AE wrote:

Answer should be C.

Let side of Square = \(a\)

If AEC is 1/4 of circumference it means that Angle ADC is 90 as the circle is cut into 4 quarters. (also evident from the fact that ADC is the angle of a square)

Since AD is the radius as well, we can find the area of the arc AEC. Similarly we can find the area of the Arc AFC. Now if you add the two areas up, you notice that the shaded region is counted twice. To remove this double counting and excess area you subtract the area of the square. This gives us the area of the shaded region.

Area of Shaded portion = \(2*\frac{\pi*a^2}{4}-a^2\)\(=\) \(\frac{\pi*a^2}{2}-a^2 =\frac{\pi*a^2-2*a^2}{2}\)

Area of Unshaded portion = Area of Square - Area of Shaded portion

= \(a^2 - (2*\frac{\pi*a^2}{4}-a^2)\)\(=\) \(2*a^2 - \frac{\pi*a^2}{2}=\frac{4*a^2 - \pi*a^2}{2}\)

\(\frac{Area of Unshaded portion}{Area of Shaded portion}=\frac{4*a^2 - \pi*a^2}{\pi*a^2-2*a^2}=\frac{4 - \pi}{\pi-2}\)

Putting \(\pi = \frac{22}{7}\) in above ratio , we get \(\frac{Area of Unshaded portion}{Area of Shaded portion}= \frac{(28-22)}{(22-14)} =\frac{6}{8}=\frac{3}{4}\)

Answer C

Now if you add the two areas up, you notice that the shaded region is counted twice. To remove this double counting and excess area you subtract the area of the square

Area of Shaded portion = \(2*\frac{\pi*a^2}{4}-a^2\)\(=\) \(\frac{\pi*a^2}{2}-a^2 =\frac{\pi*a^2-2*a^2}{2}\)

Please can you simplify this,it's quite confusing.