Let ABC be an isosceles triangle with base BC. If r is the radius of the circle inscribsed in `DeltaABC and r_(1)` is the radius of the circle ecribed opposite to the angle A, then the product `r_(1) r` can be equal to (where R is the radius of the circumcircle of `DeltaABC`)

Options

A. `R^(2) sin^(2)A`
B. `R^(2) sin^(2)2B`
C. `(1)/(2) a^(2)`
D. `(a^(2))/(4)`

Answer

A, B, D

Explanation

`r = (Delta)/(s), r_(1) = (Delta)/(s-a)`
`r_(1) r= (Delta^(2))/(s(s-a)) = (s(s-a) (s-b) (s-c))/(s(s-a))`
`= (s-b) (s-c) = (s-b)^(2) " " ( :' b = c)`
`= ((2s -2b)^(2))/(4)`
`= ((a + b+ c -2b)^(2))/(4) " " ( :' b =c)`
`= (a^(2))/(4) = (4 R^(2) sin^(2)A)/(4) = R^(2) sin^(2) A`
Also if `angleB = theta rArr angle A = pi - 2 theta`
`r_(1) r = R^(2) sin^(2) (pi - 2 theta) = R^(2) sin^(2) 2 theta = R^(2) sin^(2) 2B`