Question

Four equal circles, each of radius a, touch each other. Show that the area between them is $\frac{6}{7} a^{2}$ (Take $(\pi=\frac{22}{7})$).​

Answer 1

Answer:

It is proved that the area between them is 6a²/7.

Step-by-step explanation:

Given :

Radius of a circle = a

Side of a square = 2 × Radius of a circle

= 2 × a

Side of a square = 2a cm

Area of a square = Side²

= (2a)²

Area of a square = 4a²

Area of the quadrant of one circle = 1/4πr²

Area of the quadrant of four circles = 4 × 1/4πr² = πr²

Area of the shaded portion, A = Area of the square – Area of the quadrant of four circles

A = 4a² - 22/7 × a²

A = 4a² - 22a²/7

A = (28a² -22a²)/7

A = 6a²/7

Hence, it is proved that the area between them is 6a²/7.

HOPE THIS ANSWER WILL HELP YOU….

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