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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`

A

`(5)/(2)`

B

`(5)/(4)`

C

`(1)/(10)`

D

`(1)/(5)`

Text Solution

Verified by Experts

The correct Answer is:
C

`rArr AI^(2) = (IE.IF)/(sin^(2)(A//2))`
`rArr (ID.IE.IF)/(IA.IB.IC) = sin.(A)/(2) sin.(B)/(2)sin.(C)/(2) = (r)/(4R) = (1)/(10)`
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