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

Let \(\triangle \mathrm{ABC}\) be on isosceles triangle such that
\(\mathrm{AB}=\mathrm{AC}=x\)
\(\therefore \quad \angle \mathrm{ABC}=\angle \mathrm{ACB}=\theta\)
Draw seg \(\mathrm{AD} \perp\) side BC at point D .
\(\therefore \quad \triangle \mathrm{ABD}\) is a right angled triangle such that
\(\mathrm{AD}=x \sin \theta\) and \(\mathrm{BD}=x \cos \theta\)
Similarly, in \(\triangle \mathrm{ACD}\),
\(\mathrm{DC}=x \cos \theta\)
\(\therefore \quad\) In \(\triangle A B C\),
\(\begin{aligned}
& \text { Height }=\mathrm{AD}=x \sin \theta \\
& \text { Base }=\mathrm{BC}=x \cos \theta+x \cos \theta=2 x \cos \theta
\end{aligned}\)
\(\begin{aligned}
\therefore \quad \mathrm{A}(\triangle \mathrm{ABC}) & =\frac{1}{2} \times x \sin \theta \times 2 x \cos \theta \\
& =\frac{x^2}{2}(2 \sin \theta \cos \theta) \\
& =\frac{x^2}{2} \sin 2 \theta
\end{aligned}\)
Since, \(-1 \leq \sin 2 \cdot \theta \leq 1\), for maximum value of \(\sin 2 \theta\), Maximum area \(=\frac{x^2}{2}\) sq. units.