# Finding the radius of a circle inside of a triangle

What is the measure of the radius of the circle inscribed in a triangle whose sides measure $8$, $15$ and $17$ units?

I can easily understand that it is a right angle triangle because of the given edges. but I don't find any easy formula to find the radius of the circle. • Maybe you can use the formula $\frac{a}{sin\ A}=\frac{b}{sin\ B}=\frac{c}{sin\ C}=r$ where $r$ is the radius of the incircle. – Poypoyan Dec 19 '15 at 12:15
• that will be useful to find the other angles but how to get the radius? – Rayan Ahmed Dec 19 '15 at 12:16
• Draw other two radii and connect the point $O$ to the vertices. Do you see more right triangles? – A.Γ. Dec 19 '15 at 12:17
• @RayanAhmed Oops...Sorry. That's $2R$ ($R$ is circumcircle) not $r$. – Poypoyan Dec 19 '15 at 12:22

Hint:

use the fact that the area $A$ (of the triangle) is given by: $A=\frac{pr}{2}$ where $p$ is the perimeter and $r$ the incircle radius. This formula can easily be proved ( divide the triangle in three triangle with a common vertex at $O$) and is valid for a convex polygon..

• Area of what????? – Akash Patalwanshi Mar 2 '18 at 3:25
• @AkashPatalwanshi: it seems obvious, but I added. – Emilio Novati Mar 2 '18 at 8:49

You can easily calculate the area of the triangle.

Then divide the triangle into three smaller ones: $AOB, BOC, COA$. Notice that their areas are respectively $AB\cdot r/2, BC\cdot r/2, CA\cdot r/2$. Add them up and compare to the area of $ABC$ you calculated earlier.

Notice, $8^2+15^2=17^2$ hence the triangle is right angled triangle.

area of right triangle $$\Delta=\frac{1}{2}\times 8\times 15=60$$

semi-perimeter of right triangle $$s=\frac{8+15+17}{2}=20$$

Radius of inscribed circle is given as $$r=\frac{\Delta}{s}=\frac{60}{20}=3$$

Spotting that the triangle is right-angled is a great help, but the problem can be done without.

Let the sides of the triangle be $a,b,c$ and define the semi-perimeter $s=\frac {a+b+c}2$ the inradius as $r$ to be found and the area of the triangle as $A$.

Then we have both $A=rs$ and Heron's formula for the area of a triangle given the sides $A=\sqrt {s(s-a)(s-b)(s-c)}$ from which we get $$r=\sqrt{\frac {(s-a)(s-b)(s-c)}{s}}=s\sqrt {\left(1-\frac as\right)\left(1-\frac bs\right)\left(1-\frac cs\right)}$$

The nswer is $$3$$, because one formula for the inradius is given by:
$$\frac{\text{area of triangle}}{\text{semi-perimeter of triangle}}$$

• Nice trick to be used in compitative exams – Sahil Sep 23 '18 at 13:32
• This is indeed a nice trick, but this answer duplicates an existing one. – Blue Sep 23 '18 at 13:34