If in the triangle ABC, $B = 45^{0}, t h e n a^{4} + b^{4} + c^{4}$ is equal to

2a² (b² +c²)

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2c² (a² + b²)

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2b²(a² + c²)

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2 (a²b² + b²c² + 3a²c²)

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Solution

The correct option is C
2b²(a² + c²)

$c o s B = \frac{c^{2} + a^{2} - b^{2}}{2 a c} \Rightarrow \frac{1}{\sqrt{2}} = \frac{c^{2} + a^{2} - b^{2}}{2 a c} \Rightarrow \sqrt{2} a c = c^{2} + a^{2} - b^{2} On squaring both sides, we get 2 a^{2} c^{2} = c^{4} + a^{4} + b^{4} + 2 c^{2} a^{2} - 2 (c^{2} + a^{2}) b^{2} \Rightarrow c^{4} + a^{4} + b^{4} = 2 (c^{2} + a^{2}) b^{2}$

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Similar questions

A B C

2 a^{2} b^{2} + 2 b^{2} c^{2} = a^{4} + b^{4} + c^{4},

B

△ A B C, a^{4} + b^{4} + c^{4} = 2 a^{2} b^{2} + b^{2} c^{2} + 2 c^{2} a^{2},

sin A

a^{4} - 2 (b^{2} + c^{2}) a^{2} + b^{4} + b^{2} c^{2} + c^{4} = 0

∠ A

60^{\circ}

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△ A B C

2 a^{2} b^{2} + 2 b^{2} c^{2} = a^{4} + b^{4} + c^{4}, ∠ B =

△ A B C,

a^{4} + b^{4} + c^{4} = 2 c^{2} (a^{2} + b^{2}),

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