In a triangle the sum of two sides is x and the product of the same two sides is y .If `x^(2)-c^(2)=y`, where c is the third side of the triangle , then the ratio of the in-radius to the circum-radius of the triangle is

Options

A. `(3y)/(2x(x+c))`
B. `(3y)/(2c(x+c))`
C. `(3y)/(4x(x+c))`
D. `(3y)/(4c(x+c))`

Explanation

Let the lengths of two sides of the triangle whose third side is of length c be a and b respectively.
It is given that
`a+b=x,ab=y` and `x^(2)-c^(2)=y`
`rArr(a+b)^(2)-c^(2)=ab`
`rArra^(2)+b^(2)-c^(2)=-ab`
`rArr(a^(2)+b^(2)-c^(2))/(2ab)=-1/2`
`rArrcosC=-1/2rArrC=(2pi)/3`
Let `Delta` be the area of `DeltaABC`. Then,
`Delta=1/2absinCrArrDelta=1/2ysin(2pi)/3=(sqrt3y)/4`
Let r and R be respectively the in-radius and circumradius of `DeltaABC`. Then,
`r=(Delta)/s` and `R=(abc)/(4Delta)`
`rArrr/R=(4Delta^(2))/(s(abc))`
`rArrr/R=(4((sqrt3)/4y)^(2))/(((x+c)/2)(y,c))[becauses=(a+b+c)/2=(x+c)/2` and `ab=y]`
`rArrr/R=(3y)/(2c(x+c))`