Question

An isosceles triangle of vertical angle $2 \theta$ is inscribed in a circle of radius $a$.
Statement I The area of triangle is maximum when $\theta=\frac{\pi}{6}$.
Statement II The area of triangle is minimum when $\theta=\frac{\pi}{6}$.

• A. Only statement I is true
• B. Only statement II is true
• C. Both the statements are true
• D. Both the statements are false

Answer 1

Answer: (A)

Let $A B C$ be an isosceles triangle inscribed in the circle with radius a such that $A B=A C$.
$A D=A O+O D=a+a \cos 2 \theta$ and $B C=2 B D=2 a \sin 2 \theta$
Therefore, area of the $\Delta A B C$ i.e. $\Delta=\frac{1}{2} B C . A D$
$ =\frac{1}{2} 2 a \sin 2 \theta \cdot(a+a \cos 2 \theta) $
$=a^2 \sin 2 \theta(1+\cos 2 \theta) $
$\Rightarrow \Delta =a^2 \sin 2 \theta+\frac{1}{2} a^2 \sin 4 \theta$

Therefore, $\frac{d \Delta}{d \theta}=2 a^2 \cos 2 \theta+2 a^2 \cos 4 \theta$
$=2 a^2(\cos 2 \theta+\cos 4 \theta) $
$\frac{d \Delta}{d \theta} =0 \Rightarrow \cos 2 \theta=-\cos 4 \theta=\cos (\pi-4 \theta)$
Therefore, $2 \theta=\pi-4 \theta \Rightarrow \theta=\frac{\pi}{6}$
$\frac{d^2 \Delta}{d \theta^2}=2 a^2(-2 \sin 2 \theta-4 \sin 4 \theta)<0\left(\text { at } \theta=\frac{\pi}{6}\right) \text {. }$
Therefore, area of triangle is maximum when $\theta=\frac{\pi}{6}$.