# A triangle has corners at (2 , 5 ), (4 ,8 ), and (9 ,6 ). What is the radius of the triangle's inscribed circle?

May 28, 2018

See below.

#### Explanation:

From diagram:

$O D = O E = O F = \text{radius of circle}$

$\Delta A O B , \Delta A O C , \Delta B O C$ all have bases that are the sides of $\Delta A B C$.

They also all have heights $r$ ( the radius of the circle.)

So it follows:

$\text{Area of " DeltaABC= "Area of } \Delta A O B + \Delta A O C + \Delta B O C$

$\text{Area of " DeltaABC= } \frac{1}{2} A B \cdot r + \frac{1}{2} A C \cdot r + \frac{1}{2} B C \cdot r$

$\text{Area of " DeltaABC= } \frac{1}{2} \cdot r \left(A B + A C + B C\right)$

$\text{Area of " DeltaABC= } \frac{1}{2} \cdot r \times$perimeter of ABC

Let:

$A = \left(2 , 5\right)$ , $B = \left(4 , 8\right)$, $C = \left(9.6\right)$

Using the distance formula:

$| A B | = \sqrt{{\left(4 - 2\right)}^{2} + {\left(8 - 5\right)}^{2}} = \sqrt{13}$

$| B C | = \sqrt{{\left(9 - 4\right)}^{2} + {\left(6 - 8\right)}^{2}} = \sqrt{29}$

$| A C | = \sqrt{{\left(9 - 2\right)}^{2} + {\left(6 - 5\right)}^{2}} = \sqrt{50} = 5 \sqrt{2}$

Now we need to find area of $\Delta A B C$.

There are different ways we can do this. Having the lengths of the sides we could use Heron's formula, or we could find the height of ABC using line equations. Since we have surds for the lengths of the sides, Heron's formula will be pretty messy and a calculator or computer will be really helpful.

Heron's formula is given as:

$\text{Area} = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$

Where $a , b , c$ are the sides of ABC and s is the semi-perimeter.

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

I won't include the calculation for this, it's too big and too messy. I will just give the result.

$\text{Area} = \frac{19}{2}$

$\frac{19}{2} = \frac{1}{2} \times r \times \left(\sqrt{13} + \sqrt{29} + 5 \sqrt{2}\right)$
$r = \frac{19}{\sqrt{13} + \sqrt{29} + 5 \sqrt{2}} \approx 1.182932116$