Question

Of all isosceles triangles inscribed in a given circle the triangle having maximum area is .....triangle

• A. Equilateral
• B. Scalene
• C. Right angled isosceles
• D. isosceles

Answer 1

Answer: (A)

Equilateral

In right $△ODC$,
$OD=acos2\theta$ and $DC=asin2\theta$
So, $AD=a+acos2\theta$
and $BC=2DC=2asin2\theta$
Area of $△ABC=\frac{1}{2}BC.AD$
$A\left(\theta \right)=a^{2}sin2\theta +\frac{a^2}{2}sin4\theta$
$A^{\prime }\left(\theta \right)=2a^{2}cos2\theta +2a^{2}cos4\theta$
For maxima or minima,
$A^{\prime }\left(\theta \right)=0$
$cos2\theta +cos4\theta =0$
$\Rightarrow \theta =\frac{\pi }{2},\frac{\pi }{6}$
$A^{\prime \prime }\left(\theta \right)=-6\sqrt{3}{a}^2$ at $\theta =\frac{\pi }{6}$
Hence, area is maximum at $\theta =\frac{\pi }{6}$
So, vertical angle $BAC=2\left(\frac{\pi }{6}\right)=\frac{\pi }{3}$
Since, ABC is isosceles, so $\angle ABC=\angle ACB=\frac{\pi }{3}$
Hence , $△ABC$ is equilateral.