In Fig. 4, ABCD is a square of side 14 cm. Semi-circles are drawn with each side of square as diameter. Find the area of the shaded region.

(use `pi=22/7`)

#### Solution

Let the four shaded regions be I, II, III and IV and the centres of the semicircles be P, Q, R and S, as shown in the figure.

It is given that the side of the square is 14 cm.

Now,

Area of region I + Area of region III = Area of the square − Areas of the semicircles with centres S and Q

=14×14−2×`1/2`×π×7^{2} (∵ Radius of the semicircle=7 cm)

`=192-49xx22/7`

=196−154

=42 cm^{2}

Similarly,

Area of region II + Area of region IV = Area of the square − Areas of the semicircles with centres P and R.

=14×14−2×`1/2`×π×7^{2 } (∵ Radius of the semicircle=7 cm)

=196−49×`22/7`

=196−154

=42 cm^{2}

Thus^{,}Area of the shaded region = Area of region I + Area of region III + Area of region II + Area of region IV

= 42 cm^{2}^{ }+ 42 cm^{2}^{ }

= 84 cm^{2}^{ }