In the given figure `DeltaABC` is equilateral on side AB produced. We choose a point such that A lies between P and B. We now denote 'a' as the length of sides of `DeltaABC`, `r_1` as the radius of incircle `DeltaPAC` and `r_2` as the ex-radius of `DeltaPBC` with respect to side BC. Then `r_1 + r_2` is equal to

Options

A. (a) `(1)/(2)`
B. (b)`(3)/(2) a`
C. (c) `(sqrt3)/(2) a`
D. (d) `a sqrt2`

Answer

C

Explanation

From the given information, we get the following figure

`angleCAP = 120^(@)`
So, in `DeltaPAC`, using cosine rule, we get
`PC^(2) " " a^(2) + a^(2) -2 (a)^(2) ((-1)/(2)) = 3a^(2)`
In `DeltaPAC`,
Perimeter `= s_(1) = (2a + sqrt3a)/(2)`
Using formula for in radius, `(A)/(2)`, we get
`r_(1) = (sqrt3a)/(2) xx tan 15^(@) = (sqrt3a)/(2) xx (2 -sqrt3)`
In `DeltaPBC, " let " r_(2)` be exradius
Perimeter, `s_(2) = (2a +a + sqrt3a)/(2)`
`r_() = (sqrt3a)/(2) (1+sqrt3) tan 15^(@) ("using " r_(1) = s tan.(A)/(2))`
`= (sqrt3a)/(2) (1+sqrt3) (2-sqrt3)`
`:. r_(1) + r_(2) = (sqrt3a)/(2) (2-sqrt3) (1+sqrt3 + 1) = (sqrt3a)/(2)`