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Question

A ΔABC with sides a,b,c opposite to the angles A,B,C satisfies the inequality a2+b2+c2sin2A+sin2B+sin2C9.
Given abc, and a attends its maximum possible value, then

A
Maximum area of the triangle is 94
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B
Maximum area of the triangle is 34
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C
Circumradius of the triangle is 3
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D
Inradius of the triangle when its area is maximum is 32(21)
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Solution

The correct options are
A Maximum area of the triangle is 94
D Inradius of the triangle when its area is maximum is 32(21)
If R be the circumradius of ΔABC, then
asinA=bsinB=csinC=2R

a=2RsinA,b=2RsinB,c=2RsinC

a2+b2+c2sin2A+sin2B+sin2C9

(2RsinA)2+(2RsinB)2+(2RsinC)2sin2A+sin2B+sin2C9

(2R)29
2R3

Now, a=2RsinA
a3sinA
amax=3 at sinA=1
A=90B+C=90

Area of ΔABC =12bcsinA=12(2RsinB)(2RsinC)=94(2sinBcosB) (B+C=90)=94sin2B
Area is maximum when sin2B=1 and maximum area =94
sin2B=1B=45C=45

Since, ΔABC is a right angled triangle,
Circumradius, R=32

Inradius, r=4Rsin(A/2).sin(B/2).sin(C/2)
=2×3×12×sin2(B/2) (B=C)
=62×(1cosB2)=32(21)

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