The distance of incentre of the right-angled triangle ABC (right angled at A) from B and C are `sqrt10 and sqrt5`, respectively. Then find the perimeter of the triangle.

Answer

12

Explanation

In the figure, `BL = BM =x`
`CM = CN = y`
and `AN = AL =r` (in radius)
In `DeltaILB, "sin"(B)/(2) = (r)/(IB) = (r)/(sqrt10)`
In `DeltaINC, "sin"(C)/(2) = (r)/(IC) = (r)/(sqrt5)`
`rArr "sin" (B)/(2) = (1)/(sqrt2) "sin"(C)/(2)`...(i)
Also, `(B)/(2) = (pi)/(4) - (C)/(2)`
`rArr "sin"(B)/(2) = (1)/(sqrt2) ("cos"(C)/(2) - "sin"(C)/(2))` ...(ii)
From (i) and (ii), we get
`"cos"(C)/(2) = 2 "sin"(C)/(2)`
`:. "tan"(C)/(2) = (1)/(2) = (r)/(y) rArr y = 2r`
Now, `y^(2) + r^(2) = 5`
`rArr 5r^(2) = 5`
`rArr r = 1, y = 2 and x =3`
`:. AB+BC + CA = 2(r + x + y) = 12`