How to Find the Radius of the Inscribed Circle in a Triangle

Geometry Lessons

The short answer

r = A / s

Divide the area by the semiperimeter, meaning half the sum of the three sides. One division, and the same two quantities you already need for Heron formula.

The inscribed circle is the largest circle that fits inside a triangle, touching all three sides at once. Its radius looks like it should need construction, and it turns out to be a single division involving two numbers you probably already have.

Why area over semiperimeter

Put the centre of the incircle in place and draw a line from it to each of the three corners. The triangle is now cut into three smaller ones, each using one of the original sides as its base.

Here is the part that makes it work. The height of each small triangle is the same, because the incircle touches every side and the perpendicular distance from the centre to each side is exactly r. So the three areas are half a times r, half b times r and half c times r.

Add them and the total is r times half of a plus b plus c, which is r times s. That total has to be the area of the whole triangle, so A equals r times s, and dividing gives r equals A over s. The semiperimeter is not an arbitrary quantity here, it is what falls out of the algebra.

1

Find the semiperimeter

Add the three sides and halve. The same s that Heron formula uses.

2

Find the area

By Heron from the sides, or by any method you already have.

3

Divide the area by s

That is the inradius. No square root at this step.

A worked example

Take the 13, 14, 15 triangle. The sides add to 42, so s is 21. Heron gives the area as 84. Dividing, r is 84 over 21, which is exactly 4.

Check it backwards. If r is 4 and s is 21, then r times s should return the area: 4 times 21 is 84, which matches. That reverse check costs nothing and catches an arithmetic slip immediately.

A 13, 14, 15 triangle with a shaded circle inside touching all three sides, its radius of 4 drawn as a dashed perpendicular to the base, beside the working dividing the area of 84 by the semiperimeter of 21.
The radius is drawn square to the base because that is the only direction that measures the true distance from the centre to a side. It meets each of the other two sides at the same distance.

Semiperimeter, not perimeter

Dividing by the full perimeter instead of half of it is the standard slip, and it returns exactly half the correct radius. The answer still looks like a plausible length, and nothing on the page flags it.

A / perimetergives84 / 422

the perimeter was never halved, so the radius comes out half size

A / sgives84 / 214

the derivation produces s, not the full perimeter

The centre sits at the incenter

The centre of the inscribed circle is where the three angle bisectors meet, a point called the incenter. It is the counterpart to the circumcenter, which comes from perpendicular bisectors instead.

One useful difference is that the incenter always sits inside the triangle, on every triangle without exception. The circumcenter does not, and lands outside whenever the triangle is obtuse. That follows from what each point measures: the incenter is equidistant from the three sides, and the circumcenter from the three corners.

Mistakes that catch people out

  • Dividing by the perimeter. Use the semiperimeter. Forgetting to halve gives half the radius.
  • Confusing the inradius with the circumradius. The inradius fits inside and is always smaller. The circumradius passes through the corners.
  • Measuring the radius to a corner. It runs perpendicular to a side, which is the shortest distance, not to any vertex.
  • Using the wrong area. The area must be of the whole triangle, not of one of the three pieces the construction creates.
  • Confusing the incenter with the centroid. They coincide only on an equilateral triangle.

Practice questions

  1. Sides of 3, 4 and 5. Find the inradius.
  2. Sides of 5, 12 and 13. Find the inradius.
  3. Sides of 6, 6 and 6. Find the inradius.
  4. Sides of 7, 8 and 9. Find the inradius.
  5. A triangle has an inradius of 3 and a semiperimeter of 20. Find its area.
Show the answers
  1. s = 6, A = 6, so r = 6 / 6 = 1
  2. s = 15, A = 30, so r = 30 / 15 = 2
  3. s = 9, A = 15.588, so r = 15.588 / 9 = 1.732
  4. s = 12, A = 26.833, so r = 26.833 / 12 = 2.236
  5. A = r x s = 3 x 20 = 60 square units

Common questions

How do I find the radius of the inscribed circle?

Divide the area by the semiperimeter. The semiperimeter is half the sum of the three sides, so r equals A over s.

What is the inscribed circle?

The largest circle that fits inside the triangle, touching all three sides without crossing any of them. It is also called the incircle.

Why does dividing area by semiperimeter work?

Join the centre to each corner and the triangle splits into three, each with height r. Their areas total r times s, and that has to equal the whole area.

Where is the centre of the inscribed circle?

At the incenter, where the three angle bisectors meet. It always sits inside the triangle, unlike the circumcenter.

Does the incircle touch each side once?

Yes, at exactly one point per side. A circle touching a line at a single point is tangent to it, which is what fitting snugly means here.

Can I use this if I only know the three sides?

Yes. Get the area with Heron formula from the same three sides, then divide by the semiperimeter. No angles or coordinates are needed.

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