If $a^{2} + b^{2} + c^{2} = 9 R^{2}$ then the triangle is?

Isosceles

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Right angled

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Isosceles right angled

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Equilateral

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Solution

The correct option is D Equilateral
According to sine rule, $a = 2 R s i n A b = 2 R s i n B c = 2 R s i n C$

Now,
$a^{2} + b^{2} + c^{2} = 4 R^{2} (s i n^{2} A + s i n^{2} B + s i n^{2} C)$

$\Rightarrow 4 (s i n^{2} A + s i n^{2} B + s i n^{2} C) = 4 \times (3 \times (\frac{\sqrt{3}}{2})^{2}) = 9$ at $A = B = C = 60^{o}$

Therefore, It should be an Equilateral Triangle
So, Answer is $D$

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