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

Solution

## The correct option is C EquilateralIn right $$\triangle ODC$$,$$OD=acos2{\theta}$$ and $$DC=asin2{\theta}$$So, $$AD= a+acos2{\theta}$$and $$BC=2DC=2asin2{\theta}$$Area of $$\triangle ABC =\dfrac{1}{2} BC.AD$$$$A(\theta)=a^{2}sin2{\theta}+\dfrac{a^{2}}{2}sin4{\theta}$$$$A'(\theta)=2a^{2}cos{2\theta}+2a^{2}cos{4\theta}$$For maxima or minima,$$A'(\theta)=0$$$$cos{2\theta}+cos{4\theta}=0$$$$\Rightarrow {\theta}=\dfrac{\pi}{2},\dfrac{\pi}{6}$$$$A''(\theta)=-6{\sqrt{3}}a^2$$ at $${\theta=\dfrac{\pi}{6}}$$Hence, area is maximum at $${\theta=\dfrac{\pi}{6}}$$So, vertical angle $$BAC=2(\dfrac{\pi}{6})=\dfrac{\pi}{3}$$Since, ABC is isosceles, so $$\angle ABC = \angle ACB=\dfrac{\pi}{3}$$Hence , $$\triangle ABC$$ is equilateral. Mathematics

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