Perimeter and area of a circle
- 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)
Diameter= 2*radius —–(2)
From 1 and 2 we can write,
Π=Circumferance/2Xradius —- (3)
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)
breadth = r —-(2)
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.