In the given figure, the shape of the top of a table is that of a sector of a circle with centre O and $\angl…

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.

Accepted Answer

## Model Answer Given: radius $r = 42$ cm, $\angle AOB = 90°$ Length of arc AB $= \dfrac{	heta}{360} 	imes 2\pi r = \dfrac{90}{360} 	imes 2 	imes \dfrac{22}{7} 	imes 42 = 66$ cm Perimeter of top of table = Arc AB + OA + OB $= 66 + 42 + 42 = \mathbf{150 	ext{ cm}}$ Source: Areas of Sector and Segment of a Circle, Chapter 11 ## Explanation The perimeter of a sector consists of **two radii + arc length** — students often forget to add the two straight edges (OA and OB). Use the arc length formula $\dfrac{	heta}{360} 	imes 2\pi r$ with $	heta = 90°$, $r = 42$ cm, and $\pi = \dfrac{22}{7}$.

Mathematics — Standard (041) — CBSE Class 10 board question