Lesson Proof of the formula for the area of a triangle via the radius of the inscribed circle

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Proof of the formula for the area of a triangle via the radius of the inscribed circle

In this lesson you will learn the proof of the formula for the area of a triangle via the radius of the inscribed circle:

,

where

is the radius of the inscribed circle in the triangle,

is the triangle semiperimeter

, and

and

are the measures of the triangle sides.

Let

ABC be a triangle with the side measures

and

(Figure 1).

As you know, the angle bisectors of any triangle are concurrent (see the lesson
Angle bisectors of a triangle are concurrent under the topic Triangles of the
section Geometry in this site). Their common intersection point is equidistant
from the sides of the triangle and is the center of the inscribed circle.

In Figure 1 the angle bisectors are shown in green; P is their intersection
point, and the inscribed circle is shown in red.

The angle bisectors AP, BP and CP divide the triangle DELTA

ABC in three
smaller triangles DELTA

APB,

APC and DELTA

BPC (Figure 2). Therefore, the
area of the triangle DELTA

ABC is the sum of the areas of the triangles DELTA

APB,

APC and DELTA

BPC:

Figure 1. To the formulation
of the statement

Figure 2. To the proof
of the statement

.

The area of each of the three smaller triangle is half the product of the measure of the corresponding side of the original triangle and the radius of the inscribed circle, because the radius drawn to the tangent point is perpendicular to that side and, hence, is the altitude in the smaller triangle. In other words,

.

It is exactly what has to be proved.

Example 1

Find the radius of the inscribed circle in the triangle with the side measures of 4 cm, 13 cm and 15 cm.

Solution

The semiperimeter of the triangle is

.
The area of the triangle, according to the Heron's formula (see the lessons
- Proof of the Heron's formula for the area of a triangle and
- One more proof of the Heron's formula for the area of a triangle
in this site), is

sqrt%2816%2A%2816-4%29%2A%2816-13%29%2A%2816-15%29%29

.

Now, using the formula

proved above, you can calculate the radius of the inscribed circle. It is

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