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(C) Let $	riangle ABC$ be an isosceles triangle such that $AB = AC = x$.Let $\angle ABC = \angle ACB = 	heta$.Draw segment $AD \perp$ side $BC$ at p

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$\frac{1}{2} x^2$

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(C) Let $	riangle ABC$ be an isosceles triangle such that $AB = AC = x$.Let $\angle ABC = \angle ACB = 	heta$.Draw segment $AD \perp$ side $BC$ at point $D$.In $	riangle ABD$,$AD = x \sin 	heta$ and $BD = x \cos 	heta$.Similarly,in $	riangle ACD$,$DC = x \cos 	heta$.Therefore,in $	riangle ABC$,the height is $AD = x \sin 	heta$ and the base is $BC = BD + DC = x \cos 	heta + x \cos 	heta = 2x \cos 	heta$.The area $A$ of $	riangle ABC$ is given by:$A = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$$A = \frac{1}{2} 	imes (2x \cos 	heta) 	imes (x \sin 	heta)$$A = x^2 \sin 	heta \cos 	heta = \frac{x^2}{2} \sin(2	heta)$.Since the maximum value of $\sin(2	heta)$ is $1$ (when $2	heta = 90^\circ$ or $	heta = 45^\circ$),the maximum area is $\frac{x^2}{2}$ sq. units.

$A$ triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having a fence are of the same length $x$. The maximum area (in sq. units) enclosed by the park is

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