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**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**

- 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.**

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 is **π r ^{2}**

where **r is the radius of the circle.**

You may be having doubt how it is?

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

**breadth = r —-(2)**

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

**Area = 2π r × r =2π r^{2}**

**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, r _{1} = 19 cm**

** r _{2} = 9 cm**

** ∴ Circumference of circle-I = 2π r _{1} = 2π (19) cm**

** Circumference of circle-II = 2π r _{1} = 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, r _{1} = 8 cm**

** ∴ Area of circle-I = πr _{1} ^{2} = π(8)^{2} cm^{2}**

** Area of circle-II = πr _{22} = π(6)2 cm^{2}**

** ****Let the area of the circle-III be R**

** ∴ Area of circle-III = πr _{2}**

** Now, according to the condition,**

** πr _{1}^{2} + πr_{2}^{2} = πr_{2}**

** i.e. π(8) ^{2} + π(6)^{2} = πr_{2}**

** ⇒ π(8 ^{2} + 6^{2}) = πr_{2}**

** ⇒ 8 ^{2} + 6^{2} = r_{2}**

** ⇒ 64 + 36 = r _{2}**

** ⇒ 100 = r _{2}**

** ⇒ 10 ^{2} = r_{2} ⇒ R = 10**

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

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