How to Find the Apothem of a Triangle
The term apothem most often appears when we study regular polygons—shapes whose sides and angles are all equal. In practice, in that context, the apothem is the perpendicular distance from the polygon’s center to any of its sides, which is also the radius of the inscribed circle. So although a general triangle is not a regular polygon, the concept of an apothem can still be useful: for any triangle the distance from its incenter (the point where the three angle‑bisectors meet) to each side is the same. This common distance is called the inradius, and it serves as the triangle’s apothem That's the part that actually makes a difference..
Below you will find a step‑by‑step guide, the underlying geometry, worked examples, and answers to frequently asked questions. By the end of the article you’ll be able to compute the apothem of any triangle quickly and confidently That alone is useful..
What Is an Apothem?
Apothem (from the Greek apóthema, meaning “that which is put away”) is defined as:
The length of a segment drawn from the center of a regular polygon perpendicular to one of its sides That's the part that actually makes a difference..
Because the segment is perpendicular, it also represents the radius of the polygon’s incircle (the circle that touches every side). For a regular n-gon with side length s, the apothem a can be expressed as
[ a = \frac{s}{2\tan(\pi/n)} . ]
When we move to triangles, the only shape that possesses a true “center” equidistant from all three sides is the incenter. Practically speaking, the distance from the incenter to each side is the triangle’s inradius r. Because of this, for any triangle we can treat the inradius as its apothem.
Why the Inradius Equals the Apothem of a Triangle
Consider a triangle △ABC with sides a, b, c opposite vertices A, B, C. That's why let I be the incenter, the intersection of the internal angle bisectors. Draw perpendiculars from I to each side; they meet BC, CA, and AB at points D, E, F respectively That's the part that actually makes a difference. Turns out it matters..
[ ID = IE = IF = r, ]
where r is the inradius. Because each of these segments is perpendicular to a side, they satisfy the geometric definition of an apothem. Hence:
[ \text{Apothem of a triangle} = r = \frac{2\Delta}{p}, ]
where (\Delta) is the area of the triangle and (p = \frac{a+b+c}{2}) is the semiperimeter Worth keeping that in mind..
Deriving the Formula
Starting from the area expressed via the inradius:
[ \Delta = \frac{1}{2} \times (\text{perimeter}) \times r = \frac{1}{2} (2p) r = p r . ]
Solving for r gives the familiar relation:
[ \boxed{r = \frac{\Delta}{p}} . ]
If you prefer to work with side lengths only, you can substitute Heron’s formula for the area:
[ \Delta = \sqrt{p(p-a)(p-b)(p-c)} . ]
Thus the apothem (inradius) becomes
[ r = \frac{\sqrt{p(p-a)(p-b)(p-c)}}{p} = \sqrt{\frac{(p-a)(p-b)(p-c)}{p}} . ]
For an equilateral triangle where all sides equal s, the formula simplifies dramatically:
[ p = \frac{3s}{2},\qquad \Delta = \frac{\sqrt{3}}{4}s^{2}, \qquad r = \frac{\Delta}{p}= \frac{\frac{\sqrt{3}}{4}s^{2}}{\frac{3s}{2}} = \frac{s\sqrt{3}}{6}. ]
So the apothem of an equilateral triangle is (\displaystyle \frac{\sqrt{3}}{6}s) Not complicated — just consistent. But it adds up..
Step‑by‑Step Procedure to Find the Apothem
Whether you have side lengths, angles, or the area, follow these steps:
-
Identify what you know
- Three side lengths (a, b, c) → go to step 2A.
- One side and two angles (or two sides and an included angle) → compute the missing side(s) using the Law of Sines or Cosines, then proceed as in 2A.
- Area ((\Delta)) and semiperimeter (p) → go directly to step 3.
-
Compute the semiperimeter
[ p = \frac{a+b+c}{2}. ] -
Find the area (if not already given)
- Heron’s formula: (\displaystyle \Delta = \sqrt{p(p-a)(p-b)(p-c)}).
- Alternative: (\displaystyle \Delta = \frac{1}{2}ab\sin C) (if you know two sides and the included angle).
-
Calculate the apothem (inradius)
[ r = \frac{\Delta}{p}. ] -
Check your work (optional but recommended)
- Verify that (r) is positive and less than the smallest altitude.
- For an equilateral triangle, confirm that (r = \frac{s\sqrt{3}}{6}).
Worked Examples
Example 1: Equilateral Triangle (side = 10 cm)
- Semiperimeter: (p = \frac{3\times10}{2}=15) cm.
- Area: (\displaystyle \Delta = \frac{\sqrt{3}}{4}\times10^{2}=25\sqrt{3}) cm².
- Apothem: (r = \frac{
\frac{25\sqrt{3}}{15} = \frac{5\sqrt{3}}{3} \approx 2.89 \text{ cm}. ]
Example 2: Scalene Triangle (sides 13 cm, 14 cm, 15 cm)
-
Semiperimeter:
[ p = \frac{13+14+15}{2} = 21 \text{ cm}. ] -
Area (Heron’s formula):
[ \Delta = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84 \text{ cm}^2. ] -
Apothem:
[ r = \frac{\Delta}{p} = \frac{84}{21} = 4 \text{ cm}. ]
Verification: The altitudes are (h_a = \frac{2\Delta}{a} = \frac{168}{13} \approx 12.9), (h_b = 12), (h_c = 11.2). The inradius (r=4) is indeed smaller than all of them, as expected.
Example 3: Triangle Given Two Sides and Included Angle (SAS)
Given (a = 8), (b = 10), and included angle (C = 60^\circ):
-
Find third side (c) (Law of Cosines):
[ c^2 = 8^2 + 10^2 - 2(8)(10)\cos 60^\circ = 164 - 160\left(\frac{1}{2}\right) = 84 \quad\Rightarrow\quad c = 2\sqrt{21} \approx 9.17. ] -
Semiperimeter:
[ p = \frac{8 + 10 + 2\sqrt{21}}{2} = 9 + \sqrt{21} \approx 13.58. ] -
Area (SAS formula):
[ \Delta = \frac{1}{2}ab\sin C = \frac{1}{2}(8)(10)\sin 60^\circ = 40 \cdot \frac{\sqrt{3}}{2} = 20\sqrt{3} \approx 34.64. ] -
Apothem:
[ r = \frac{\Delta}{p} = \frac{20\sqrt{3}}{9 + \sqrt{21}} \approx \frac{34.64}{13.58} \approx 2.55. ]
Geometric Significance and Connections
The apothem (inradius) is far more than a computational exercise; it is the key to the triangle’s incircle—the unique circle tangent to all three sides. This circle is the solution to the classic optimization problem: find the largest circle that fits inside a given triangle.
The incenter (I) (the center of the incircle) is the concurrency point of the three angle bisectors. This gives the apothem a dual nature:
- As a distance: (r = \frac{\Delta}{p}) (area/semiperimeter).
- As a trigonometric function: (r = (p-a)\tan\frac{A}{2} = (p-b)\tan\frac{B}{2} = (p-c)\tan\frac{C}{2}).
You'll probably want to bookmark this section.
The inradius also anchors Euler’s inequality for triangles: [ R \ge 2r, ] where (R) is the circumradius. Equality holds only for the equilateral triangle, reinforcing the special symmetry where the apothem is exactly half the circumradius.
In coordinate geometry, if the vertices are ((x_1,y_1), (x_2,y_2), (x_3,y_3)), the incenter coordinates are the weighted average of the vertices using side lengths as weights: [ I = \left( \frac{ax_1+bx_2+cx_3}{a+b+c},; \frac{ay_1+by_2+cy_3}{a+b+c} \right), ] and the apothem is simply the perpendicular distance from (I) to any side line That alone is useful..
Common Pitfalls
| Pitfall | Why It’s Wrong | Correct Approach |
|---|---|---|
| Confusing apothem with altitude | The apothem is the distance from the incenter to a side; an altitude runs from a vertex to the opposite side. They coincide only in equilateral triangles. | Remember: (r = \frac{\Delta}{p}); altitude (h_a = \frac{2\Delta}{a}). |
