Question 15- In the given figure, …

In right-angled {tex}\triangle {/tex}BAC,
By using pythagoras theorem, we get
CB² = AC² + AB²
{tex}= 24^2 + 7^2\\= 576 + 49\\= 625{/tex}
{tex}\Rightarrow C B = \sqrt { 625 }{/tex}
{tex}= 25 \ cm{/tex}
{tex}\Rightarrow O C = \frac { 1 } { 2 } C B{/tex}
{tex}= \frac { 25 } { 2 } \mathrm { cm }{/tex}
So, radius of the circle {tex}= 12.5 cm{/tex}
Now, Area of {tex}\triangle {/tex}BAC
{tex}= \frac { 1 } { 2 } \times A C \times A B{/tex}
{tex}= \frac { 1 } { 2 } \times 24 \times 7 {/tex}
= 84 cm²
Area of the circle= {tex}3.14 \times 12.5 \times 12.5{/tex}
= 490.625 cm²
Area of quadrant COD
{tex}= \frac { 1 } { 4 } \times 3.14 \times 12.5 \times 12.5{/tex}
= 122.66 cm²
Now, area of the shaded region
= Area of the circle - Area of {tex}\triangle {/tex}BAC - Area of quadrant COD
= {tex}490.625 - 84 - 122.66{/tex}
= 283.96 cm²