Question

A triangular park is enclosed on two sides of 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?

• A. $\frac{3}{2}{x}^2$
• B. $\sqrt{\frac{x^3}{8}}$
• C. $\frac{1}{2}{x}^2$
• D. $\pi x^2$

Answer 1

The correct option is C $\frac{1}{2}{x}^2$

Let ABC be the triangle such that AB=AC=x
And let $\angle ACB=sin\theta$
Draw AD perpendicular to BC.
Area of park =2 Area of $△ADC$
In right triangle ADC,
$\sin \theta =\frac{AD}{x}$
and $\cos \theta =\frac{DC}{x}$
Now, Area of $△ADC=\frac{1}{2}AD\times DC$
$=\frac{1}{2}{x}^2\sin \theta \cos \theta$
$=\frac{1}{4}{x}^2\sin 2\theta$
Area of park $=\frac{1}{2}{x}^2\sin 2\theta$
For maximum area , $\sin 2\theta =1$
So, Area of park $=\frac{1}{2}{x}^2$