Question

The sum of the hypotenuse and a side of a right abgled triangle is constant . If the area of the triangle is maximum then the angle between the hypotenuse and the given side

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is C $\frac{\pi }{3}$

Let $AC=y$ and $BC=x$ and $x+y=K$ ($K$ is a constant) suppose $\theta$ is the angle between $BC$ and $AC$. Let $A$ be the area of a triangle. Then,

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

$\Rightarrow A^2=\frac{1}{4}{x}^2(y^2-x^2)⟶\left(1\right)$

$\Rightarrow \frac{x^2}{4}[{(K-x)}^2-x^2]=A^2$

$\Rightarrow A^2=\frac{x^2}{4}\{{(K-x)}^2-x^2\}$

$\Rightarrow A^2=\frac{K^2{x}^2-{2Kx}^3}{4}$

Differentiating $\left(1\right)$ w.r.t $x$, we get;

$\frac{dA}{dx}=\frac{K^{2}x-3{Kx}^2}{4A}⟶\left(2\right)$

Putting $\frac{dA}{dx}=0$, We get;

$0=\frac{K^{2}x-3{Kx}^2}{4A}$

$\Rightarrow x=\frac{K}{3}$

Again, differentiating $\left(2\right)$, We get,

$2{\left(\frac{dA}{dx}\right)}^2+2A\frac{d^{2}A}{{dx}^2}=\frac{{2K}^2-12Kx}{4}⟶\left(3\right)$

Put $\frac{dA}{dx}=0$ and $x=\frac{K}{3}$ in $\left(3\right)$, We get;

$\frac{d^{2}A}{{dx}^2}=\frac{-K^2}{4A}<0$

Thus, $A$ is maximum when $x=\frac{K}{x}$

When $x=\frac{4}{3},y=K-\frac{K}{3}=\frac{2K}{3}$

$\therefore$ $\cos \theta =\frac{x}{y}$

$\Rightarrow \cos \theta =\frac{K/3}{2K/3}=\frac{1}{2}$

$\Rightarrow \cos \theta =\cos \frac{\Pi }{3}$

$\Rightarrow \theta =\frac{\Pi }{3}$