Question

In Fig. 13.121, ABCD is a square of side 10 cm and a circle is inscribed in it. The area of the shaded part is __________.

Answer 1

Let O be the centre of the circle. Suppose the circle touches the sides BC and CD of the square at E and F, respectively.



OE = OF (Radius of the circle)

Now, ∠OEC = ∠OFC = 90º (Radius is perpendicular to the tangent at the point of contact)

Also, OE = OF = CF = CE

∴ OECF is a square.

We know that if a circle is inscribed in a square, then the diameter of the circle is equal to the side of the square.

$\therefore \mathrm{OE}=\frac{\mathrm{BC}}{2}=\frac{10}{2}=5\mathrm{cm}$

⇒ Length of each side of the square OECF = 5 cm

Now,

Area of the shaded portion

= $\frac{1}{2}$(Area of the square OECF − Area of the quadrant OEFO)

$$\begin{gathered} =\frac{1}{2}\left[{\left(\mathrm{OE}\right)}^2-\frac{1}{4}\times \mathrm{\pi }{\left(\mathrm{OE}\right)}^2\right] \\ =\frac{1}{2}\left[{\left(5\right)}^2-\frac{1}{4}\times \mathrm{\pi }{\left(5\right)}^2\right] \\ =\frac{25}{2}\left(1-\frac{\mathrm{\pi }}{4}\right) \\ =\frac{25}{2}\left(\frac{4-\mathrm{\pi }}{4}\right) \end{gathered}$$
$=\frac{100-25\mathrm{\pi }}{8}{\mathrm{cm}}^2$

In Fig. 13.121, ABCD is a square of side 10 cm and a circle is inscribed in it. The area of the shaded part is $\underline{\frac{100-25\mathrm{\pi }}{8}{\mathrm{cm}}^2}$.