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Question

For triangle ABC, R=5/2 and r=1. Let I be the incenter of the triangle and D,E and F be the feet of the perpendiculars from I to BC, CA and AB, respectively.
The value of ID×IE×IFIA×IB×IC is equal to ?

A
52
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B
54
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C
110
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D
15
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Solution

The correct option is C 110
From the figure, IDIC=sinC2
Similarly, IEIA=sinA2
and IFIB=sinB2
Thus, ID×IE×IFIA×IB×IC=sinA2sinC2sinB2
=r4R=14×52=110

151114_117054_ans_3054183aba544236b3199dd0836c5eaf.png

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