In the adjoining figure, ABCD is a square of side 10 cm and two semicircles with side of the square as diameter. A quarter circle is also seen in figure with side of square as radius. Find the area of the square region excluding the shaded part (deepak). (Take `pi = 3.14`)

Explanation

First of all we will find the area of I and II i.e., Batti of Deepak, draw `OM bot AB and ON bot BC`.
`:. Ar (I) = ar ("sector" NBKO) - ar (Delta BNO)`
`= pi (5)^(2) xx (90^(@))/(360^(@)) - (1)/(2) xx 5 xx 5`
`rArr ar (II) = ar (I)`
`:. ar(I + II) = 2 ((25pi)/(4) - (25)/(2)) = (25)/(2) (pi - 2)`
`= (25)/(2) (3.14 - 2) = 14.25 cm^(2)`
Now, area of deepak = ar (quadrant APCB) - [ar (semicircle `IV + I + II`) + ar (semicircle `V + I + II`) `- 2 xx ar (I + II)`)
`= pi (10)^(2) xx (90)/(360) - [2 xx (pi(5)^(2))/(2) - 2 xx 14.25]`
`= 25 pi - (25 pi - 28.5) = 28.5 cm^(2)`
Area of unshaded portion = Area of square - Area of deepak
`=(10)^(2) - 28.5 = 100 - 28.5 = 71.5 cm^(2)`