Four circular cardboard pieces of radii 7 cm are placed on a paper in such a way that each piece touches other two pieces. Find the area of the portion enclosed between these pieces.

#### Solution

The four circles are placed such that each piece touches the other two pieces

By joining the centres of the circles by a line segment, we get a square ABDC with sides,

AB = BD = DC = CA = 2(Radius)

= 2(7) cm

= 14 cm

Area of the square = (Side)^{2}

= (14)^{2}

= 196 cm^{2}

ABDC is a square

So, each angle has a measure of 90°.

∠A = ∠B = ∠D = ∠C = 90 = `pi/2` radius

= θ ....(Let)

Radius of each sector = 7 cm

Area of the sector with central angle A = `(1/2) r^2 theta`

= `11/2 r^2 theta`

= `1/2 xx 49 xx pi/2`

= `1/2 xx 49 xx 22/(2 xx 7)`

= `77/2 cm^2`

As the central angles and the radius of each sector are same, area of each sector is `77/2` cm^2`.

Area of the shaded portion = Area of square – Area of the four sectors

= `196 - (4 xx 77/2)`

= `196 - 154`

= 42 cm^2`

Therefore, required area of the portion enclosed between these pieces is 42 cm^{2}.