Four special segments live in every triangle. The exam favourite is the median — because the centroid cuts every one of them in a 2 : 1 ratio.
The problem
Median AM of a triangle is 18 cm long, and the medians meet at centroid G. How long is AG?
The centroid divides each median 2 : 1 from the vertex: AG = (23)(18) = 12, and GM = 6. A is the short piece — the 2 : 1 runs long-side-first from the vertex, not from the midpoint. B halves, but the centroid is not the midpoint of a median.
| Segment | From / to | Key property |
|---|---|---|
| median | vertex → midpoint of opposite side | splits the side in half |
| altitude | vertex → opposite side, at 90° | the height for area |
| angle bisector | vertex → opposite side | splits the ANGLE in half |
| perpendicular bisector | midpoint of a side, at 90° | needs no vertex at all |
Median halves the side; bisector halves the angle. Exams bank on students swapping those two words.
Drawn to scale: the gold median AM is 18 cm, and G sits two thirds of the way down it from A. The tick marks show M is the midpoint of BC.
AG : GM = 2 : 1
the long piece touches the vertex
AG = 23 · 18 = 12, GM = 6
thirds of the whole median
Read the ratio from the midpoint and you get AG = (13)(18) = 6 — that is GM, the short piece. Treat G as the midpoint of the median and you get 18 ÷ 2 = 9. Both are offered on the exam. The long piece always touches the vertex.
Physical anchor: the centroid is the triangle's balance point — cut the triangle from cardboard and it balances on a pencil tip there. Balance points sit closer to the heavy side, which is why G sits nearer the midpoint than the vertex.
G = x₁ + x₂ + x₃3, y₁ + y₂ + y₃3
average the three vertices
(0,0), (6,0), (0,9) → G = (2, 3)
one of the cleanest formulas in the chapter
| Segments | Meet at the… |
|---|---|
| medians | centroid — the balance point |
| altitudes | orthocenter |
| angle bisectors | incenter — centre of the inscribed circle |
| perpendicular bisectors | circumcenter — centre of the circle through all three vertices |
△ABC has AB = AC = 25 and BC = 14. G is the centroid. Find AG. In an isosceles triangle the median from the apex is also the altitude — so Pythagoras gives its length.
BM = 14 ÷ 2 = 7
M is the midpoint of BC
AM = √25² − 7² = √576 = 24
AM ⊥ BC because AB = AC
AG = ⅔ × 24 = 16
long piece at the vertex
The right triangle is ABM, with leg BM = 7. Using 14 as the leg gives √429, which matches nothing. And the median from B in this triangle is not an altitude — only the one from the apex is.
The pieces can also come as expressions. If AG = 5x − 3 and GM = x + 3, then AG = 2 × GM:
5x − 3 = 2(x + 3) = 2x + 6
the 2 : 1 as an equation
x = 3: AG = 12, GM = 6, AM = 18
add both pieces for the median
A right triangle has legs 6 and 8. Find the median from the right angle to the hypotenuse, and how far the centroid is from the right-angle vertex.
hypotenuse = 10
6-8-10
median = 10 ÷ 2 = 5
its midpoint is the centre of the circle through all three vertices
distance to G = ⅔ × 5 = 103
long piece at the vertex
Different: in a right triangle the median to the hypotenuse is half the hypotenuse — no Pythagoras on the median needed.
The “half the hypotenuse” rule is for the median drawn from the 90° vertex. The medians from the other two vertices end on a leg and need their own right triangle.
Find the centroid of the triangle with vertices (1, 4), (−3, 2) and (6, −1).
x = (1 − 3 + 6) ÷ 3 = 43
mind the −3
y = (4 + 2 − 1) ÷ 3 = 53
mind the −1
G = (43, 53)
leave the thirds
Different: the sums are not multiples of 3, so the centroid has fractional coordinates. Do not round to (1, 2).
In △ABC, BC = 10 and the altitude from A is 9. The altitude from C to AB is 6. Find AB. Every side times its own altitude gives the same area.
area = ½ × 10 × 9 = 45
base BC with its altitude
½ × AB × 6 = 45
base AB with its altitude
AB = 15
90 ÷ 6
Backwards with coordinates: the three vertices add to 3 × G. If G = (1, 3), A = (0, 0) and B = (5, 2), then C = (3 − 0 − 5, 9 − 0 − 2) = (−2, 7).
Expect the centroid inside a bigger problem: a median you must first find with Pythagoras (isosceles apex), pieces given as expressions, a missing vertex from the centroid, or an altitude found from the area. Write AG = 2 × GM or “area = ½ × side × its altitude” as an equation and solve. Ten-second check: AG is twice GM, and the longest side has the shortest altitude.
The problem
In a triangle, the distance from centroid G to the midpoint M of a side is 5 cm. How long is the whole median through M?
GM is the SHORT piece — one third of the median — so the whole is 3 × 5 = 15. A doubles it, treating GM as half; that is exactly the 2 : 1-direction slip the intro question warned about, coming from the other end.
