Question

Let a right angled triangle with maximum possible area is drawn keeping the sum of lengths of the hypotenuse and a side constant. If a square is drawn taking the hypotenuse as the side and the ratio of the area of triangle to square is $r$, then which of the following is (are) CORRECT?

• A. $r=\frac{\sqrt{3}}{8}$
• B. The other angles of the triangle are $30°,60°.$
• C. The other angles of the triangle are $45°,45°$
• D. $r=\frac{3}{8}$

Answer 1

Answer: (A), (B)

The correct options are
A $r=\frac{\sqrt{3}}{8}$
B The other angles of the triangle are $30°,60°.$

Let the sides of triangle be $x$ and $y$.


Given that sum of one side and hypotenuse is constant.
Let $h+x=k\cdots \left(1\right)$
$\Rightarrow x+\sqrt{x^2+y^2}=k$
where $k$ is constant.
$\Rightarrow x^2+y^2=(k-x{)}^2\Rightarrow y^2=k^2-2kx$

Area of the triangle,
$A=\frac{1}{2}\times x\times y$
$\Rightarrow A=\frac{1}{2}x\sqrt{k^2-2kx}\Rightarrow \frac{dA}{dx}=\frac{\sqrt{k^2-2kx}}{2}-\frac{2kx}{4\sqrt{k^2-2kx}}\Rightarrow \frac{dA}{dx}=\frac{k^2-2kx-kx}{2\sqrt{k^2-2kx}}\Rightarrow \frac{dA}{dx}=-\frac{3k(x-\frac{k}{3})}{2\sqrt{k^2-2kx}}\Rightarrow \frac{dA}{dx}=0\Rightarrow x=\frac{k}{3}$
As $x>\frac{k}{3}\to \frac{dA}{dx}<0$
$x<\frac{k}{3}\to \frac{dA}{dx}>0$
So, area is maximum for $x=\frac{k}{3}$

Now, using equation $\left(1\right)$,
$h=\frac{2k}{3}$
$y=\sqrt{\frac{4k^2}{9}-\frac{k^2}{9}}=\frac{k}{\sqrt{3}}$
$\tan \theta =\sqrt{3}\Rightarrow \theta ={60}^{\circ }$
Area of the triangle $=\frac{k^2}{6\sqrt{3}}$
Area of the square $=h^2=\frac{4k^2}{9}$
Therefore, $r=\frac{\frac{k^2}{6\sqrt{3}}}{\frac{4k^2}{9}}=\frac{\sqrt{3}}{8}$