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Question

Δ2a2+b2+c2{1r21+1r22+1r23+1r2}=

A
1
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B
0
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C
4Rr2
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D
1r
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Solution

The correct option is B 1
FORMULA USED IN SOLVING PROBLEM

  1. s=a+b+c2
  2. r=Δs (∆ = area of triangle)
  3. r1=Δsa,r2=Δsb,r3=Δsc

Putting equation 2 and 3 in given expression , we get
=(sa)2+(sb)2+(sc)2+(s)2a2+b2+c2

Putting equation 1 in the expression , we get
=14×((a+b+c)2+(ab+c)2+(a+bc)2a2+b2+c2)

On expansion equation get simplified to
=1

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