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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 fence are of same length x. The maximum area enclosed by the park is
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The correct answer is:
$\frac{1}{2} x^2$
$\frac{1}{2} x^2$
Area $=\frac{1}{2} x^2 \sin \theta$
$A_{\max }=\frac{1}{2} x^2\left(\right.$ at $\left.\sin \theta=1, \quad \theta=\frac{\pi}{2}\right)$
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