Chapter 21.7 - Length of Arc - Solved examples

In the previous section we have seen the central angle of an arc for each 1 cm length. In this section we will see some solved examples.

Solved example 21.16

In a circle, the length of an arc is 3π cm. The central angle of this arc is 40^o. What is the perimeter of the circle? What is it's radius?
Solution:

1. Both values of the arc is given to us:
• Length of arc = 3π cm
• Central angle of arc = 40^o
2. We need an equation which gives the relation between the two. We can use theorem 21.1
According to the theorem, For every 1^o central angle, the length of arc will be (^πr⁄₁₈₀) cm 
3. So for 40^o, the length of arc will be (^πr⁄₁₈₀) × 40 = (^πr⁄_4.5) cm
4. We can equate it to the given length. So we get: (^πr⁄_4.5) = 3π cm ⇒ r = 3 × 4.5 = 13.5 cm
5. Perimeter = 2πr = 2 × 13.5 × π =  27π cm

Solved example 21.17
In a circle, the length of an arc is 4 cm. It's central angle is 25^o
(i) In the same circle, what is the length of an arc whose central angle is 75^o?
(ii) In a circle of radius one and a half times the radius of this circle, what is the length of an arc whose central angle is 75^o?
Solution:
1. Both values of the arc is given to us:
• Length of arc = 4 cm
• Central angle of arc = 25^o
2. We need an equation which gives the relation between the two. We can use theorem 21.1
According to the theorem, For every 1^o central angle, the length of arc will be (^πr⁄₁₈₀) cm 
3. So for 25^o, the length of arc will be (^πr⁄₁₈₀) × 25 = (^πr⁄_7.2) cm
4. We can equate it to the given length. So we get: (^πr⁄_7.2) = 4 cm ⇒ πr = 4 × 7.2 = 28.8 cm
• We need not divide 28.8 by π to get the actual radius because, we will be using 'πr' as a whole in our calculations.
Part (i)
1. We use theorem 21.1 again:
For 75^o, the length of arc =  (^πr⁄₁₈₀) × 75 = (^28.8⁄₁₈₀) × 75 = 12 cm
Part (ii)
1. The radius is one and a half times. So new πr = 28.8 ×1.5 = 43.2 cm
For 75^o, the length of arc =  (^πr⁄₁₈₀) × 75 = (^43.2⁄₁₈₀) × 75 = 18 cm

Solved example 21.18
From a bangle of radius 3 cm, a piece is to be cut out to make a ring of radius ¹⁄₂ cm. 
(i) What is the central angle of the piece to be cut out?
(ii) The remaining part of the bangle was bent to make a smaller bangle. What is it's radius?
Solution:
1. When a piece is cut out from a bangle, it would be an arc. What is the length of this arc?
Ans: Enough to make a ring of radius ¹⁄₂ cm 
2. So, if r is the radius of the ring, it's perimeter 2πr must be equal to the length of the arc cut out
3. But radius of the ring is given as ¹⁄₂ cm. So length of cut out arc = 2π × ¹⁄₂ = π cm
4. So we have the length of arc. From that, we need to find the central angle. We can use theorem 21.2.
5. For every 1 cm of an arc on a circle of radius r, the central angle will be (¹⁸⁰⁄_πr)^o
6. So for π cm, the angle will be (¹⁸⁰⁄_πr) × π = (¹⁸⁰⁄_r)^o .
7. But r is given as 3 cm. So angle = (¹⁸⁰⁄₃) = 60^o. This is the answer for part (i)
8. π cm is cut out. So remaining arc length = 2πr - π = (2r-1)π = (2×3 - 1)π = 5π
9. So the smaller bangle is made using an arc of length 5π. That means, perimeter of the smaller bangle = 5π
10. Let r₁ be the radius of the smaller bangle. Then it's perimeter = 2πr₁
11. So we get: 2πr₁= 5π ⇒ r₁ = 2.5 cm. This is the answer for part (ii)

Solved example 21.19

In fig.21.33(a) below, parts of a circle centred at each vertex of an equilateral triangle, and passing through the other two vertices is shown. 

Fig.21.33

What is the perimeter of this fig.?

Solution:

1. In fig.21.33(b), more details are added. ABC is the equilateral triangle. 

2. A circle is drawn centred at each vertex. What is the radius of those circles?

Ans: Any of the above three circles centred on a vertex, passes through the other two vertices. For example, in the fig.b, the circle centred at A passes through B and C.

Also, it is an equilateral triangle. So radius of any circle is the side of the triangle, and is equal to 4 cm.

3. An arc is taken out between the 'other two vertices', from each circle. 

4. For an equilateral triangle, all three angles are 60^o

5. So we have 3 equal arcs

• Each of them have a radius of 4 cm. (∵ they are part of a circle with 4 cm radius)

• Each of them have central angle 60^o.  

6. We want the length of these arcs. We can use  theorem 21.2.

According to the theorem, For every 1^o central angle, the length of arc will be (^πr⁄₁₈₀) cm 
7. So for 60^o, the length of arc will be (^π×4⁄₁₈₀) × 60 = (⁴⁄₃)π cm
8. Thus the total perimeter = 3×(⁴⁄₃)π = 4π cm

Solved example 21.20

Parts of a circle are drawn, centred at each vertex of a regular octagon, and a fig. is cut out as in fig.21.34(b) below. Calculate the perimeter of fig.b

Fig.21.34

Solution:
1. The measure of each interior angle of a regular polygon is (^[180(n-2)]⁄_n). Where n is the number of sides.
2. We are given a regular octagon. It has 8 sides. So each interior angle is equal to:
(^[180(8-2)]⁄₈) = (^180×6⁄₈) = 135^o.
3. We want to find the perimeter in fig.b. It consists of equal arcs. How are these arcs formed?
We get the answer from fig.c
Equal circles are centred at each vertex. Then the portions outside the octagon are removed.
Radius of each circle is 1 cm

4. So we have 8 equal arcs

• Each of them have a radius of 1 cm. (∵ they are part of a circle with 1 cm radius)

• Each of them have central angle 135^o.  

5. We want the length of these arcs. We can use  theorem 21.2.

According to the theorem, For every 1^o central angle, the length of arc will be (^πr⁄₁₈₀) cm 
6. So for 135^o, the length of arc will be (^π×1⁄₁₈₀) × 135 = (³⁄₄)π cm
7. Thus the total perimeter = 8×(³⁄₄)π = 6π cm

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In the next section we will see Area of Sector.

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Chapter 21.6 - Central angle from Length of Arc

In the previous section we have seen the length of an arc for each 1^o turn. In this section we will see the converse. That is., we can find the 'angle turned' for each 1 cm of the arc. We can find this from the same four cases. We will use the same fig.21.30. For convenience, it is shown again below:

Fig.21.30

Case 1:

1. In fig.21.30(b), when a point travels from P to Q along the minor arc PQ, the distance covered is one fourth of the total perimeter of the ‘circle with radius r cm’.

2. We know that the total perimeter = 2πr cm. So the distance covered = (¹⁄₄ × 2πr) cm

3. Also, when the above distance is covered, we can say, the point ‘turns’ through an angle of 90^o

4. So we can write: (¹⁄₄ × 2πr) cm → 90^o

5. So 1 cm → [90 ÷ (¹⁄₄ × 2πr)]

6. [90 ÷ (¹⁄₄ × 2πr)] = ^(90×4)⁄_2πr = ¹⁸⁰⁄_πr

7. So 1 cm → ¹⁸⁰⁄_πr. That means, when the point travels a distance of 1 cm along the arc, the angle turned is (¹⁸⁰⁄_πr)^o 
8. Another form of writing this is: In a circle of radius r, an arc of length 1 cm will subtend an angle of (¹⁸⁰⁄_πr)^o at the centre.

Case 2:

1. When a point travels from P to R along the arc PQR, the distance covered is one half of the total perimeter of the ‘circle with radius r cm’.

2. We know that the total perimeter = 2πr. So the distance covered = (¹⁄₂ × 2πr) cm

3. Also, when the above distance is covered, we can say, the point ‘turns’ through an angle of (90+90) = 180^o

4. So we can write: (¹⁄₂ × 2πr) cm → 180^o

5. So 1 cm → [180 ÷ (¹⁄₂ × 2πr)] 

6. [180 ÷ (¹⁄₂ × 2πr)] =  ^(180×2)⁄_2πr = ¹⁸⁰⁄_πr
7. So 1 cm → ¹⁸⁰⁄_πr. That means, when the point travels a distance of 1 cm along the arc, the angle turned is (¹⁸⁰⁄_πr)^o 
8. Another form of writing this is: In a circle of radius r, an arc of length 1 cm will subtend an angle of (¹⁸⁰⁄_πr)^o at the centre.
Case 3:

1. When a point travels from P to S along the arc PQRS, the distance covered is three fourth of the total perimeter of the ‘circle with radius r cm’.

2. We know that the total perimeter = 2πr. So the distance covered = (³⁄₄ × 2πr)

3. Also, when the above distance is covered, we can say, the point ‘turns’ through an angle of (90+90+90) = 270^o

4. So we can write: (³⁄₄ × 2πr) cm → 270^o

5. So 1 cm → [270 ÷ (³⁄₄ × 2πr)]

6. [270 ÷ (³⁄₄ × 2πr)] =  ^(270×4)⁄_6πr = ¹⁸⁰⁄_πr
7. So 1 cm → ¹⁸⁰⁄_πr. That means, when the point travels a distance of 1 cm along the arc, the angle turned is (¹⁸⁰⁄_πr)^o 
8. Another form of writing this is: In a circle of radius r, an arc of length 1 cm will subtend an angle of (¹⁸⁰⁄_πr)^o at the centre.
Case 4:

1. When a point travels from P, and returns back to P along the arc PQRSP, the distance covered is one full of the total perimeter of the ‘circle with radius r cm’.

2. We know that the total perimeter = 2πr. So the distance covered = (2πr) cm

3. Also, when the above distance is covered, we can say, the point ‘turns’ through an angle of (90+90+90+90) = 360^o

4. So we can write: (2πr) cm → 360^o

5. So 1 cm → [360 ÷ (2πr)]

6. [360 ÷ (2πr)] == ¹⁸⁰⁄_πr
7. So 1 cm → ¹⁸⁰⁄_πr. That means, when the point travels a distance of 1 cm along the arc, the angle turned is (¹⁸⁰⁄_πr)^o 
8. Another form of writing this is: In a circle of radius r, an arc of length 1 cm will subtend an angle of (¹⁸⁰⁄_πr)^o at the centre.
■ In all the four cases, we get the same result. We can write it in the form of a theorem:

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Theorem 21.2:
• A point is situated on the circumference of a circle with radius r
• It travels a distance of 1 cm along the circumference of the circle
• Then the angle turned by the point is (¹⁸⁰⁄_πr)^o.
• Another form of writing this is: If the radius of a circle is r cm, then an arc of length l cm will subtend an angle of (¹⁸⁰⁄_πr)^o at the centre.
• So, if the length of an arc (in a circle of radius r) is 'l' cm, the angle subtended by that arc at the centre
= (¹⁸⁰⁄_πr × l)^o = (^180l⁄_πr)^o. 

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Based on the above theorem, we can write:
In fig.21.30(a), if the length of the arc AB is l, then x = (^180l⁄_πr)^o. 

A sample calculation:
In a circle of radius 3 cm, length of an arc is 2.5 cm. What is the central angle of the arc?
Solution:
1. According to theorem 21.2 above, In a circle of radius r, for every 1 cm length of an arc, the central angle will be equal to (¹⁸⁰⁄_πr)^o.
2. So in a circle of radius 3 cm, an arc of length 2.5 cm will make a central angle of
[(¹⁸⁰⁄_3π) × 2.5] = [(⁶⁰⁄_π) × 2.5] = (¹⁵⁰⁄_π) = 47.77^o

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The two theorems can be represented in diagrams as shown in the fig.21.31 below. Fig.a represents theorem 21.1 and fig.b represents theorem 21.2

Fig.21.31

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Now we will see a very interesting case:

• In the fig.21.32, two circles with radii r₁ cm and r₂ cm are drawn. 

• An arc AB is marked in the inner circle. And an arc PQ is marked in the outer circle.

Fig.21.32

• Both arcs have the same central angle x^o. Using this central angle, we can calculate their lengths:

1. We have: Length of arc AB (Theorem 21.1) = (^πr₁x⁄₁₈₀) cm

2. Length of arc PQ = (^πr₂x⁄₁₈₀) cm

3. We find that the lengths are different. Let us compare them: (^πr₁x⁄₁₈₀) and (^πr₂x⁄₁₈₀) cm

4. The only difference is in the radius. r₂ is greater than r₁. So the length of arc PQ will be greater than the length of arc AB.

So we can write: 
■ Two arcs may have the same central angle. But they may have different lengths. Out of the two arcs, the one which gave a greater radius will have a greater length.

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In the next section we will see some solved examples.

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Chapter 21.5 - Length of Arc

In the previous section we completed the discussion on Area of circles. In this section we will see Length of Arcs.

Consider the circle in fig.21.29(a) below. It has the centre at ‘O’, and a radius ‘r’ cm. 

Fig.21.29

• A point travels from point A to point B along the circumference of the circle. We want to find the distance travelled by the point.

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■ Any part of a circle between two points on it, is called an arc. 
• So AB is an arc. We denote it as 'arc AB'. 
• But we notice that, just saying arc AB can cause confusion. For example, in fig.b, the green portion, as well as the yellow portion, can be called arc AB. To avoid such a confusion, we mark two more points ‘C’ and ‘D’ on the circle. Then we get two separate arcs: arc ADB and arc ACD. 
• arc ACD is called the major arc and arc ABD is called the minor arc.

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We want the length of the minor arc AB in cm. For that, we need the help of some angle measures. 
In fig.21.30(a), the starting point A is joined to the centre O. 

Fig.21.30

The end point B is also joined to the centre O.

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■ Line joining a point on the circle or arc to the centre is called a radial line

■ The angle between the two radial lines drawn through the end points of an arc is called the central angle of the arc

■ This angle is also known as the ‘angle subtended by the arc at the centre’

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• Let x^o be the central angle of our arc AB. We can say that, when a point travels from A to B along the circumference of the circle, it ‘turns’ through an angle of x^o. We are going to find the length of the arc AB with the help of this x^o

• Look at fig.21.29(b). The circle with radius ‘r’ is divided into four equal parts using two green lines. So P, Q, R and S are the quadrant points. Let us see four cases:

Case 1:

1. When a point travels from P to Q along the minor arc PQ, the distance covered is one fourth of the total perimeter of the ‘circle with radius r cm’.

2. We know that the total perimeter = 2πr. So the distance covered = (¹⁄₄ × 2πr)

3. Also, when the above distance is covered, we can say, the point ‘turns’ through an angle of 90^o

4. So we can write: 90^o → (¹⁄₄ × 2πr) 

5. So 1^o → [(¹⁄₄ × 2πr) ÷ 90]

6. [(¹⁄₄ × 2πr) ÷ 90] = ^2πr⁄_(4×90) = ^πr⁄₁₈₀

7. So 1^o → ^πr⁄₁₈₀. That means, when the point turns through 1^o, the distance covered along the circumference is ^πr⁄₁₈₀ cm. 
8. Another form of writing this is: If an arc of a circle of radius r, subtends an angle of 1^o at the centre, the length of that arc is ^πr⁄₁₈₀ cm.   
Case 2:

1. When a point travels from P to R along the arc PQR, the distance covered is one half of the total perimeter of the ‘circle with radius r cm’.

2. We know that the total perimeter = 2πr. So the distance covered = (¹⁄₂ × 2πr)

3. Also, when the above distance is covered, we can say, the point ‘turns’ through an angle of (90+90) = 180^o

4. So we can write: 180^o → (¹⁄₂ × 2πr) 

5. So 1^o → [(¹⁄₂ × 2πr) ÷ 180]

6. [(¹⁄₂ × 2πr) ÷ 180] = ^2πr⁄_(2×180) = ^πr⁄₁₈₀
7. So 1^o → ^πr⁄₁₈₀. That means, when the point turns through 1^o, the distance covered along the circumference is ^πr⁄₁₈₀ cm
8. Another form of writing this is: If an arc of a circle of radius r, subtends an angle of 1^o at the centre, the length of that arc is ^πr⁄₁₈₀ cm.
Case 3:

1. When a point travels from P to S along the arc PQRS, the distance covered is three fourth of the total perimeter of the ‘circle with radius r cm’.

2. We know that the total perimeter = 2πr. So the distance covered = (³⁄₄ × 2πr)

3. Also, when the above distance is covered, we can say, the point ‘turns’ through an angle of (90+90+90) = 270^o

4. So we can write: 270^o → (³⁄₄ × 2πr) 

5. So 1^o → [(³⁄₄ × 2πr) ÷ 270]

6. [(³⁄₄ × 2πr) ÷ 270] = ^6πr⁄_(4×270) = ^πr⁄₁₈₀
7. So 1^o → ^πr⁄₁₈₀. That means, when the point turns through 1^o, the distance covered along the circumference is ^πr⁄₁₈₀ cm
8. Another form of writing this is: If an arc of a circle of radius r, subtends an angle of 1^o at the centre, the length of that arc is ^πr⁄₁₈₀ cm.
Case 4:

1. When a point travels from P, and returns back to P along the arc PQRSP, the distance covered is one full of the total perimeter of the ‘circle with radius r cm’.

2. We know that the total perimeter = 2πr. So the distance covered = (2πr)

3. Also, when the above distance is covered, we can say, the point ‘turns’ through an angle of (90+90+90+90) = 360^o

4. So we can write: 360^o → (2πr) 

5. So 1^o → [(2πr) ÷ 360]

6. [(2πr) ÷ 360] = ^2πr⁄₍₃₆₀₎ = ^πr⁄₁₈₀
7. So 1^o → ^πr⁄₁₈₀. That means, when the point turns through 1^o, the distance covered along the circumference is ^πr⁄₁₈₀ cm
8. Another form of writing this is: If an arc of a circle of radius r, subtends an angle of 1^o at the centre, the length of that arc is ^πr⁄₁₈₀ cm.
■ In all the four cases, we get the same result. We can write it in the form of a theorem:

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Theorem 21.1:
• A point is situated on the circumference of a circle with radius r
• It turns through an angle of 1^o about the centre of the circle
• Then the distance travelled by it along the circumference of the circle is ^πr⁄₁₈₀ cm
• Another form of writing this is: If an arc of a circle of radius r, subtends an angle of 1^o at the centre, the length of that arc is ^πr⁄₁₈₀ cm.
• So, if an arc in a circle of radius r, subtends an angle of x^o at the centre, the length of that arc
= (^πr⁄₁₈₀ × x) = ^πrx⁄₁₈₀ cm. 

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Based on the above theorem, we can write:
In fig.21.30(a), the length of the arc AB = ^πrx⁄₁₈₀ cm

A sample calculation:
In a circle of radius 3 cm, an arc subtends an angle of 60^o at the centre. What is the length of the arc?
Solution:
1. According to theorem 21.1 above, for every 1^o turn, the length of arc on a circle of radius r will be equal to ^πr⁄₁₈₀ cm..
2. So for 60, the length of arc on a circle of radius 3 cm = ^(3π×60)⁄₁₈₀ cm = π cm = 3.14 cm

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So we have seen the length of an arc for each 1^o turn. We can find the converse also. That is., we can find the 'angle turned' for each 1 cm of the arc. We can find this from the same four cases above. We will see it in the next section.

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Chapter 21.4 - Area of Circles - Solved examples

In the previous section we derived the formula for the Area of a circle. We also saw a solved example. In this section we will see a few more solved examples.

Solved example 21.12

(i) Fig.21.24(a) below shows the circle through the vertices of a square. Calculate the area of the circle

(ii) Fig.21.24(b) below shows the circle through the vertices of a rectangle. Calculate the area of the circle

Fig.21.24

Solution:

Part (i):

1. In fig.21.25(a) below, we can see that, the diagonal splits the square into two right triangles.

Fig.21.25

2. Applying Pythagoras theorem we get:

diagonal² = 3² + 3²
⇒ diagonal² = 9 + 9  = 18
⇒ diagonal = √18 = √[9×2] = √9 × √2 = 3√2
3. But diagonal is same as the diameter of the circle. So we get diameter d = 3√2 cm. So radius = (³⁄₂)√2
4. So area of the circle = πr² = π[(³⁄₂)√2]² = (⁹⁄₂)π
Part (ii)

1. In fig.21.25(b) above, we can see that, the diagonal splits the rectangle into two right triangles.

2. Applying Pythagoras theorem we get:
diagonal² = 4² + 2²
⇒ diagonal² = 16 + 4  = 20
⇒ diagonal = √20 = √[4×5] = √4 × √2 = 2√2
3. But diagonal is same as the diameter of the circle. So we get diameter d = 2√2 cm. So radius = √2
4. So area of the circle = πr² = π(√2)² = 2π

Solved example 21.13

Draw a square, and draw circles centred on each of it's four corners (fig.21.26.a). The radius of each of the circle must be equal to half the side of the square. 

Fig.21.26

Draw a second square (fig.21.26.b) formed by four of the first square. Draw a circle inside the second square. Prove that area of the large circle is equal to the sum of the areas of the four small circles.

Solution:

1. Let the radius of the small circles in fig.a be 'r'. Then the side of the square in fig.a will be equal to 2r

2. So the side of the large square in fig.b will be equal to 4r. 

3. Thus the diameter of the large circle in fig.b = 4r. 

4. So radius of the large circle in fig.b = 2r

5. Area of the large circle in fig.b = π(2r)² = 4πr²

6. Area of each of the small circles in fig.a = πr²

7. Total area of the 4 small circles in fig.a = 4 × πr² = 4πr²

8. Result in (5) = result in (7). Hence proved

Solved example 21.14

In fig.21.27 below, the squares in figs. (a) and (b) are of the same size. 

Fig.21.27

Prove that the green regions are of the same area.

Solution:

1. There are 4 equal circles in fig.a. One at each corner of the square. These are shown in fig.c. Let the radius of these circles be 'r'

2. So the area of each of these circles = πr²

3. Each of these circles contribute only one fourth of it's area in side the square.

4. That means., contribution from each circle = ¹⁄₄ × πr²

5. There are 4 such contributions. So total contribution = 4 × ¹⁄₄ × πr² = πr² 

6. So the green area in fig.c = area of square – total contribution from circles 

= (2r × 2r) - πr² = 4r² - πr² = (4 - π)r²

7. Now we take up the square in fig.b

8. The size of the square is the same. So area of the square in fig.b = 4r²

9. Area of the circle in fig.b:

Diameter of the circle = 2r ⇒ radius = r ⇒ area = πr²

10. So green area in fig.b = (8) - (9) = 4r² - πr² = (4 - π)r² 

11. Result in (6) = result in (10). Hence proved

Solved example 21.15

In the fig.21.28(a) below, parts of circles are drawn inside a square.

Fig.21.28

Prove that, the area of the green region is half the area of the square

Solution:

1. The green region in fig.a is split up into two parts in fig.b. The two parts are distinguished by giving a lighter green shade to the upper part.

2. Now we can see that

• half of a circle occupies the upper part

• two 'quarter circles' occupy the lower part.

3. Let the radius of each of the lower circles be 'r'. 

4. Then diameter of the upper circle = 2r. So the radius of the upper circle is also the same 'r'

5. The upper and lower parts are shown separately in fig.c

6. The green region in the upper part in fig.c = half of 'area of a circle with radius r' = ¹⁄₂ × πr²

7. The yellow region in the lower part in fig.c = two times 'quarter of a circle with radius r' 

= 2 × ¹⁄₄ × πr² = ¹⁄₂ × πr² 

8. So green region in the lower part in fig.c = (r × 2r) - (¹⁄₂ × πr²) = (2r²) - (¹⁄₂ × πr²)

9. So total green region in fig.c = total green region in fig.a 

= (6) + (8) = (¹⁄₂ × πr²) + (2r²) - (¹⁄₂ × πr²) = 2r²

10. Total area of the square in fig.a = 2r × 2r = 4r²

11. Half of the above area = 2r²

12. Result in (9) = result in (11). Hence proved

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In the next section we will see Length of Arcs.

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