Question

The sum of lengths of the hypotenuse and another side of a right angled triangle is given. The area of the triangle will be maximum if the angle between them is:

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{5\pi }{12}$

Answer 1

Answer: (C)

$\frac{\pi }{3}$

Assume $a,b,\sqrt{a^2+b^2}$ are side of given right triangle
$b+\sqrt{a^2+b^2}=k\left(constant\right)\Rightarrow a^2+b^2=k^2+b^2-2kb\Rightarrow a^2=k(k-2b)$
Let A be the area of this triangle
$\Rightarrow A^2=1/4a^2{b}^2=1/4k{b}^2(k-2b)$
For maximum area A
$\frac{d{A}^2}{db}=0=1/4k\left[b^2\right(-2)+(k-2b\left)2b\right]\Rightarrow k=3b$
If $\theta$ is the angle between $b$ and hypotenus then
$cos\theta =\frac{b}{\sqrt{b^2+a^2}}=\frac{k/3}{k-b}=\frac{k/3}{2k/3}=1/2$
Hence $\theta =\frac{\pi }{3}$