Question

The area of an equilateral triangle ABC is $17320.5c{m}^2$. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see Fig. 12.28). Find the area of the shaded region.
$(Use\pi =3.14and\sqrt{3}=1.73205)$

Answer 1

ABC is an equilateral triangle.
$\therefore \angle A=\angle B=\angle C={60}^{\circ }$
There are three sectors each making ${60}^{\circ }$.

Area of equilateral $\Delta$
$=\frac{\sqrt{3}}{4}\times (side{)}^2$

$Areaof\Delta ABC=17320.5c{m}^2$
$\Rightarrow \frac{\sqrt{3}}{4}\times (side{)}^2=17320.5\Rightarrow (side{)}^2=17320.5\times \frac{4}{1.73205}\Rightarrow (side{)}^2=4\times {10}^4\Rightarrow side=200cm$

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

Area of the sector making angle $\theta$
$=\left(\frac{\theta }{{360}^{\circ }}\right)\times \pi r^2$

Area of the sector
$=\left(\frac{{60}^{\circ }}{{360}^{\circ }}\right)\times \pi r^{2}c{m}^2$
$=\frac{1}{6}\times 3.14\times (100{)}^{2}c{m}^2$
$=\frac{15700}{3}c{m}^2$

Area of 3 sectors
$=3\times \frac{15700}{3}=15700c{m}^2$

Area of the shaded region = Area of equilateral triangle ABC - Area of 3 sectors
$=17320.5c{m}^2-15700c{m}^2=1620.5c{m}^2$