Question

Prove that the perimeter of a right angled triangle of given hypotenuse is maximum

Answer 1

Answer:

The complete question is

Prove that the perimeter of a right angled triangle of given hypotenuse is maximum when the triangle is isosceles.

Solution:

Let the hypotenuse of the triangle is H. Let one of the angle of the triangle is θ

Then

The other two sides of the triangle will be Hsinθ and Hcosθ

The perimeter of the triangle P  = H + Hsinθ + Hcosθ

For maxima or minima

$dP/d\theta=0$

or $\frac{d}{d\theta} (H+H\sin\theta+H\cos\theta)=0$

$\implies \frac{dH}{d\theta}+H\frac{d \sin\theta}{d\theta}+H\frac{d \cos\theta}{d\theta}=0$

$\implies 0+H\cos\theta-H\sin\theta=0$

$H\cos\theta=H\sin\theta$

$\implies \tan\theta=1$

$\implies \theta=45^\circ$

For $\theta=45^\circ$

i.e. the triangle has two of its sides equal or in other words, the triangle is isosceles.

$\frac{d^2P}{d\theta^2} < 0 \implies \text{maxima}$

Therefore, the perimeter of a right angled triangle with given hypotenuse is maximum when the triangle is isosceles.