Question

If 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 $\frac{\pi }{3}$.

Answer 1

Let the sum be $b+hyp=x$.

Third side, $h=\sqrt{hy{p}^2-b^2}$

Area = $\frac{1}{2}\times b\times h$

$⟹\frac{1}{2}\times \sqrt{hy{p}^2-b^2}\times b$

$⟹\frac{1}{2}\times \sqrt{(x-b{)}^2-b^2}\times b$

$⟹\frac{1}{2}\times \sqrt{x^2-2bx}\times b$

squaring both sides,we get

$A^2=\frac{1}{4}{b}^2(x^2-2bx)$

For maximum area, $\frac{dA}{db}=0$

$2A\frac{dA}{db}=\frac{2b{x}^2}{4}-\frac{3b^{2}x}{2}$

$⟹b{x}^2-3b^{2}x=0$

$b=\frac{x}{3}$

$hyp=\frac{2x}{3}$

$\cos \theta =\frac{\frac{x}{3}}{\frac{2x}{3}}$

$\cos \theta =\frac{1}{2}$

$\theta =\frac{\pi }{3}$