Question

Question 97

Find the area enclosed by the following figure:

Answer 1

The given shape contains a semi-circle and a triangle.
Area of semi-circle $=\frac{1}{2}\pi r^2=\frac{1}{2}\times \frac{22}{7}\times 10\times 10=\frac{1100}{7}c{m}^2$
Area of triangle $=\frac{1}{2}\times$ Base $\times Height=\frac{1}{2}\times 20\times 7=70c{m}^2$
Hence, total area enclosed by the shape $=\frac{1100}{7}+70=\frac{1100+490}{7}$
$=\frac{1590}{7}=227.14c{m}^2=227c{m}^2$ (approx)