Circles Inscribed In Triangles

I watched an interesting video that shows the area of a circle inscribed in a 3 : 4 : 5 right-angled triangle is equal to \( \pi \).

This got me thinking about right-angled triangle in general and I noted in the comments to the video that for a right-angled triangle with sides of \(a, b, c\) where \(c\) is the hypotenuse, the radius \(r\) of the inscribed circle is given by:$$r=\frac{a+b-c}{2}$$One can see that when \(a=3\), \(b=4\) and \(c=5\), we get:$$ r =\frac{3+4-5}{2} = \frac{2}{2}=1$$The area of the inscribed circle is thus \( \pi\) square units. So I investigated the general case, for any triangle with sides of \(a\), \(b\) and \(c\), and this is the explanation and formula that Gemini came up with:

Inscribed Circles in Any General Triangle

The area of the inscribed circle (the incircle) can be expressed purely in terms of the side lengths $a$, $b$, and $c$ for any triangle, whether it is right-angled, acute, or obtuse.

To do this, we use two foundational concepts in geometry: Heron's Formula and the relationship between a triangle's area and its inradius.

Step 1: Define the Semi-perimeter

For any triangle with sides $a$, $b$, and $c$, we first calculate the semi-perimeter (half of the perimeter), usually denoted as $s$:

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

Step 2: The Radius Formula

The radius $r$ of an inscribed circle in any triangle is equal to the area of the triangle divided by its semi-perimeter. Using Heron's Formula for the area of the triangle, this gives us:

$$r = \frac{\sqrt{s(s-a)(s-b)(s-c)}}{s}$$

Which simplifies algebraically to:

$$r = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$$

Step 3: Calculate the Area of the Incircle

Now that we have the radius purely in terms of the sides (via $s$), we just plug it into the circle area formula ($A = \pi \, r^2 \,$). The square root neatly disappears:

$$A = \pi \left( \frac{(s-a)(s-b)(s-c)}{s} \right)$$

Summary of the General Formula

If you want to write the area of the inscribed circle entirely in a single expression using just $a$, $b$, and $c$ (without using the $s$ substitution), it looks like this:

$$A = \pi  \left( \frac{(b+c-a)(a+c-b)(a+b-c)}{4(a+b+c)} \right)$$

Both the formula using $s$ and the expanded algebraic formula above will work perfectly for any valid triangle.