# Isosceles triangle with sides (a,b and c) has circle inside with radius R, Prove R= sqrt([(s-a)(s-b)(s-c)]/s); s=semipereimetr

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You should remember Heron's formula that helps you to evaluate the area of triangle such that:

`A = sqrt(s(s-a)(s-b)(s-c))`

s represents the semiperimeter of triangle

`s = (a+b+c)/2`

a,b,c represent the side of triangle ABC

Inscribing a circle inside the triangle, you may evaluate the area of triangle ABC calculating the sum of areas of triangles formed joining the center of circle with vertices of triangle such that:

`A_(Delta ABC) = A_(Delta AOB) + A_(Delta AOC) + A_(Delta BOC)`

`A_(Delta ABC) = (OD*c)/2 + (OE*a)/2 + (OF*b)/2`

Notice that the heights OD, OE, OF represent the radius of the inscribed circle such that:

`A_(Delta ABC) = (R*c)/2 + (R*a)/2 + (R*b)/2`

Factoring out `R/2` yields:

`A_(Delta ABC) = R*(a+b+c)/2`

Notice that you may substitute s for `(a+b+c)/2` such that:

`A_(Delta ABC) = R*s`

Hence, you may write the area of triangle ABC using two formulas:

`{(A_(Delta ABC) = sqrt(s(s-a)(s-b)(s-c))),(A_(Delta ABC) = R*s):} =>sqrt(s(s-a)(s-b)(s-c)) = R*s`

You may divide by s such that:

`R = (sqrt(s(s-a)(s-b)(s-c)))/s => R = (sqrt s*sqrt((s-a)(s-b)(s-c)))/(sqrt s* sqrt s)`

Reducing by `sqrt s` yields:

`R = sqrt(((s-a)(s-b)(s-c))/s)`

**Hence, expressing the area of triangle ABC in two ways yields that the radius of inscribed circle is given by the following formula `R = sqrt(((s-a)(s-b)(s-c))/s).` **