Question

In fig. 9.33 and equilateral triangle of side 6 cm and its circumcircle is shown. Find the area of shaded portion.
($\mathrm{\pi }=3.14,\sqrt{3}=1.73$)

Answer 1

We know that the perpendicular and the median of an equilateral triangle coincide with each other.
Draw one of its perpendicular.

Now, AC = 6 cm and DC = $\frac{\mathrm{BC}}{2}=\frac{6}{2}=3\mathrm{cm}$
In right triangle ADC, we have:
AC² = AD² + DC² (By Pythagoras theorem)
$\Rightarrow$ (6)² = AD² + (3)²
$\Rightarrow$ AD² = 36 - 9
$\Rightarrow$ AD² = 27
$\Rightarrow$ AD = $3\sqrt{3}$ cm

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{equilateral}\mathrm{triangle}=\frac{\sqrt{3}}{4}(\mathrm{AC}{)}^2 \\ =\frac{1.73}{4}(6{)}^2{\mathrm{cm}}^2 \\ =15.57{\mathrm{cm}}^2 \end{gathered}$$

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{circle}=\pi r^2 \\ =3.14\times (2\sqrt{3}{)}^2{\mathrm{cm}}^2 \\ =37.68{\mathrm{cm}}^2 \end{gathered}$$

Area of the shaded portion = area of the circle – area of the equilateral triangle
= (37.68 – 15.57) cm²
= 22.11 cm²