Question

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 fences are of same length $x$. The maximum area enclosed by the park is

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

Answer 1

The correct option is B

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

Explanation for the correct option:

Step 1: Find the relation between angle and side:

Assume the two equal sides $AB$ and $AC$ have length $x$.

Also, assume $\angle ACB=\theta$

Draw a perpendicular line $AD$.

So, $\sin \theta =\frac{AD}{x}$

And $\cos \theta =\frac{DC}{x}$

Step 2: Find the area of the smaller triangle:

Use the area of triangle formula to find the area of $∆ADC$.

$\begin{array}{rcl}Ar& =& \frac{1}{2}\times AD\times DC\\ & =& \frac{1}{2}\times x\sin \theta \times x\cos \theta \\ & =& \frac{x^2}{2}\sin \theta \cos \theta \end{array}$ $\left[\because ar∆ABC=\frac{1}{2}·bh\right]$

Step 3: Find the area of the whole triangle:

Use the obtained area to get the area of the bigger triangle.

$Ar∆ABC=2Ar∆ADC$

Find the area of $∆ABC$ using the above relation.

$\begin{array}{rcl}Ar∆ABC& =& 2·\frac{x^2}{2}\sin \theta \cos \theta \\ & =& \frac{x^2}{2}\sin 2\theta \end{array}$

Step 4: Maximize the area of $∆ABC$:

Maximize the area of the triangle by maximizing $\begin{array}{rcl}& & \sin 2\theta \end{array}$.

Maximum value of $\sin 2\theta =1$

So, maximum value of $\begin{array}{rcl}Ar∆ABC& =& \frac{x^2}{2}\end{array}$

Hence, option (B) is the correct answer.