Question

In a circular table corner of radius $32cm$, a design is formed leaving an equilateral triangle $ABC$ in the middle as shown in Fig. Find the area of the design

Answer 1

In $△BOD$ By Trigonometry,
$BD=32\sin {60}^o$ $=27.71cm$
$BC=55.43cm$
$OD=32\cos {60}^o=16cm$
$\therefore$ Area of $△$ $BOC$ $=\frac{1}{2}\times 16\times 55.43$
$=443.4c{m}^2$
$\therefore$ Area of Equilateral triangle $=3\times A△BOC$ $=3\times 443.4$ $=1330.2c{m}^2$
Area of circle$=\pi \times r^2$ $=3.14\times 32\times 32$ $=3215.36c{m}^2$
$\therefore$ Area of shaded region $=3215.36-1330.2$ $=1885.16c{m}^2$