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For triangle ABC,R=5/2 and r=1. Let I be...

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->BC,CA and AB,` respectively. The value of `(ID*IE*IF)/(IA*IB*IC)` is equal to (a) `5/2` (b) `5/4` (c) `1/10` (d) `1/5`

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The correct Answer is:
10
`((IA)(IB)(IC))/((ID)(IE)(IF)) = ((r "cosec"(A)/(2)) (r "cosec"(B)/(2)) (r "cosec"(C)/(2)))/(r^(3))`
`= (1)/("sin"(A)/(2) "sin"(B)/(2) "sin"(C)/(2)) = (4R)/(r) = 10`
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