In the below figure, $ O A C B $ is a quadrant of a circle with centre $ O $ and radius $ 3.5 \mathrm{~cm} $. If $ O D=2 \mathrm{~cm} $, find the area of the quadrant $ O A C B $."

Given:

\( O A C B \) is a quadrant of a circle with centre \( O \) and radius \( 3.5 \mathrm{~cm} \).

\( O D=2 \mathrm{~cm} \).

To do: 

We have to find the area of the quadrant \( O A C B \).

Solution:

Radius of the outer quadrant $R = 3.5\ cm$

Radius of the inner quadrant $r= 2\ cm$

This implies,

Area of the quadrant $OACB=\frac{1}{4} \pi \mathrm{R}^{2}$

$=\frac{1}{4} \times \frac{22}{7}(3.5)^{2}$

$=\frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5$

$=9.625 \mathrm{~cm}^{2}$

The area of the quadrant \( O A C B \) is $9.625\ cm^2$.

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Updated on: 10-Oct-2022

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