Finding the radius of a circle inside of a triangle

Question

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

Emilio Novati · Accepted Answer · 2018-03-02T08:47:36.620

9

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..

edited Mar 02 '18 at 08:47

answered Dec 19 '15 at 12:16

Emilio Novati

62,675

score 5 · Answer 2 · answered Dec 19 '15 at 12:16

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.

score 3 · Answer 3 · answered Dec 19 '15 at 12:30

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 $$r=\sqrt{\frac {(s-a)(s-b)(s-c)}{s}}=s\sqrt {\left(1-\frac as\right)\left(1-\frac bs\right)\left(1-\frac cs\right)}$$

Harish Chandra Rajpoot · Answer 4 · 2015-12-19T12:35:37.207

1

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 $$r=\frac{\Delta}{s}=\frac{60}{20}=3$$

edited Dec 19 '15 at 12:35

answered Dec 19 '15 at 12:34

Harish Chandra Rajpoot

37,450

score 1 · Answer 5 · answered Oct 23 '22 at 17:43

In fact, once you have spotted that it is a right triangle there is a simple formula for the diameter of the inscribed circle.
$d = a+b-c$
So your radius is:
$r = (8+15-17)/2 = 3$

It follows from two ways to compute the area of the triangle:
$A = ab/2$
$4A = 2ab = (a+b)^2 - a^2 - b^2 = (a+b)^2 - c^2 = (a+b+c)(a+b-c)$
On the other side
$2A = radius * perimeter = r (a+b+c)$
Combining the two you have
$4A = 2r(a+b+c) = (a+b+c)(a+b-c)$
$\Rightarrow 2r = (a+b-c)$

that's a cool proof (especially the $2A$ aspect) and should be approved as and voted for answer. — FirstName LastName, Feb 14 '23 at 23:51

score -2 · Answer 6 · edited Sep 23 '18 at 13:51

-2

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

edited Sep 23 '18 at 13:51

amWhy

209,954

answered Sep 23 '18 at 13:30

Sahil

1

Nice trick to be used in compitative exams – Sahil Sep 23 '18 at 13:32
6

This is indeed a nice trick, but this answer duplicates an existing one. – Blue Sep 23 '18 at 13:34

Finding the radius of a circle inside of a triangle

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