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AP EAPCET 20-Aug-2021 Shift 1 Paper

Section: Mathematics

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Question : 26 of 160

Marks: +1, -0

In a

\Delta ABC, 2 \Delta^{2} = \frac{a^{2} b^{2} c^{2}}{a^{2}+b^{2}+c^{2}}

then the triangle is ________

Equilateral
Isosceles
Right angled
Acute angled triangle

Solution:

2 \Delta^{2} (a^{2}+b^{2}+c^{2}) = a^{2} b^{2} c^{2}

a^{2}+b^{2}+c^{2} = \left(\frac{abc}{\Delta}\right)^{2} \cdot \frac{1}{2} = 8 R^{2}

[\because \frac{abc}{\Delta} = 4 R

where

R

-circumradius;

a = 2 R \sin A, b = 2 R \sin B, c = 2 R \sin C]

\Rightarrow 4 R^{2} (\sin^{2} A + \sin^{2} B + \sin^{2} C) = 8 R^{2}

\Rightarrow \sin^{2} A + \sin^{2} B + \sin^{2} C = 2

\Rightarrow (\cos 2A + \cos 2B + \cos 2C) = -1

\Rightarrow -1 - 4 \cos A \cos B \cos C = -1

\Rightarrow \cos A \cos B \cos C = 0

So, any one among

A, B

and

C

has to be

\frac{\pi}{2}

So, right angled triangle.

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