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# 12. AREAS RELATED TO CIRCLES

## 1 Introduction

You are already familiar with some methods of finding perimeters and areas of simple plane figures such as rectangles, squares, parallelograms, triangles and circles from your earlier classes. Many objects that we come across in our daily life are related to the circular shape in some form or the other. Cycle wheels, wheel barrow (thela), dartboard, round cake, papad, drain cover, various designs, bangles, brooches, circular paths, washers, flower beds, etc. are some examples of such objects (see Fig. 12.1). So, the problem of finding perimeters and areas related to circular figures is of great practical importance. In this chapter, we shall begin our discussion with a review of the concepts of perimeter (circumference) and area of a circle and apply this knowledge in finding the areas of two special ‘parts’ of a circular region (or briefly of a circle) known as sector and segment. We shall also see how to find the areas of some combinations of plane figures involving circles or their parts.

**Fig. 12.1**

## 2 Perimeter and Area of a Circle — A Review

Recall that the distance covered by travelling once around a circle is its perimeter, usually called its circumference. You also know from your earlier classes, that circumference of a circle bears a constant ratio with its diameter. This constant ratio is denoted by the Greek letter π (read as ‘pi’). In other words,

You may also recall that area of a circle is πr2, where r is the radius of the circle. Recall that you have verified it in Class VII, by cutting a circle into a number of sectors and rearranging them as shown in Fig. 12.2.

**Fig. 12.2**

You can see that the shape in Fig. 12.2 (ii) is nearly a rectangle of length × r = πr2. Let us recall the concepts learnt in earlier classes, through an example.

The cost of fencing a circular field at the rate of Rs 24 per metre is
Rs 5280. The field is to be ploughed at the rate of Rs 0.50 per m^{2} . Find the cost of
ploughing the field (Take π = 22/7)

Length of the fence (in metres) = Total cost/Rate = 5280/24 = 220

So, circumference of the field = 220 m

Therefore, if r metres is the radius of the field, then

2πr = 220

or, 2 × 22/7 × r = 220

or, r = (220*7)/2*22=35

i.e., radius of the field is 35 m.

Therefore, area of the field = πr^{2} =× 35 × 35 m^{2} = 22 × 5 × 35 m^{2}

Now, cost of ploughing 1 m^{2} of the field = Rs 0.50

So, total cost of ploughing the field = Rs 22 × 5 × 35 × 0.50 = Rs 1925

## 3 Areas of Sector and Segment of a Circle

You have already come across the terms sector and segment of a circle in your earlier classes. Recall that the portion (or part) of the circular region enclosed by two radii and the corresponding arc is called a sector of the circle and the portion (or part) of the circular region enclosed between a chord and the corresponding arc is called a segment of the circle. Thus, in Fig. 12.4, shaded region OAPB is a sector of the circle with centre O. ∠ AOB is called the angle of the sector. Note that in this figure, unshaded region OAQB is also a sector of the circle. For obvious reasons, OAPB is called the minor sector and OAQB is called the major sector. You can also see that angle of the major sector is 360° – ∠ AOB.

**Fig. 12.4**

Now, look at Fig. 12.5 in which AB is a chord of the circle with centre O. So, shaded region APB is a segment of the circle. You can also note that unshaded region AQB is another segment of the circle formed by the chord AB. For obvious reasons, APB is called the minor segment and AQB is called the major segment.

**Fig. 12.5**

## Remark :

When we write ‘segment’ and ‘sector’ we will mean the ‘minor segment’ and the ‘minor sector’ respectively, unless stated otherwise.

Now with this knowledge, let us try to find some relations (or formulae) to calculate their areas.

Let OAPB be a sector of a circle with centre O and radius r (see Fig. 12.6). Let the degree

measure of ∠ AOB be θ.

**Fig. 12.6**

In a way, we can consider this circular region to be a sector forming an angle of 360° (i.e., of degree measure 360) at the centre O. Now by applying the Unitary
Method, we can arrive at the area of the sector OAPB as follows:

When degree measure of the angle at the centre is 360, area of the sector = πr^{2}

So, when the degree measure of the angle at the centre is 1, area of the sector =πr^{2}/360

Therefore, when the degree measure of the angle at the centre is θ, area of the sector=(πr^{2}/360)*θ=(θ/360)*πr^{2}

Thus, we obtain the following relation (or formula) for area of a sector of a circle:

**Area of the sector of angle θ =(θ/360)*πr ^{2}**

where r is the radius of the circle and θ the angle of the sector in degrees. Now, a natural question arises : Can we find the length of the arc APB corresponding to this sector? Yes. Again, by applying the Unitary Method and taking the whole length of the circle (of angle 360°) as 2πr, we can obtain the required length of the arc APB as (θ/360)*2πr

So, **length of an arc of a sector of angle θ =(θ/360)*2πr**

**Fig. 12.7**

Now let us take the case of the area of the segment APB of a circle with centre O and radius r (see Fig. 12.7). You can see that :

Area of the segment APB = Area of the sector OAPB – Area of Δ OAB

=(θ/360)*πr^{2} – area of Δ OAB

**Note :** From Fig. 12.6 and Fig. 12.7 respectively, you can observe that :
Area of the major sector OAQB = πr^{2} – Area of the minor sector OAPB
and Area of major segment AQB = πr^{2} – Area of the minor segment APB
Let us now take some examples to understand these concepts (or results).

Find the area of the sector of a circle with radius 4 cm and of angle 30°. Also, find the area of the corresponding major sector (Use π = 3.14).

**Fig. 12.8**

Given sector is OAPB (see Fig. 12.8).

Area of the sector =(θ/360)*πr^{2}

= (30/360) *3.14 x 4 x 4 cm^{2}

=12.56/3 cm^{2}= 4.19cm^{2}

Area of the corresponding major sector

= πr^{2} – area of sector OAPB

= (3.14 × 16 – 4.19) cm^{2}

= 46.05 cm^{2} = 46.1 cm^{2}(approx.)

**Alternatively**, area of the major sector =((360-θ))/360*πr^{2}

= ((360-30))/360*3.14*16 cm^{2}

= 330/360*3.14*16 cm^{2}

= 46.1 cm^{2}(approx.)

Find the area of the segment AYB shown in Fig. 12.9, if radius of the circle is 21 cm and ∠ AOB = 120°. (Use π =22/7)

**Fig. 12.9**

**Fig. 12.1**

Area of the segment AYB= Area of sector OAYB – Area of Δ OAB (1)

Now, area of the sector OAYB =120/360*22/7*21*21 cm^{2}= 462 cm^{2} (2)

For finding the area of Δ OAB, draw OM ⊥ AB as shown in Fig. 12.10. Note that OA = OB. Therefore, by RHS congruence, Δ AMO congurent to Δ BMO.

So, M is the mid-point of AB and ∠AOM = ∠BOM = 1/2*120°=60°.

Let OM = x cm

So, from Δ OMA, OM/OA = cos60°

or, (1/2)(x/21) = (1/2) (cos60°=(1/2))

or, x=21/2

So, OM =21/2 cm

Also, AM/OA = sin60° = (√3/2)

So, AM=21(√3/2)cm

Therefore, AB = 2AM=(2*21√3)/2cm=21√3cm

So, area of Δ OAB =(1/2)AB*OM=(1/2)*21√3*(21/2)cm^{2}

= (441/4)√3/2cm^{2} (3)

Therefore, area of the segment AYB = (462-(441/4)√3)cm^{2} [From (1), (2) and (3)]

=(21/4)(88-21√3) cm^{2}

## 4 Areas of Combinations of Plane Figures

So far, we have calculated the areas of different figures separately. Let us now try to calculate the areas of some combinations of plane figures. We come across these types of figures in our daily life and also in the form of various interesting designs. Flower beds, drain covers, window designs, designs on table covers, are some of such examples. We illustrate the process of calculating areas of these figures through some examples.

In Fig. 12.15, two circular flower beds have been shown on two sides of a square lawn ABCD of side 56 m. If the centre of each circular flower bed is the point of intersection O of the diagonals of the square lawn, find the sum of the areas of the lawn and the flower beds.

**Fig. 12.15**

Area of the square lawn ABCD = 56 × 56 m^{2} (1)

Let OA = OB = x metres

So, x^{2} + x^{2} = 56^{2}

or, 2x^{2} = 56 × 56

or, x^{2} =28×56 (2)

Now, area of sector OAB = (90/360) * Πx^{2} =(1/4)*Πx^{2}

= (1/4)*(22/7)*28*56 m^{2} [From (2)] (3)

Also, area of Δ OAB = (1/4)*56*56m^{2}

So, area of flower bed AB = [{(1/4)*(22/7)*28*56} - {(1/4)*56*56)}]m^{2}
[From (3) and (4)]

=(1/4)*28*56{(22/7)-2}m^{2}

=(1/4)*28*56*(8/7) m^{2} (5)

Similarly, area of the other flower bed
=(1/4)*28*56*(8/7) m^{2} (6)

Therefore, total area =[{56*56}+{(1/4)*28*56*(8/7)}+{(1/4)*28*56*(8/7)}]m^{2}
[From (1), (5) and (6)]

= 28 * 56(2+(2/7)+(2/7))m^{2}

= 28 * 56*(18/7)m^{2} = 4032m^{2}

**Alternative Solution :**

Total area = Area of sector OAB + Area of sector ODC + Area of Δ OAD + Area of Δ OBC
=[{(90/360)*(22/7)*28*56}+{(90/360)*(22/7)*28*56}+{(1/4)*56*56}+{(1/4)*56*56}]m^{2}
=[(1/4)*28*56{(22/7)+(22/7)+2+2}]m^{2}
=[{(7*56)/7}(22+22+14+14)]m^{2}
= 56 × 72 m^{2} = 4032 m^{2}

Find the area of the shaded region in Fig. 12.16, where ABCD is a square of side 14 cm.

**Fig. 12.16**

Area of square ABCD
= 14 × 14 cm^{2} 14*14 = 196 cm^{2}

Diameter of each circle = (14/2)cm = 7cm

So, radius of each circle =(7/2)cm

So, area of one circle = πr^{2} =(22/7)*(7/2)*(7/2)cm^{2}

154/4cm^{2}=77/2cm^{2}

Therefore, area of the four circles = 4x(77/2)cm^{2} =154cm^{2}

Hence, area of the shaded region = (196 – 154) cm^{2} = 42 cm^{2} .

Find the area of the shaded design in Fig. 12.17, where ABCD is a square of side 10 cm and semicircles are drawn with each side of the square as diameter. (Use π = 3.14)

**Fig. 12.17**

**Fig. 12.18**

Let us mark the four unshaded regions as I, II, III and IV (see Fig. 12.18).

Area of I + Area of III = Area of ABCD – Areas of two semicircles of each of radius 5 cm

[{10*10}-{2*(1/2)*Π*5^{2}}]= cm^{2} = (100 – 3.14 × 25) cm^{2}

= (100 – 78.5) cm^{2} = 21.5 cm^{2}

Similarly, Area of II + Area of IV = 21.5 cm^{2}

So, area of the shaded design = Area of ABCD – Area of (I + II + III + IV)
= (100 – 2 × 21.5) cm^{2} = (100 – 43) cm^{2} = 57 cm^{2}