Question

If the sum of the lengths of the hypotenuse and one side of a right-angled triangle is given, the area of the triangle is maximum when the angle between these sides is

• A. ${60}^{\circ }$
• B. ${90}^{\circ }$
• C. ${30}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

The correct option is A ${60}^{\circ }$

Let $ABC$ be a right-angled triangle in which side $BC=x\left(\text{say}\right)$ and hypotenuse $AC=y\left(\text{say}\right).$
Given $x+y=k\left(\text{const.}\right)$
$\Rightarrow y=k-x$
Now, the area of the triangle $ABC$ is given by
$A=\frac{1}{2}BC\cdot AB=\frac{1}{2}x\sqrt{(y^2-x^2)}=\frac{1}{2}x\sqrt{\left[\right(k-x{)}^2-x^2]}$
Let $u=A^2=\frac{1}{4}{x}^2(k^2-2kx)$
$\Rightarrow \frac{du}{dx}=\frac{1}{2}k(kx-3x^2)$ and $\frac{d^{2}u}{d{x}^2}=\frac{1}{2}k(k-6x)$
For maximum or minimum of $u,\frac{du}{dx}=0\Rightarrow x=\frac{k}{3}(\because x\ne 0)$
when $x=\frac{k}{3},\frac{d^{2}u}{d{x}^2}=-\frac{1}{2}{k}^2<0$
$\Rightarrow u$ i.e., $A$ is maximum when $x=\frac{k}{3}$ and $y=k-x=\frac{2k}{3}.$
Now, $\cos \theta =\frac{BC}{AC}=\frac{x}{y}=\frac{1}{2}\Rightarrow \theta =\frac{\pi }{3}.$
Hence, the required angle is $\frac{\pi }{3}.$