Question

The sum of lengths of the hypotenuse and a side of a right angled triangle is given. Show that the area of the triangle is maximum when the angle between them is ${60}^{\circ }$.

Answer 1

Let the hypotenuse of the right angled triangle be $x$ and the height be $y$

Hence its base$=\sqrt{x^2-y^2}$ by pythagoras theorem

Hence its area$=\frac{1}{2}\times$base$\times$height

$=\frac{1}{2}\times \sqrt{x^2-y^2}\times y$

But it is given that $x+y=p$(say)

Substituting this in the area we get

Area$=\frac{1}{2}\times \sqrt{{(p-y)}^2-y^2}\times y$

$=\frac{y}{2}\times \sqrt{p^2+y^2-2py-y^2}$

$=\frac{y}{2}\times \sqrt{p^2-2py}$

Squaring on both the sides we get

$A^2=\frac{y^2}{4}(p^2-2py)$ where $A$ is the area of the given triangle.

$\Rightarrow A^2=\frac{y^2{p}^2}{4}-\frac{p{y}^3}{2}$

For maximum or minimum area $\frac{dA}{dy}=0$

Hence $A^2=\frac{y^2{p}^2}{4}-\frac{p{y}^3}{2}$

Differentiating both sides w.r.t $y$ we get

$\Rightarrow 2A\frac{dA}{dy}=2y\frac{p^2}{4}-\frac{p}{2}3y^2$

$\Rightarrow 2y\frac{p^2}{4}-\frac{p}{2}3y^2=0$

$\Rightarrow p^{2}y-3p{y}^2=0$

$\Rightarrow py(p-3y)=0$

$\Rightarrow y=0,y=\frac{p}{3}$

When $y=\frac{p}{3}\Rightarrow x+y=p$

$\Rightarrow x=p-y=p-\frac{p}{3}=\frac{2p}{3}$

$\Rightarrow \cos \theta =\frac{y}{x}=\frac{\frac{p}{3}}{\frac{2p}{3}}=\frac{1}{2}$

$\Rightarrow \theta =\frac{\pi }{3}={60}^{\circ }$

Hence the area is maximum if the angle between the hypotenuse and the side is ${60}^{\circ }$