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Q.

In triangle , if r12+r22+r32+r2=kR2a2+b2+c2

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a

15

b

20

c

16

d

14

answer is C.

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Detailed Solution

r12+r22+r32+r2=r1+r222r1r2+r3r2+2r3r=r1+r22+r3r22r1r2+2r3r=r1+r2+r3r22r1+r2r3r2r1r2+2r3r=(4R)22Δsa+ΔsbΔscΔs2ΔsaΔsb+2ΔscΔs=16R22Δ2sb+sa(sa)(sb)ss+cs(sc)2s(sc)+2(sa)(sb)=16R22Δ2(c)(c)Δ22s2+2sc+2s2

2sa2sb+2ab=16R22c22s(a+bc)+2ab=16R22c2(a+b+c)(a+bc)+2ab=16R22c2(a+b)2+c2+2ab=16R2a2b2c2=16R2a2+b2+c2

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