Question

The area of an equilateral $\Delta$ABC is 17320.5 $c{m}^2$. A circle is drawn taking the vertex of the triangle as centre. The radius of the circle is half the length of the side of triangle. Find the area of the shaded region in $c{m}^2$ . (Use $\pi =3.14,\sqrt{3}=1.73205$)

• A. 1906.2 $c{m}^2$
• B. 1648.5 $c{m}^2$
• C. 1620.5 $c{m}^2$
• D. 1530.6 $c{m}^2$

Answer 1

The correct option is C 1620.5 $c{m}^2$


Area of shaded region = Area of $\Delta$ABC - 3 (Area of sector BPR)

Let 'a' be the side of the equilateral $\Delta$ABC.

Using area of an equilateral triangle = $\frac{\sqrt{3}}{4}{a}^2$,

$\frac{\sqrt{3}}{4}{a}^2=17320.5$

Solving, $a^2=\frac{17320.5\times 4}{\sqrt{3}}$

$⟹a^2=\frac{17320.5\times 4}{1.73205}$

$⟹a^2=\frac{17320.5\times 4}{17320.5\times {10}^{-4}}$

$⟹a=2\times {10}^2$

$⟹a=200cm$.

Radius of the circles
= $\frac{1}{2}\times 200=100cm$

Now, using area of a sector when the degree measure of the angle at the centre is $\theta =\frac{\theta }{360}\pi r^2$

$\therefore$ Required area
$=17320.5-3[\frac{60}{360}\times 3.14\times {100}^2]$

Hence, required area
$=1620.5c{m}^2$