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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