What Is the Median of a Triangle?
The median of a triangle is a line segment that joins a vertex to the midpoint of the side opposite that vertex. Because it always ends at the midpoint, a median bisects the opposite side into two equal pieces.
Every triangle has exactly three medians, one drawn from each of its three vertices. Unlike an altitude, a median is defined by a point (the midpoint), not by an angle, so it does not have to meet the opposite side at a right angle. All three medians always lie inside the triangle, whatever its shape.
Properties of the Median of a Triangle
The join-to-the-midpoint rule forces a clean and surprisingly rich set of properties:
Three medians, all inside. Every triangle has three, and none ever falls outside the figure.
Each median bisects its side. It splits the opposite side into two equal lengths.
Each median halves the area. A single median divides the triangle into two smaller triangles of equal area, because they share the same height and have equal bases.
The three medians meet at the centroid. This single common point exists for every triangle.
The three medians cut the triangle into six equal-area pieces. Together they slice it into six small triangles, all of the same area.
What Is the Centroid (and the 2:1 Rule)?
A natural question once you have three medians: where do they cross? The three medians of any triangle always meet at one point called the centroid (often written $G$), and it is the triangle's centre of mass, the balance point from the hook above.
The centroid does something precise to every median. It divides each one in the ratio 2 : 1, measured from the vertex:
$$\text{vertex-to-centroid} : \text{centroid-to-midpoint} = 2 : 1.$$
So the centroid sits two-thirds of the way along every median, counting from the vertex. If a median is 9 cm long, the centroid is 6 cm from the vertex and 3 cm from the midpoint.
For a triangle placed on coordinates, the centroid is just the average of the three vertices:
$$G = \left( \frac{x_1 + x_2 + x_3}{3}, ; \frac{y_1 + y_2 + y_3}{3} \right).$$
The Length of a Median (Apollonius's Theorem)
Knowing where a median goes is often not enough; sometimes you need its actual length. The length follows from Apollonius's theorem, which relates a median to the three side lengths. For a triangle with sides $a$, $b$, $c$, the median $m_a$ drawn to side $a$ has length:
$$m_a = \frac{1}{2}\sqrt{2b^2 + 2c^2 - a^2}.$$
Here $a$ is the side the median lands on, and $b$ and $c$ are the other two sides. The formula says the median depends on all three sides, with the side it bisects entering with a minus sign — which makes sense, since a longer opposite side pulls the midpoint farther away and shortens the reach. The medians to the other two sides follow the same pattern with the letters rotated.
Median vs Altitude: What Is the Difference?
The median is most often confused with the altitude, so it is worth fixing the distinction once. Both run from a vertex to the opposite side, but a median is defined by a point (it must reach the midpoint), while an altitude is defined by an angle (it must be perpendicular).
Feature | Median | Altitude |
|---|---|---|
Goes from a vertex to | the midpoint of the opposite side | the opposite side, at 90° |
Always bisects the base? | Yes, by definition | No |
Always perpendicular? | No, not usually | Yes |
Stays inside the triangle? | Yes, always | No (outside for obtuse) |
Three of them meet at | the centroid | the orthocentre |
They coincide only in symmetric cases, such as the median from the apex of an isosceles triangle, which is also the altitude to the base. (See our sibling article on the altitude of a triangle for the full treatment of altitudes.)
Examples of Median of a Triangle
With the centroid rule and the length formula in hand, here is the median doing real work. The problems build from the 2:1 split up to Apollonius's theorem and coordinates.
Example 1 - A median of a triangle is 12 cm long. How far is the centroid from the vertex, and from the midpoint?
The centroid divides the median in the ratio 2:1 from the vertex, so it is two-thirds of the way along:
$$\text{vertex-to-centroid} = \tfrac{2}{3}(12) = 8 \text{ cm}, \qquad \text{centroid-to-midpoint} = \tfrac{1}{3}(12) = 4 \text{ cm}.$$
Final answer: 8 cm from the vertex, 4 cm from the midpoint.
Example 2 - The centroid of a triangle is 10 cm from a vertex along one median. Find the full length of that median
A tempting first move is to read "2:1" as "the centroid is halfway, so double it", giving a median of 20 cm. Check that against the ratio: 2:1 from the vertex means the vertex-to-centroid piece is two of three equal parts, not half. If the median were 20 cm, the centroid at 10 cm would sit at the halfway mark, a 1:1 split, not 2:1.
Done correctly: the 10 cm is two of the three parts, so one part is 5 cm, and the whole median is three parts:
$$m = \frac{3}{2} \times 10 = 15 \text{ cm}.$$
Final answer: 15 cm.
Example 3 - In triangle ABC, sides are $b = 6$ cm, $c = 8$ cm, and $a = 10$ cm. Find the length of the median to side $a$
Apply Apollonius's theorem:
$$m_a = \tfrac{1}{2}\sqrt{2b^2 + 2c^2 - a^2} = \tfrac{1}{2}\sqrt{2(36) + 2(64) - 100} = \tfrac{1}{2}\sqrt{72 + 128 - 100} = \tfrac{1}{2}\sqrt{100} = 5 \text{ cm}.$$
Final answer: 5 cm. (This is a right triangle with the right angle opposite the 10 cm side, so the median to the hypotenuse equals half the hypotenuse, a neat check.)
Example 4 - A triangle has vertices $A(4, 10)$, $B(8, 2)$, and $C(-8, 4)$. Find the centroid
Average the three vertices coordinate-by-coordinate:
$$G = \left( \frac{4 + 8 + (-8)}{3}, ; \frac{10 + 2 + 4}{3} \right) = \left( \frac{4}{3}, ; \frac{16}{3} \right).$$
Final answer: $G = \left(\tfrac{4}{3}, \tfrac{16}{3}\right) \approx (1.33,\ 5.33)$.
Example 5 - Find the median to the longest side of a triangle with sides 5 cm, 7 cm, and 8 cm
The longest side is $a = 8$ cm, with the other two $b = 5$ and $c = 7$:
$$m_a = \tfrac{1}{2}\sqrt{2(25) + 2(49) - 64} = \tfrac{1}{2}\sqrt{50 + 98 - 64} = \tfrac{1}{2}\sqrt{84} \approx \tfrac{1}{2}(9.17) \approx 4.58 \text{ cm}.$$
Final answer: about 4.58 cm.
Example 6 - An equilateral triangle has side 6 cm. Find the length of any median
For an equilateral triangle every median is also an altitude, so all three are equal. Using Apollonius with $a = b = c = 6$:
$$m = \tfrac{1}{2}\sqrt{2(36) + 2(36) - 36} = \tfrac{1}{2}\sqrt{108} = 3\sqrt{3} \approx 5.20 \text{ cm}.$$
Final answer: $3\sqrt{3} \approx 5.20$ cm (the same as its altitude, as expected).
Why the Median of a Triangle Matters
The median is more than a textbook line, it is the geometry of balance and the structure behind several deeper results.
Centre of mass. The centroid is where a flat triangular object balances and where its weight effectively acts. Engineers use the centroid of cross-sections to predict how beams bend and where loads concentrate.
The Euler line and triangle centres. The centroid sits with the orthocentre and circumcentre on a single straight line, the Euler line, one of the elegant surprises of triangle geometry.
Equal-area division. Because a median splits a triangle into two equal areas, it is the natural tool for fairly dividing a triangular region, a property used in land partitioning and in computer graphics when subdividing meshes.
Coordinate geometry shortcuts. The "average the vertices" centroid formula turns a balance-point question into one line of arithmetic, the same averaging idea that underlies finding the mean position of any set of points.
For a Grade 10 student, the median opens the door to the whole family of triangle centres, master the 2:1 centroid rule and the centroid, orthocentre, and circumcentre stop feeling like three unrelated facts.
Where Students Trip Up on Medians
Mistake 1: Reading the 2:1 ratio as "halfway"
Where it slips in: The student treats the centroid as the midpoint of the median, splitting it 1:1.
Don't do this: Place the centroid at half the median's length.
The correct way: The centroid is two-thirds of the way from the vertex, a 2:1 split into three equal parts. The vertex side gets two parts, the midpoint side gets one. The memorizer who recalls "2:1" but not which end is which should remember: closer to the side it does not bisect would be wrong, the larger piece is always on the vertex side.
Mistake 2: Confusing the median with the altitude
Where it slips in: The student assumes a median is perpendicular to the base, or uses the median length as a height.
Don't do this: Treat median and altitude as the same segment.
The correct way: A median goes to the midpoint but is usually not perpendicular; an altitude is perpendicular but usually does not hit the midpoint. They coincide only in symmetric triangles (isosceles apex, equilateral). The second-guesser who knows both still has to check which rule the problem needs.
Mistake 3: Mis-assigning the sides in Apollonius's theorem
Where it slips in: The student puts the wrong side under the minus sign, using a side adjacent to the median instead of the one it bisects.
Don't do this: Subtract $b^2$ or $c^2$ when finding $m_a$.
The correct way: The side the median lands on (the one it bisects) is the one subtracted: $m_a = \tfrac{1}{2}\sqrt{2b^2 + 2c^2 - a^2}$, where $a$ is that side. Label the side opposite the chosen vertex as $a$ before plugging in.
Key Takeaways
The median of a triangle joins a vertex to the midpoint of the opposite side, bisecting it; every triangle has three.
A single median splits the triangle into two equal areas; the three together make six equal-area pieces.
The three medians meet at the centroid, which divides each median 2:1 from the vertex (two-thirds of the way along).
The median's length comes from Apollonius's theorem, $m_a = \tfrac{1}{2}\sqrt{2b^2 + 2c^2 - a^2}$, with $a$ the bisected side.
The most common error is reading the 2:1 split as "halfway"; the larger piece is always on the vertex side.
Practice These Problems to Solidify Your Understanding
A median is 18 cm long. How far is the centroid from the vertex and from the midpoint?
A triangle has sides $a = 14$, $b = 13$, $c = 15$. Find the median to side $a$.
A triangle has vertices $A(0, 0)$, $B(6, 0)$, and $C(3, 9)$. Find its centroid.
Answer to Question 1: 12 cm from the vertex, 6 cm from the midpoint. Answer to Question 2: $m_a = \tfrac{1}{2}\sqrt{2(169) + 2(225) - 196} = \tfrac{1}{2}\sqrt{592} \approx 12.17$ cm. Answer to Question 3: $G = (3, 3)$. If Question 1 gave 9 cm and 9 cm, check that you used the 2:1 split, not a halfway split (see Mistake 1).
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