In a triangle ABC,if 2 Delta^2 = frac{a^2 b^2 | vedclass

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(C) Given $2 \Delta^2 = \frac{a^2 b^2 c^2}{a^2+b^2+c^2}$.Using the relation $\Delta = \frac{abc}{4R}$,we have $a^2 b^2 c^2 = (4R \Delta)^2 = 16 R^2 \D

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Right angled

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(C) Given $2 \Delta^2 = \frac{a^2 b^2 c^2}{a^2+b^2+c^2}$.Using the relation $\Delta = \frac{abc}{4R}$,we have $a^2 b^2 c^2 = (4R \Delta)^2 = 16 R^2 \Delta^2$.Substituting this into the given equation: $2 \Delta^2 = \frac{16 R^2 \Delta^2}{a^2+b^2+c^2}$.Dividing by $2 \Delta^2$,we get $a^2+b^2+c^2 = 8 R^2$.Using $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$,we get $4R^2(\sin^2 A + \sin^2 B + \sin^2 C) = 8R^2$.Thus,$\sin^2 A + \sin^2 B + \sin^2 C = 2$.Using the identity $\sin^2 A + \sin^2 B + \sin^2 C = 2 + 2 \cos A \cos B \cos C$,we have $2 + 2 \cos A \cos B \cos C = 2$.This implies $2 \cos A \cos B \cos C = 0$,so $\cos A = 0$ or $\cos B = 0$ or $\cos C = 0$.Therefore,one of the angles must be $90^{\circ}$,which means the triangle is right-angled.

In a $\triangle ABC$,if $2 \Delta^2 = \frac{a^2 b^2 c^2}{a^2+b^2+c^2}$,then the triangle is

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