Question

If the sum of the lengths of the hypotenuse and another side of a right-angled triangle is given, then the area of the triangle is a maximum when the angle between these sides is $\theta$. The value of $\frac{12\theta }{\pi }$ is-

Answer 1

Let S be sum & x be one side
Let "b" be the base of the triangle
$\begin{array}{c}\tan \theta =\frac{b}{x}\\ x\tan \theta =b\\ b=\sqrt{{\left(s-x\right)}^2-x^2}\\ \sin \theta =\frac{b}{\left(s-x\right)}\\ Area=\frac{1}{2}\times \sqrt{{\left(s-x\right)}^2-x^2}\\ \frac{dA}{dx}=\frac{1}{2}\left[\sqrt{{\left(s-x\right)}^2-x^2}+\frac{x\left(2\left(s-x\right)-2x\right)}{2\sqrt{{\left(s-x\right)}^2-x^2}}\right]=0\\ {\left(s-x\right)}^2-x^2+x(s-x)-x^2=0\\ =s^2-sx-2x^2=0\\ 2x^2+sx-s^2=0\\ x=\frac{-s\pm \sqrt{s^2+8s^2}}{2}\\ =\frac{s}{2}\\ \cos \theta =\frac{s}{2.s}=\frac{1}{2}\\ \theta ={60}^{\circ }=\frac{\pi }{3}\\ \frac{12\theta }{\pi }=\frac{12\times \frac{\pi }{3}}{\pi }=4\end{array}$