# Perimeter and Areas Related to Circle

## Perimeter and area of a circle

PI-

• The number π is a mathematical constant.
• It is defined as the ratio of a circle’s circumference to its diameter. And its value is
• From the above definition, we can derive the following equation for
• Circumference/Diameter = π,
• The great Indian mathematician Aryabhata (C.E. 476 – 550) gave an approximate value of π. He stated that π=22/7 = 3.1416 approx

## Circumference of a circle

• The perimeter of a circle is the distance covered by going around its boundary once.
• The perimeter of a circle has a special name: Circumference, which is π times the diameter
• which is given by the formula 2πr.

### You may be having doubt how it is 2πr?

Here is proof:

As we know π= Circumference/Diameter —–(1)

From 1 and 2 we can write,

From above equation 3, we get,

Circumference or perimeter= 2πr.

### Area of a Circle

Area of a circle is π r2, where π = or 22/7 ≈ 3.14

where r is the radius of the circle.

You may be having doubt how it is?

### Proof of Area of a circle.

Consider the following diagram,

You can see that the shape in Fig. 2 is nearly a rectangle with

Length = 1/2*2π r —-(1)

and

This suggests that the area of the rectangle = length*breath

Area =  2π r × r =2π r2

Let’s Solve some Example problems based on circumference and perimeter:

Q.1.   The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles.

Sol. We have, r1 = 19 cm

r2 = 9 cm

∴ Circumference of circle-I = 2π r1 = 2π (19) cm

Circumference of circle-II = 2π r1 = 2π (19) cm

Sum of the circumferences of circle-I and circle-II = 2π (19) + 2π (9)

= 2π (19 + 9) cm

= 2π (28) cm

Let R be the radius of the circle-III.

∴ Circumference of circle-III = 2π R

According to the condition,

2π R = 2π (28)

Thus, the radius of the new circle = 28 cm.

Q.2.   The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having an area equal to the sum of the areas of the two circles.

Sol. We have,

Radius of circle-I, r1 = 8 cm

∴ Area of circle-I = πr1 2 = π(8)2 cm2

Area of circle-II = πr22 = π(6)2 cm2

Let the area of the circle-III be R

∴ Area of circle-III = πr2

Now, according to the condition,

πr12 + πr22 = πr2

i.e. π(8)2 + π(6)2 = πr2

⇒ π(82 + 62) = πr2

⇒ 82 + 62 = r2

⇒ 64 + 36 = r2

⇒ 100 = r2

⇒ 102 = r2 ⇒ R = 10

Thus, the radius of the new circle = 10 cm.

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