Question

A triangle has corners at `(3 , 2 )`, `(6 ,7 )`, and `(2 ,4 )`. What is the radius of the triangle's inscribed circle?

Answer 1

`0.842`

Explanation:

The area `\Delta` of triangle with vertices `(x_1, y_1)\equiv(3, 2)`, `(x_2, y_2)\equiv(6, 7)` & `(x_3, y_3)\equiv(2, 4)` is given by following formula

`\Delta=1/2|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|`

`=1/2|3(7-4)+6(4-2)+2(2-7)|`

`=5.5`

Now, the lengths of all three sides say `a, b` & `c` of given triangle are computed by using distance formula as follows

`a=\sqrt{(3-6)^2+(2-7)^2}=\sqrt34`

`b=\sqrt{(6-2)^2+(7-4)^2}=5`

`c=\sqrt{(3-2)^2+(2-4)^2}=\sqrt5`

hence, the semi-perimeter `s` of given triangle is computed as follows

`s=\frac{a+b+c}{2}`

`=\frac{\sqrt34+5+\sqrt5}{2}=6.533`

hence, the radius of inscribed circle is given as

`\frac{\Delta}{s}`

`=\frac{5.5}{6.533}`

`=0.842`