# What is the radius of the in-circle of the triangle with vertices (1, 4), (3, 8) and (4, 2).

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### 1 Answer

The radius of the in-circle of a circle with sides a, b, c is given by `r = sqrt(((s - a)(s - b)(s - c))/s)` where a, b , c are the sides of the triangle and s is the semi-perimeter `(a +b+c)/2`

For the triangle that has been given:

the length of the side joining (1, 4) and (3, 8) is `sqrt(3 - 1)^2 + (8 - 4)^2) = sqrt 20`

the length of the side joining (3, 8) and (4, 2) is `sqrt(4 - 3)^2 +(8 - 2)^2) = sqrt 37`

the length of the side joining (4, 2) and (1, 4) is `sqrt(4 - 1)^2 + (4 - 2)^2) = sqrt 13`

s = `(sqrt 20 + sqrt 37 + sqrt 13)/2 = 7.0802`

Substituting these in the expression given above, the in-radius is

`r = sqrt(((7.0802 - sqrt 20)(7.0802 - sqrt 13)(7.0802 - sqrt 37))/7.0802)`

=> r = 1.1299

**The required radius of the in-circle is 1.1299**

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