A triangular park is enclosed on two sides by a fence and on the third side a straight river bank. The two…

1. $\frac{3}{2} x^2$
2. $\sqrt{\frac{x^3}{8}}$
3. $\frac{1}{2} x^2$
4. $\pi x^2$

Solution

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.

Asked in: MHT CET 2024 (03 May Shift 1)

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