Four equal circles, each of radius a, touch each other. Show that the

1.	Four equal circles, each of radius a, touch each other. Show that the area between them is \(\frac{6}7a^2\)(Take π = 3.14).
Answer» Radius of each circle = a meter If we join the centre of each circle it makes a square of side = 2a Area of square = (2a)² = 4a² m² Area of each quadrant of circle = \(\frac{πr^2}4\) = \(\frac{πa^2}4\) m² Area of 4 quadrants = \(4\times\frac{πa^2}4\) = πa² m² So, Area between circles = 4a² – πa² =\(4a^2 - \frac{22}7a^2\)=\(\frac{28a^2 - 22a^2}7\) =\(\frac{6a^2}7\)m²

Four equal circles, each of radius a, touch each other. Show that the area between them is \(\frac{6}7a^2\)(Take π = 3.14).

Answer»

Radius of each circle = a meter

If we join the centre of each circle it makes a square of side = 2a

Area of square = (2a)² = 4a² m²

Area of each quadrant of circle = \(\frac{πr^2}4\) = \(\frac{πa^2}4\) m²

Area of 4 quadrants = \(4\times\frac{πa^2}4\) = πa² m²

So,

Area between circles = 4a² – πa²

=\(4a^2 - \frac{22}7a^2\)=\(\frac{28a^2 - 22a^2}7\) =\(\frac{6a^2}7\)m²