Question

In the given figure, the shape of the top of a table is that of a sector of a circle with centre O and $\angle AOB={90}^{\circ }$. If AO = OB = 42 cm then find the perimeter of the top of the table.

Answer 1

We have r = 42 cm and $\theta ={360}^o-{90}^o={270}^o$

Perimeter of the top of the table = Length of the major arc AB + Length of OA + Length of OB

$=\frac{\theta }{{360}^o}\times 2\pi r+42+42=\frac{{270}^o}{{360}^o}\times 2\times \frac{22}{7}\times 42+84=198+84=282cm$

Hence, the perimeter of the top of the table is 282 cm.