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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Chapter 21.3 - Derivation of the formula for Area of a circle

In the previous section we saw some solved examples related to the Perimeter of circle. In this section we will see Area of circle.

Consider fig.21.19 below. Some polygons are drawn inside circles. All the circles in the fig.21.19 are given yellow colour, and they all have the same diameter.

Fig.21.19

The polygons are given red colour. We can see that, when the number of sides 'n' increases, the yellow colour decreases. That means, the polygon covers more and more area of the circle. In other words, the area of the polygon becomes closer and closer to the area of the circle.

Just as we saw in the case of perimeter, if 'n' is very large, the area of the polygon will be approximately equal to the area of the circle. So we want to find the area of the regular polygon which has a very large 'n'. For finding it, we will first see the area of a regular polygon with a smaller 'n'. Say n = 5. That is., a pentagon. Fig.21.20(a) below shows a regular pentagon drawn inside a circle. 

Fig.21.20

The centre of the circle is joined to all the five corners. Thus we get 5 equal triangles. How are they equal?
They are equal because, they all have the same base and same sides. So we need only find the area of any one of those triangles.
1. Let the base of the triangle be 's'
2. Let the height be 'h'
[Note that, 's' forms a chord. The perpendicular bisector of this chord will pass through the centre of the circle. Thus we can easily draw the height 'h']
3. Now we have the base and height of the triangle. So area = ¹⁄₂ × s × h  
4. Area of the pentagon = Total area of the five triangles = 5 × ¹⁄₂ × s × h  
5. Let us group the '5' and 's' together. We will get:
6. Area of the pentagon = ¹⁄₂ × (5 × s) × h 
7. But (5 × s) is the perimeter of the pentagon. Let us denote this perimeter as p₅ (The subscript '5' denotes the five sides of the pentagon). So we get:
8. Area of the pentagon = ¹⁄₂ × p₅ × h 

Next consider the octagon in fig.21.20(b). We will get a similar result. Let us write the steps:
1. Let the base of the triangle be 's'
2. Let the height be 'h'
[Note that, 's' forms a chord. The perpendicular bisector of this chord will pass through the centre of the circle. Thus we can easily draw the height 'h']
3. Now we have the base and height of the triangle. So area = ¹⁄₂ × s × h  
4. Area of the octagon = Total area of the eight triangles = 8 × ¹⁄₂ × s × h  
5. Let us group the '8' and 's' together. We will get:
6. Area of the pentagon = ¹⁄₂ × (8 × s) × h 
7. But (8 × s) is the perimeter of the pentagon. Let us denote this perimeter as p₈ (The subscript '8' denotes the eight sides of the pentagon). So we get:
8. Area of the octagon = ¹⁄₂ × p₈ × h

In a similar way, we will get the area of a regular polygon of 'n' sides as: ¹⁄₂ × p_n × h  
1. If 'n' is very large, we have seen that, the perimeter of the regular polygon will be equal to the perimeter of the circle. If 'r' is the radius of the circle, it's perimeter = 2πr
2. So we can write:
When 'n' is very large, 
Area of the regular polygon = Area of the circle = ¹⁄₂ × 2πr × h = πrh
3. So now we have to find 'h'
Look again at the pentagon and octagon in fig.21.20 above. When the pentagon (n=5) became an octagon (n=8), the sides came closer to the periphery of the circle. If 'n' is increased further, 
• the sides will become closer and closer to the periphery
• the length of the chords 's' becomes smaller and smaller
• the height of the triangles 'h' will become larger and larger.
4. So, when 'n' is very large, the chord will lie almost on the periphery, and 'h' will be almost equal to 'r'. So we can put 'r' in the place of 'h'
5. Thus, from (2) we get: Area of the circle = πrh = πr²

A sample calculation:
Area of a circle with radius 4 cm = π × 4² = π × 16 = 16π cm²

Area of a ring

Consider a ring shown in the fig.21.21(a) below. It has an outer radius of r₁ and inner radius of r₂. How do we find the area of the ring?

Fig.21.21

In the fig.b, the inside hollow area is filled up by a green circle. Now it is clear that the required area is the difference between the 'area of the outer circle' and 'area of the inner circle'
1. Area of the outer circle = πr₁²
2. Area of the inner circle = πr₂² 
3. Area of the ring = πr₁² - πr₂² = π(r₁² - r₂²)

A sample calculation:
Area of a ring with outer diameter 2.5 cm and inner diameter 2.0 cm = π(r₁² - r₂²) = π(2.5² - 2.0²) 
=  π(6.25 - 4.0) =  π(2.25) = 2.25π cm²

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Now we will see some solved examples:
Solved example 21.11

(i) In fig.21.22(a) below, find the difference between the area of the circle and area of the square up to two decimal places

(ii) In fig.21.22(b) below, find the difference between the area of the circle and area of the hexagon up to two decimal places

Fig.21.22

Solution:

Part (i):

1. In fig.21.23(a) below, we can see that, the diagonal splits the square into two right triangles.

Fig.21.23

2. Let the sides of the square be ‘s’. Then, applying Pythagoras theorem, we get:

4² = s² + s²
⇒ 16 = 2(s)² ⇒ s²= 8
3. But s² is the area of the square. So we get:
4. Area of the square = s² = 8 cm²

5. Now we want the area of the circle. The diameter is given as 4 cm. So radius = 2 cm

6. Thus area = πr² = π2² = 4π = 4 × 3.14 = 12.56 cm²

7. Thus difference in area = 12.56 - 8 = 4.56 cm²

Part (ii):

1. In fig.21.23(b), all the three diagonals of the hexagon are drawn. This gives 6 equilateral triangles

2. The sides of the hexagon are chords of the circle. Perpendicular bisector of any chord will pass through the centre of the circle

3. One such perpendicular bisector is shown as dashed green line. It is perpendicular to the base, and so will become the altitude ‘h’ of the equilateral triangle. We have to find the value of this ‘h’

4. The altitude splits the equilateral triangle into two right triangles. Consider any one of the two.

Applying Pythagoras theorem, we get:

2² = 1² + h²
⇒ h² = (2)² - (1)² = 4 - 1 = 3
⇒ h = √3

5. Now, area of the equilateral triangle = ¹⁄₂ × b × h = ¹⁄₂ × 2 × √3 = √3 cm² 

6. So total area of the hexagon = 6 × √3 = 6 × 1.7321 = 10.39  cm²

7. In (6) of part (i), we have already calculated the area of a circle of 4 cm diameter as 12.56 cm²

8. So the difference = 12.56 - 10.39 = 2.17 cm².

• The answer for part (i) is 4.56

• The answer for part (i) is 2.17

• The diameters of the two circles are the same. 

• So we find that, when the number of sides of the regular polygon increases, it's area becomes closer to the area of the circle.

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In the next section we will see a few more solved examples.

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Chapter 21.2 - Perimeter of Circles - Solved examples

In the previous section we derived the formula for the perimeter of any circle. In this section we will see some solved examples.

Solved example 21.4
A wire was bent into a circle of 4 cm diameter. What would be the diameter of a circle made by bending a wire of half the length?
Solution:
1. Perimeter of a circle of 4 cm diameter = πd = π × 4 = 4π
2. So length of the wire = 4π
3. Half length of wire = ¹⁄₂ × 4π = 2π
4. Let d₁ be the diameter of the new circle. 
5. Then perimeter of the new circle = πd₁= 2π
From this we get d₁ = 2 cm

Solved example 21.5

The perimeter of a circle of diameter 2 m was measured and found out to be 6.28 m. How do we compute the perimeter of a circle of diameter 3 m, with out measuring?

Solution:
1. We are given two circles. One with diameter 2 m, and the other with diameter 3 m
2. Imagine that they are placed in such a way that, their centres are at the same point (see fig.below)

3. Then, we have:
• Perimeter of outer circle = k × perimeter of inner circle
• Where k = ^{outer diameter}⁄_{inner diameter}.
4. So we get: k = ³⁄₂.
5. Perimeter of the inner circle with 2 m radius is given to be 6.28 m
6. So Perimeter of outer circle with 3 m radius = ³⁄₂ × 6.28 = 9.42 m

Solved example 21.6

In the fig.21.12 below, the circles have the same centre and the line drawn is the diameter of the large circle.

Fig.21.12

What is the difference in the perimeter of the large and small circles?

Solution:

1. It is given that the line is the diameter of the large circle. So the line passes through the centre of the large circle.

2. Also it is given that, the circles have the same centre. So the line is the inner portion of the line is the diameter of the inner circle

3. Let d_i and d_o be the diameters of the inner and outer circles respectively.

4. Then d_o = (d_i +2) cm

5. Perimeter of the inner circle = πd = πd_i

6. Perimeter of the outer circle = πd = πd_o = π(d_i + 2)

7. So difference = π(d_i + 2) - πd_i = πd_i + 2π - πd_i = 2π 

Solved example 21.7

The perimeter of a regular hexagon, with vertices on a circle is 24 cm

(i) What is the perimeter of a square with vertices on this circle?

(ii) What is the perimeter of a square with vertices on a circle with double the diameter?

(iii) What is the perimeter of an equilateral triangle with vertices on a circle with half the diameter of the first circle?

Solution:

Fig.21.13(a) below shows the regular hexagon drawn inside a circle. It's perimeter is 24 cm.

Fig.21.13

So length of one side will be = ²⁴⁄₆ = 4 cm

1. Draw the three diagonals of the hexagon
2. The diagonals will interset at the centre O of the circle
3. The diagonals will give 6 equilateral triangles
4. Consider any one of those 6 triangles. Say OAB
5. OB is half of diameter. 
6. But OB = length of one side of the hexagon = 4 cm
7. So total diameter = 4×2 = 8 cm
Part (i): 
1. Now, in fig.b, a square is drawn inside the same circle

2. Draw the two diagonals of the square.

3. Length of each of these diagonals are same as the diameter of the circle = 8 cm [from (7)]

4. The diagonals bisect each other at right angles. So OP = OQ = 4 cm

5. As the diagonals are perpendicular to each other, angle at O = 90^o

6. Consider any of the right triangles. Say ⊿POQ. Applying Pythagoras theorem we get:
PQ² = OP² + OQ²
⇒ PQ² = (4)² + (4)² = 16 + 16 = 32
⇒ PQ = √32 = √[16×2] = √16 × √2 = 4√2
Thus we get length of one side
7. Perimeter of the square = 4 × 4√2 = 16√2
Part (ii):
1. Now, the circle in fig.b, has a diameter of 16 cm

2. Draw the two diagonals of the square.

3. Length of each of these diagonals are same as the diameter of the circle = 16 cm

4. The diagonals bisect each other at right angles. So OP = OQ = 8 cm

5. As the diagonals are perpendicular to each other, angle at O = 90^o

6. Consider any of the right triangles. Say ⊿POQ. Applying Pythagoras theorem we get:
PQ² = OP² + OQ²
⇒ PQ² = (8)² + (8)² = 64 + 64 = 128
⇒ PQ = √128 = √[64×2] = √64 × √2 = 8√2
Thus we get length of one side
7. Perimeter of the square = 4 × 8√2 = 32√2
Part (iii)
In this part, we are placing an equilateral triangle inside a circle with a diameter half that of the first circle. Half of 8 is 4. So the diameter is 4 cm. This is shown in the fig.21.14 below:

Fig.21.14

We have only two information:

• The triangle is equilateral

• The diameter of the circle is 4 cm

With these, we have to calculate the perimeter of the triangle.

1. Let us draw two medians as in fig.b

• Median AE (E is the midpoint of BC)

• Median CD (D is the midpoint of AB)

Now look at the two medians carefully.

• The median CD is drawn from the vertex C to the midpoint D of the opposite side AB. This median will be perpendicular to the side AB. This is because ABC is an equilateral triangle. Also, CD passes through the midpoint D. So CD is the perpendicular bisector of AB

• Similarly, AE is the perpendicular bisector of BC.

So we can write:

The 'medians' in any equilateral triangle will serve another purpose also:

They will bisect the side perpendicularly. In other words, they are perpendicular bisectors also. 

• We know that the point of intersection of any two perpendicular bisectors of a triangle is it’s circumcentre. (details here)

2. So O is the circumcentre of the given equilateral triangle. Let us mark the radius. OB is the radius marked with a dashed green arrow. OB = half of diameter = ¹⁄₂ of 4 cm = 2 cm

3. Now we use one important property of medians: The point of intersection splits the medians in the ratio 2:1, measured from the vertex. (Theorem 18.6)

4. So we can write: OC:OD = 2:1

5. That means: If we divide CD into 3 equal parts, OC will constitute 2 such parts, and OD will constitute 1 such part

6. When we divide CD into 3 equal parts, each part will be ^CD⁄₃

7. Two such parts = ^2CD⁄₃

8. So OC = ^2CD⁄₃

9. But OC = radius = 2 cm

10. Substituting this in (8) we get: 2 cm = ^2CD⁄₃. So we get CD = 3 cm

11. Now, AD = ^AB⁄₂. But AB = AC (∵ ABC is an equilateral triangle)

12. So AD = ^AC⁄₂

13. Now consider the right triangle ADC

14. Applying Pythagoras theorem, we get: AC² = AD² + CD².

⇒ AC² = (^AC⁄₂)² + (3)². ⇒ AC² = (^AC2⁄₄) + 9
⇒ AC² - (^AC2⁄₄) = 9 ⇒ (^3AC2⁄₄) = 9
⇒ AC² = 12 ⇒ AC = √[4 × 3] = √4 × √3 = 2√3
15. Thus we got one side of the equilateral triangle.
So perimeter = 3 × 2√3 = 6√3

Solved example 21.8

In the figs.21.15(a), (b) and (c) below, a regular hexagon, square and a rectangle are drawn with their vertices on a circle.

Fig.21.15

Calculate the perimeter of each circle
Solution:
Part (i): Regular hexagon
1. In fig.21.16(a) below, it's three diagonals can be drawn, giving 6 inner triangles. 

Fig.21.16

)

2. All the 6 inner triangles are equilateral, and are also equal. So we can consider any one of them. Say OAB
3. OB is half of diameter. 
4. But OB = length of one side of the hexagon = 2 cm
5. So total diameter = 2×2 = 4 cm
6. So perimeter of the circle = πd = π × 4 = 4π cm
Part (ii): Square
1. In fig.21.16(b), it's two diagonals are drawn

2. Length of each of these diagonals are same as the diameter of the circle. Let this be equal to 'd' cm

3. The diagonals bisect each other at right angles. So OP = OQ = ^d⁄₂ cm

4. As the diagonals are perpendicular to each other, angle at O = 90^o

5. Consider any of the right triangles. Say ⊿POQ. Applying Pythagoras theorem we get:
PQ² = OP² + OQ²
⇒ PQ² = (^d⁄₂)² + (^d⁄₂)² = ^d2⁄₄ + ^d2⁄₄ = ^2d2⁄₄ = ^d2⁄₂. 
⇒ 2² = ^d2⁄₂. ⇒ 4 = ^d2⁄₂ ⇒ d² = 8 
⇒ d = √8 = √[4×2] = √4 × √2 = 2√2
Thus we get diameter of the circle as 2√2 cm
6. So perimeter of the circle = πd = π × 2√2 = 2π√2 cm
Part(iii): Rectangle
1. In fig.21.16(c), One of it's diagonal is drawn. This diagonal VW is a diameter of the circle.
2. From the right triangle UVW, we get: VW² = UV² + UW²
⇒ VW² = 2² + 1.5² = 4 + 2.25 = 6.25
⇒ VW = √6.25 = 2.5
3. Thus we get the diameter as 2.5 cm
4. So perimeter of the circle = πd = π × 2.5 = 2.5π cm

Solved example 21.9
An isosceles triangle with it's vertices on a circle is shown in the fig.21.17(a) below. 

Fig.21.17

What is the perimeter of the circle?
Solution:
1. In fig.b, the triangle is named as ABC. AC and BC are the equal sides.
2. Let 'O' be the centre of the circle, and let 'x' be the radius
3. Then OB = OC = x
4. From the right triangle OBD, we get: OB² = BD² + OD².
⇒ x² = 2² + (4-x)² ⇒ x² = 2² + 16 -8x + x²
⇒ 8x = 16 + 4 ⇒ 8x = 20 ⇒ x = ²⁰⁄₈ = 2.5 cm
5. Thus we get the radius as 2.5 cm
6. So perimeter of the circle = 2πr = 2 × π × 2.5 = 5π cm

Solved example 21.10

In all the figs.21.18(a), (b) and (c) below, the centres of the circles are on the same line.

Fig.21.18

In the figs (a) and (b), the small circles are of the same diameter. Prove that, in all the figs, the perimeter of the large circle is the sum of the perimeters of the small circles
Solution:
Case 1:

1. Let the diameters of the two inner circles be d_i

2. Let the diameter of the outer circle be d_o

3. Then we get d_o = 2d_i

4. Perimeter of each of the inner circles = πd_i

5. Sum of the perimeters of the two inner circles = πd_i + πd_i = 2πd_i

6. Perimeter of the outer circle = πd_o.

7. But from (3), we have d_o = 2d_i .

8. So from (6) we get: Perimeter of the outer circle = πd_o. = π × 2d_i = 2πd_i.

9. (5) and (8) are equal. So we can write: The perimeter of the large circle is the sum of the perimeters of the two small circles 

Case 2:

1. Let the diameters of the three inner circles be d_i

2. Let the diameter of the outer circle be d_o

3. Then we get d_o = 3d_i

4. Perimeter of each of the inner circles = πd_i

5. Sum of the perimeters of the two inner circles = πd_i + πd_i + πd_i = 3πd_i

6. Perimeter of the outer circle = πd_o.

7. But from (3), we have d_o = 3d_i .

8. So from (6) we get: Perimeter of the outer circle = πd_o. = π × 3d_i = 3πd_i.

9. (5) and (8) are equal. So we can write: The perimeter of the large circle is the sum of the perimeters of the three small circles

Case 3:

1. Let the diameters of the three inner circles be d₁, d₂  and d₃
 2. Let the diameter of the outer circle be d_o

3. Then we get d_o = d₁ + d₂ + d₃ 

4. Perimeter of the inner circles will be πd₁, πd₂  and πd₃

5. Sum of the perimeters of the three inner circles = πd₁ + πd₂ + πd₃

6. Perimeter of the outer circle = πd_o.

7. But from (3), we have d_o = d₁ + d₂ + d₃ 

8. So from (6) we get: Perimeter of the outer circle = πd_o. = π × (d₁ + d₂ + d₃ ) = πd₁ + πd₂ + πd₃

9. (5) and (8) are equal. So we can write: The perimeter of the large circle is the sum of the perimeters of the three small circles

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In the next section we will see Area of circles.

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