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Question: In triangle ABC, ∠B = \(\frac{\pi}{2}\), then \(\frac{1}{r_{1}^{2}} + \frac{1}{r_{2}^{2}} + \frac{1}...

In triangle ABC, ∠B = π2\frac{\pi}{2}, then 1r12+1r22+1r32+1r2\frac{1}{r_{1}^{2}} + \frac{1}{r_{2}^{2}} + \frac{1}{r_{3}^{2}} + \frac{1}{r^{2}}is equal to

A

4b2a2c2\frac{4b^{2}}{a^{2}c^{2}}

B

2b2a2c2\frac{2b^{2}}{a^{2}c^{2}}

C

8b2a2c2\frac{8b^{2}}{a^{2}c^{2}}

D

b2a2c2\frac{b^{2}}{a^{2}c^{2}}

Answer

8b2a2c2\frac{8b^{2}}{a^{2}c^{2}}

Explanation

Solution

= 1Δ2\frac { 1 } { \Delta ^ { 2 } }((s – a)2 + (s – b)2 + (s – c)2 + s2)

= 1Δ2\frac { 1 } { \Delta ^ { 2 } }(4.s2 – 2.s (a + b + c) + a2 + b2 + c2)

= 1Δ2\frac { 1 } { \Delta ^ { 2 } }(a2 + b2 + c2)

=

= 2b214a2c2\frac { 2 b ^ { 2 } } { \frac { 1 } { 4 } a ^ { 2 } c ^ { 2 } } = 8b2a2c2\frac { 8 b ^ { 2 } } { a ^ { 2 } c ^ { 2 } }.