The perimeter of the quadrant of …

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Sia ? 4 years ago
Let r be the radius of the quadrant.
We know that, Perimeter of the quadrant {tex}= \frac { 25 r } { 7 } \mathrm { cm }{/tex}
Given that, perimeter of a quadrant {tex}= 25 cm{/tex}
{tex}\Rightarrow \frac { 25 r } { 7 } {/tex}{tex}= 25{/tex}
{tex}\Rightarrow r = 7 \mathrm { cm }{/tex}
{tex}\therefore {/tex} Area of the quadrant{tex}= \frac { 1 } { 4 } \pi r ^ { 2 }{/tex}
{tex}= \left( \frac { 1 } { 4 } \times \frac { 22 } { 7 } \times 7 \times 7 \right) \mathrm { cm } ^ { 2 }{/tex}
{tex}= \frac { 77 } { 2 } \mathrm { cm } ^ { 2 }{/tex}
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