A wire of length 20 text{ cm} is bent in the | vedclass

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(B) Let the radius of the sector be $r$ and the arc length be $s$. The perimeter of the sector is given by $P = 2r + s = 20 	ext{ cm}$.Thus,$s = 20 -

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$25$

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(B) Let the radius of the sector be $r$ and the arc length be $s$. The perimeter of the sector is given by $P = 2r + s = 20 	ext{ cm}$.Thus,$s = 20 - 2r$.The area $A$ of the sector is given by $A = \frac{1}{2} s r$.Substituting the value of $s$,we get $A = \frac{1}{2} (20 - 2r) r = 10r - r^2$.To find the maximum area,we differentiate $A$ with respect to $r$ and set it to zero:$\frac{dA}{dr} = 10 - 2r = 0 \Rightarrow r = 5 	ext{ cm}$.Substituting $r = 5$ into the area equation:$A = 10(5) - (5)^2 = 50 - 25 = 25 	ext{ cm}^2$.

$A$ wire of length $20 \text{ cm}$ is bent in the form of a sector of a circle. The maximum area that can be enclosed by the wire is (in $\text{ cm}^2$)

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