Question

In the circular table cover of radius $32$ cm , a design is formed leaving an equilateral triangle ABC in the middle as shown in the figure . Find the area of the design. (shaded region)

Answer 1

Here, $△ABC$ is an equilateral triangle. Let $O$ be the circumcenter of circumcircle.

Radius, $r=32cm$

Now, area of circle $=\pi r^2$

$=\frac{22}{7}\times 32\times 32=\frac{22528}{7}c{m}^2$

Area of $△ABC=3\times$ Area of $△BOC$

$=3\times \frac{1}{2}\times 32\times 32\sin {120}^o$

.......... $[\angle BOC=2\angle BAC=2\times {60}^o={120}^o]$

$=3\times 16\times 32\times \frac{\sqrt{3}}{2}(\because \sin {120}^o=\sin ({180}^o-{60}^o)=\sin {60}^o=\frac{\sqrt{3}}{2})$

$=3\times 16\times 16\times \sqrt{3}=768\sqrt{3}c{m}^2$

$\therefore$ Area of the design = Area ofe circle - Area of $△ABC$

$=(\frac{22528}{7}-768\sqrt{3})c{m}^2$

$=(3218.28-1330.176)c{m}^2=1888.7c{m}^2$