = (π×62⁄360) × 300 = 30π cm2
Solved example 21.36
Given a circular metal disc of perimeter 30 cm. A regular hexagon of maximum possible size is to be cut out from the disc. What will be the perimeter of the hexagon?
Solution:
Consider fig.21.51 below
1. A regular hexagon can be divided into 6 equilateral triangles by drawing it's three diagonals.
2. Each diagonal will be a diameter of the circle. So our first step is to find the diameter of the circle
3. Given that perimeter = 30 cm. That means 2πr = 30. So diameter = 2r = 30⁄π .
4. Each side of the equilateral triangle = half of diameter = radius = 1⁄2 × 30⁄π = 15⁄π .
5. So side of the regular hexagon = 15⁄π cm.
6. Thus perimeter of the regular hexagon = 6 × 15⁄π = 90⁄π cm
Solved example 21.37
In the fig.21.52 below, O is the centre of the circle, and AB is the diameter
The length of an arc with central angle 72o is 4π cm.
(i) Find the diameter of the circle.
(ii) Find the area of the green portion
Solution:
1. Given that: length of an arc with central angle 72o is 4π cm.
2. From this, we can find the radius: We can use theorem 21.1. According to the theorem, For every 1o central angle, the length of arc will be (πr⁄180) cm
3. So for 72o, the length of arc will be (π×r⁄180) × 72 = (π×r⁄5) × 2 cm
4. But this length is given as 4π cm. So we can equate the two:
(π×r⁄5) × 2 cm = 4π. ⇒ r = 10 cm
5. So diameter of the circle = 20 cm. This is the answer of part (i)
6. AB is the diameter. So central angle of the yellow portion = 180o.
7. So central angle of the green portion = [360 -(180 + 72)] = 108o.
8. So area of the green portion = (πr2⁄360) × x = (π×102⁄360) × 108 = 30π cm2. This is the answer of part (ii)
Solved example 21.25
Perimeter of two circles are in the ratio 2:3
(i) What is the ratio of their radii?
(ii) What is the ratio of their areas?
Solution:
1. Let the radii of the two circles be r1 and r2. Then the perimeters are 2πr1 and 2πr2.
2. Given that 2πr1 : 2πr2 = 2:3
3. Dividing both sides by 2π we get r1 : r2 = 2:3 [This is the answer for part(i)]
4. Ratio of areas = πr12 : πr22
5. Dividing both sides of the ratio by π, we get:
Ratio of areas = r12 : r22
6. From (3) we have: r1⁄r2 = 2⁄3 ⇒ r1 = 2r2⁄3 . Substituting this value of r1 in (5), we get:
7. Ratio of areas = (2r2⁄3)2 : r22
8. Dividing both sides of the ratio by r22 we get:
Ratio of areas = 4⁄9 :1 ⇒ Ratio of areas = 4:9 [This is the answer for part(ii)]
Solved example 21.26
A wheel of radius 20 cm rolls along. How much would it move ahead after 10 rotations?
Solution:
Consider the circle in fig.21.43 below. The circle represents a wheel
 |
| Fig.21.43 |
2. When the wheel rolls, point A moves along the circumference of the wheel. When A touches the ground again, one rotation is completed. This is shown as position (2)
3. So the distance travelled after one rotation is equal to the perimeter of the wheel
4. So the distance travelled after 10 rotations = 10 × perimeter = 10 × 2πr = 20πr
5. In this problem, r is given as 20 cm. So the distance travelled = 20 × 3.14 × 20 = 1256 cm
Solved example 21.27
Two semicircular pieces are cut off from a rectangular sheet as shown in the fig.21.44 below.
 |
| Fig.21.44 |
Solution:
1. Total area = area of the rectangle = 50 × 20 = 1000 cm2.
2. Consider any one semicircular part. It reaches from the top edge to bottom edge of the rectangle.
3. So the diameter of the semicircle = height of the rectangle = 20 cm
4. So radius = 10 cm
5. Area of the semicircular part = half the area of a circle with 10 cm radius
= 1⁄2 × πr2 = 1⁄2 × π × 102 = 50π cm2
6. There are two semicircular parts. Their total area = 2 × 50π = 100π cm2.
7. So area of the remaining part = 1000 - 100π = 1000 - (100×3.14) = 1000 - 314 = 686 cm2
Solved example 21.28
The length of an arc of a circle is 4 cm. What is the length of an arc with the same central angle, in a circle of double the radius?
Solution:
1. We are given the length of an arc. No other details are given. So let us assume the required details:
Let the radius be r, and let the central angle be x
2. So length of arc will be equal to (πr⁄180 × x) cm
3. But the length of arc is given as 4 cm. So we can write:
(πr⁄180 × x) = 4
4. Now we are given another arc. It has the same central angle. But radius = 2r. We want the length of this new arc
5. We can write: length of new arc = (π(2r)⁄180 × x)
⇒ length of new arc = 2 × (πr⁄180 × x)
6. But from (3), the quantity inside brackets is 4 cm. So we get:
7. length of new arc = 2 × 4 = 8 cm
Solved example 21.29
In the fig.21.45 below, the sector OAB has a radius of 6 cm, and central angle 60o.
 |
| Fig.21.45 |
Solution:
1. If a sector has central angle 60o, we can separate out an equilateral triangle from it.
2. Area after the removal of equilateral triangle from a sector of radius 'r' and central angle 60o
3. In our problem, r = 6 cm. So we get:
Area = [(π⁄6) - (√3⁄4)]62 = [6π - 9√3] cm2Solved example 21.30
The area of a circular table is 31400 cm2. What is it's radius? And it's perimeter?
Solution:
1. Area of a circle = πr2 = 3.14 × r2 = 31400
⇒ r2 = 31400⁄3.14 = 10000 ⇒ r = √10000 = 100 cm
2. Perimeter = 2πr = 2 × 3.14 × 100 = 628 cm
Solved example 21.31
In fig.21.46 below, OPQR is a sector of a circle with centre O and radius 8 cm.
 |
| Fig.21.46 |
Solution:
1. The central angle of the sector OPQR is given as 90o. So this sector will cover one fourth of the area of the full circle. We can write:
Area of sector OPQR = 1⁄4 × πr2 = 1⁄4 × π × 82 = 16π cm2.
2. Now, OPR is a triangle. It is a right triangle. We have it's base and height. So area of ΔOPR
= 1⁄2 × b × h = 1⁄2 × 8 × 8 = 32 cm2.
3. So area of the yellow region = (16π - 32) = 16(π - 2) cm2.
Solved example 21.32
The length of the pendulum of a clock is 12 cm. It makes an angle of 45o in one swing. What is the distance travelled by the tip of the pendulum in on swing?
Solution:
1. Fig.21.47 below shows one swing of the pendulum
 |
| Fig.21.47 |
3. So we want the length of arc PQ. We can use theorem 21.1
According to the theorem, For every 1o central angle, the length of arc will be (πr⁄180) cm
4. So for 45o, the length of arc will be (π×12⁄180) × 45 = (π×12⁄4) = 3π cm
Solved example 21.33
In the fig.21.48 below, 'O' is the centre of the circle, and OPQR is a rectangle.
 |
| Fig.21.48 |
Solution:
1. Consider a diagonal OQ of the rectangle. It's length can be obtained as follows:
OQ2 = OP2 + PQ2
⇒ OQ2 = (8)2 + (15)2 = 6•4 + 225 = 289 (∵ PQ = OR = 15 cm)
⇒ OQ = √289 = 17
2. But OQ is the radius of the circle. So we have all the details required to find the area of the yellow region:
• It's radius = 17 cm
• It's central angle = 90o
3. The yellow region is one fourth of the total area of the circle. So it's area
= 1⁄4 × πr2 = 1⁄4 × π × 172 = 72.25π cm2.
Solved example 21.34
In the fig.21.49 below, both sectors OAB and OPQ have the same centre O, and the same central angle 45o.
 |
| Fig.21.49 |
Solution:
1. If we subtract the area of the inner sector OPQ from the area of the outer sector, we will get the area of the shaded portion ABQP
2. Area of the inner sector OPQ = (πr2⁄360) × x = (π×OP2⁄360) × 45 = (π⁄8)×OP2 cm2
3. Area of the outer sector OAB = (πr2⁄360) × x = (π×OA2⁄360) × 45 = (π⁄8)×OA2 cm2.
4. So area of ABPQ = [(π⁄8)×OA2 - (π⁄8)×OP2 ] = (π⁄8)×[OA2 - OP2 ] cm2.
5. But area of ABPQ is given as 18π. So we can equate the two:
(π⁄8)×[OA2 - OP2 ] = 18π.
⇒[OA2 - OP2 ] = 18 × 8 = 144
⇒(OA-OP)(OA+OP) = 144 [∵ (a2 - b2) = (a+b)(a-b)]
6. But (OA+OP) is given as 18 cm. So we can write:
(OA-OP) × 18 = 144 ⇒ (OA-OP) = 8 cm. Now we have two equations:
7. OA + OP = 18
8. OA - OP = 8
9. From (8) we get: OA = 8 + OP
10. Substituting this value of OA in (7), we get:
8 + OP + OP = 18 ⇒ 8 + 2OP = 18 ⇒ 2OP = 10 ⇒ OP = 5 cm
11. Substituting this value of OP on (8) we get: OA = 8 + 5 = 13 cm
Solved example 21.35
Find the area of the yellow portion in fig.21.50 below
 |
| Fig.21.50 |
1. Area of the green sector = (πr2⁄360) × x = (π×62⁄360) × 60 = 6π cm2
2. Total area = πr2 = π × 62 = 36π cm2.
3. So area of the yellow portion = 36π - 6π = 30π cm2.
Another method:
1. Central angle of the yellow portion = 360 -60 = 300o.
2. So area = (πr2⁄360) × x = (π×62⁄360) × 300 = 30π cm2
Solved example 21.36
Given a circular metal disc of perimeter 30 cm. A regular hexagon of maximum possible size is to be cut out from the disc. What will be the perimeter of the hexagon?
Solution:
Consider fig.21.51 below
 |
| Fig.21.51 |
2. Each diagonal will be a diameter of the circle. So our first step is to find the diameter of the circle
3. Given that perimeter = 30 cm. That means 2πr = 30. So diameter = 2r = 30⁄π .
4. Each side of the equilateral triangle = half of diameter = radius = 1⁄2 × 30⁄π = 15⁄π .
5. So side of the regular hexagon = 15⁄π cm.
6. Thus perimeter of the regular hexagon = 6 × 15⁄π = 90⁄π cm
Solved example 21.37
In the fig.21.52 below, O is the centre of the circle, and AB is the diameter
 |
| Fig.21.52 |
(i) Find the diameter of the circle.
(ii) Find the area of the green portion
Solution:
1. Given that: length of an arc with central angle 72o is 4π cm.
2. From this, we can find the radius: We can use theorem 21.1. According to the theorem, For every 1o central angle, the length of arc will be (πr⁄180) cm
3. So for 72o, the length of arc will be (π×r⁄180) × 72 = (π×r⁄5) × 2 cm
4. But this length is given as 4π cm. So we can equate the two:
(π×r⁄5) × 2 cm = 4π. ⇒ r = 10 cm
5. So diameter of the circle = 20 cm. This is the answer of part (i)
6. AB is the diameter. So central angle of the yellow portion = 180o.
7. So central angle of the green portion = [360 -(180 + 72)] = 108o.
8. So area of the green portion = (πr2⁄360) × x = (π×102⁄360) × 108 = 30π cm2. This is the answer of part (ii)
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