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# Area of the Circle

Let us find a way to calculate the area (A) of a circle in terms of known area, that is the area of a rectangle.

Area of the Circle

Let us consider the following situation.

A bull is tied with a rope to a pole. The bull goes round to eat grass. What will be the portion of grass that the bull can graze?

Can you tell what is needed to be found in the above situation, Area or Perimeter? In this situation we need to find the area of the circular region.

Let us find a way to calculate the area (A) of a circle in terms of known area, that is the area of a rectangle.

1. Draw a circle on a sheet of paper. 2. Fold it once along its diameter to obtain two semicircles. Shade one half of the circle (Fig. 2.13) 3. Again fold the semicircles to get 4 sectors. Fig. 2.14 shows a circle divided into four sectors. The sectors are re-arranged and made into a shape as shown in Fig. 2.15. 4. Repeat this process of folding to eight folds, then it looks like a small sectors as shown in the Fig. 2.16. The sectors are re-arranged and made into a shape as shown in Fig. 2.17. 5. Press and unfold the circle. It is then divided into 16 equal sectors and then into 32 equal sectors. As the number of sectors increase the assembled shape begins to look more or less like a rectangle, as shown in Fig. 2.19. 6. The top and bottom of the rectangle is more or less equivalent to the circumference of the circle. Hence the top of the rectangle is half of the circumference = πr . The height of the rectangle is nearly equivalent to the radius of the circle. When the number of sectors is increased infinitely, the circle can be rearranged to form a rectangle of length ‘ πr ’ and breadth ‘r’.

We know that the area of the rectangle = l × b

= πr × r

= π r2

= Area of the circle

So, the area of the circle, A = π r2 sq.units .

Activity

Draw circles of different measurements on a graph sheet. Find the area by counting the number of squares enclosed by the circle. So we get a rough estimate of the area of the circle. Designs obtained by spirograph.

Shown below are some shapes using a spirograph. We can observe that each of the different designs are circular. Think

If the area and circumference of the circle are equal then what will be the value of the radius?

Example 2.9

Find the area of the circle of radius 21 cm (Use π  = 3.14 ).

Solution

Area of a circle = πr2 sq. units

= 3.14 × 21 × 21

= 1384.74

Area of the circle = 1384.74 cm2

Example 2.10

Find the area of a hula loop whose diameter is 28 cm (use π = 22/7). Solution

Given the diameter (d) = 28 cm

Radius (r) = 28/2 = 14 cm

Area of a circle   = πr2 sq.units

= 22/7 × 14 × 14

So, the area of the circle = 616 cm2.

Example 2.11

The area of the circular region is 2464 cm2. Find its radius and diameter. (use π = 22/7)

Solution

Given that the area of the circular region = 2464 cm2

πr2 = 2464

22/7 × r2 = 2464

r2 = 2464 × 7/22

r2 = 112 × 7 = 784 r2 =2×2×2×2×7×7

= 4×4×7×7

= 42 ×72

r2 = (4×7)2       [r×r = (4×7)×(4×7)]

r = 4×7

= 28 cm

Diameter (d) = 2 × r = 2 × 28 = 56 cm.

Example 2.12

A gardener walks around a circular park of distance 154 m. If he wants to level the park at the rate of ₹25 per sq.m, how much amount will he need? (use π = 22/7)

Solution

The distance covered by the man is nothing but the circumference of the circle.

Given that the distance covered = 154 m

Therefore circumference of the circle = 154 m

That is, 2πr = 154

2 × 22/7 × r = 154

r =154 × (7/44)

r = 3.5 × 7

= 24.5

Area of the park = πr2 sq.units

= (22/7) × 24.5 × 24.5

= 22 × 3.5 × 24.5

= 1886.5 m2

Cost of levelling the park per sq.m = ₹ 25.

Cost of levelling the park of 1886.5 m2 = 1886.5 × 25 = ₹ 47,162.50

Example 2.13

Find the length of the rope by which a cow must be tethered in order that it may be able to graze an area of 9856 sq.m (use π =22/7)

Solution

Given that, the area of the circle = 9856 sq.m

πr2 = 9856

(22/7)  × r2 = 9856

r2 = 9856 × 7/22

r2  = 448 × 7 = 3136 r2 = 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7

r2 = 8 × 8 × 7 × 7 = 82 × 72 = (8 × 7)2

r = 8 × 7 = 56 m

Therefore the length of the rope should be 56 m.

Example 2.14

A garden is made up of a rectangular portion and two semicircular regions on either sides. If the length and width of the rectangular portion are 16 m and 8 m respectively, calculate (π  = 3.14) (i) the perimeter of the garden

(ii) the total area of the garden

Solution

(i) The perimeter of the rectangular garden

Perimeter include two lengths each of 16 m and two semi circular arcs of diameter 8 m.

Circumference of the semicircle = πd / 2 units

= [π × 8] /2  = 4 π

= 4×3.14

= 12.56 m

Therefore, circumference of two semicircles = 2 × 12.56

= 25.12 m

Total perimeter = length + length + circumference of two semicircles

= 16 +16 + 25.12

= 32 + 25.12

= 57.12 m

(ii) The total area of the garden

Total area of the garden = Area of rectangle + Area of 2 semicircles

= Area of rectangle + Area of the circle

Here, the area of the rectangle = l × b sq.units

= 16×8

= 128 m2           ...(1)

Area of the circle = πr2 sq.units

= 3.14 ×4×4

= 3.14 × 16

= 50.24 m2        ...(2)

From (1) and (2), the total area of garden = 128 + 50.24

= 178.24 m2

Try these

Draw circles of different radii on a graph paper. Find the area by counting the number of squares covered by the circle. Also find the area by using the formula. Number of squares covered by the circle = 52 + 8 × ½

= 52 + 4 = 56

Radius of circle r = 4.2 cm

Area of the circle = πr2

= 22/7 × 4.2 4.2 sq.cm

= 55.44 sq.cm Number of squares covered by the circle = 88 +16 × ½ +10 × ¼

= 88 + 8 + 2 ½ = 98.5

Radius of the circle r = 5.6 cm

Area of the circle = πr2

= 22/7 × 5.6 4.2 sq.cm

= 98.56 sq.cm Number of squares covered by the circle =12 + 4 × ½

= 12 + 2 = 14

Area of the circle = πr2

= 22/7 × 2.1 21 sq.cm

= 13.86 sq.cm

(i) Find the area of the circle, if the radius is 4.2 cm.

Radius of the circle r = 4.2 cm

Area of the circle = πr2

= 22/7 × 4.2 42 sq.cm

= 55.44 sq.cm

(ii) Find the area of the circle if the diameter is 28 cm.

Diameter of the circle d = 28 cm

Area of the circle = πr2

= 22/7 × 14 =14 sq.cm

= 616 sq.cm

Tags : Measurements | Term 2 Chapter 2 | 7th Maths , 7th Maths : Term 2 Unit 2 : Measurements
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7th Maths : Term 2 Unit 2 : Measurements : Area of the Circle | Measurements | Term 2 Chapter 2 | 7th Maths