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 $ABC$ be an isosceles triangle inscribed in the circle with radius a such that $AB=AC$.
$AD=AO+OD=a+a\cos 2\theta$ and $BC=2BD=2a\sin 2\theta$
Therefore, area of the $\Delta ABC$ i.e. $\Delta =\frac{1}{2}BC.AD$
$=\frac{1}{2}2a\sin 2\theta \cdot (a+a\cos 2\theta )$
$=a^2\sin 2\theta (1+\cos 2\theta )$
$\Rightarrow \Delta =a^2\sin 2\theta +\frac{1}{2}{a}^2\sin 4\theta$

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