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Question

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


A
Equilateral
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B
Scalene
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C
Right angled isosceles
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D
isosceles
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Solution

The correct option is C Equilateral
In 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. 

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Mathematics

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