In a triangle XYZ, let x, y, z be the lengths of sides opposite to the angles X, Y, Z, respectively, and 2s = x + y + z. If `(s-x)/4=(s-y)/3=(s-z)/2` and area of incircle of the triangle XYZ is `(8pi)/3` . then the area of triangle XYZ is _

Answer

A, C, D

Explanation

`(s-x)/(4) = (s-y)/(3) = (s-z)/(2) = (3s -(x +y + z))/(9) = (s)/(9)`
`:. x = (5s)/(9), y = (2s)/(3), z = (7s)/(9)`
Area of incircle `pi r^(2) = (8pi)/(3) rArr r^(2) = (8)/(3)`
`rArr Delta^(2) = r^(2) s^(2) = (8s^(2))/(3)`
`rArr s.(4s)/(9) .(s)/(3).(2s)/(9) = (8)/(3) s^(2)`
`rArr s = 9`
`:. Delta = sqrt((8)/(3)) xx 9 = 6 sqrt6` sq. units
`R = (xyz)/(4Delta) = ((5s)/(9).(2s)/(3).(7s)/(9))/(4 xx 6 sqrt6) = (35)/(24) sqrt6`
`"sin"(X)/(2) "sin"(Y)/(2) "sin"(Z)/(2) = (r)/(4R) = (sqrt((8)/(3)))/(4(35 sqrt6)/(24)) = (4)/(35)`
`sin^(2) ((X +Y)/(Z)) = cos^(2) ((Z)/(2)) = (1 + cos Z)/(2) = (3)/(5)` (Using cosine rule)