1.

Incircle of DeltaABC touches the sides BC, AC and AB at D, E and F, respectively. Then answer the following question Area of DeltaDEF is

Answer»

`2r^(2) sin(2A) sin(2B) sin(2C)`
`2r^(2) cos.(A)/(2) cos.(B)/(2) cos.(C)/(2)`
`2r^(2) sin (A -B) sin (B -C) sin(C -A)`
none of these

Solution :
Points I, D, C, E are concylic. Therefore,
`angleEID = pi -C`
Also, I is CIRCUMCENTER of `DeltaDEF`. Hence,
`angleDFE = (1)/(2) angleDIE = (pi -C)/(2)`
Similarly, `angleFDE = (pi -A)/(2)`
`angleFED = (pi -B)/(2)`
`:.` Area of `DeltaDEF = 2r^(2) sin (ANGLED) sin (angleE) sin (angleF)`
`= 2r^(2) "cos"(A)/(2) "cos"(B)/(2) "cos"(C)/(2)`
Also by sine rule, in `DeltaDEF`,
`(EF)/(sin angleD) = 2r`
or `EF = 2r cos.(A)/(2)`


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