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Question

If a2r21+b2r22+c2r23=k, then

A
k<144
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B
k 144
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C
k = 144
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D
k = 16
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Solution

The correct option is D k 144
We know that Δ=abc/4R
so Δ1=OB×OC×BC4R1=aR24R1
aR1=4ΔR2 similarly
bR2=4Δ2R2,cR3=4Δ3R2

Hence aR1+bR2+cR3=4R2(Δ1+Δ2+Δ3)=4ΔR2
BOD=12BOC=A

From ΔOBD,a/22r1=tanA
ar1=4tanA, similarly br2=4tanB,cr3=4tanC
Hence ar1+br2+cr3=4(tanA+tanB+tanC)
=4tanAtanBtanC[A+B+C=180o]

a=4r1tanA=4Δ1R1R2R1r1=R2tanAΔ1

similarly R2r2=R2tanBΔ2,R3r3=R2tanCΔ3
Hence
R1r1+R2r2+R3r3=R2[tanAΔ1+tanBΔ2+tanCΔ3]

k=a2r21+b2r22+c2r23=16(tan2a+tan2B+tan2C)
tan2A+tan2B+tan2C3(tan2Atan2Btan2C)1/3 (A.M.G.M.) (1)
Similarly tanA+tanB+tanC3(tanAtanBtanC)13
tanAtanBtanC3(tanAtanBtanC)13
(tanAtanBtanC)233 (2)
From (1) and (2) we get
k=16(tan2A+tan2B+tan2C)16×3×3=144


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