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Question

The value of 1r12+1r22+1r32+1r2 is


A

a2+b2+c2āˆ†2

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B

0

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C

1

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D

āˆ†2a2+b2+c2

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Solution

The correct option is A

a2+b2+c2āˆ†2


Explanation for the correct answer:

Consider the standard notation followed in triangle geometry.

Let r be the radius of the incircle of the triangle.

Let r1,r2,r3 be the radii of the excircles drawn opposite to the vertices A,B,C respectively.

Let s be the semi-perimeter of the triangle where s=a+b+c2

Let a,b,c be the lengths of the sides of the triangles opposite to the vertices A,B,C respectively..

Let āˆ†be the area of the triangle.

We know,

āˆ†=r.s=r1s-a=r2s-b=r3s-c

āˆ“1r=sāˆ†1r1=s-aāˆ†1r2=s-bāˆ†1r3=s-cāˆ†

ā‡’ 1r12+1r22+1r32+1r2=1āˆ†2s-a2+s-b2+s-c2+s2

=1āˆ†2s2-2as+a2+s2-2bs+b2+s2-2cs+c2+s2

=1āˆ†24s2-2sa+b+c+a2+b2+c2

=1āˆ†24s2-4s2+a2+b2+c2 (āˆµa+b+c=2s)

āˆ“ 1r12+1r22+1r32+1r2=a2+b2+c2āˆ†2

Hence, option(A) is the correct answer.


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