Proof that a Triangle's Area is base × height ÷ 2

Area of a Triangle

Introduction (middle-school level)

Goal of this page

Prove that the area of a triangle equals $\dfrac{1}{2}\times$ base $\times$ height, using the congruent right triangles obtained by cutting a rectangle along a diagonal. We also see how to make the proof work for every triangle, obtuse ones included.

1. What we want to prove

Take one side of a triangle as the base, and the length of the perpendicular dropped from the opposite vertex to the base as the height. Then the area $S$ of the triangle is

$$S = \frac{\text{base} \times \text{height}}{2}$$

The "height" is not the length of a slanted side, but the length of the perpendicular dropped from the vertex to the base (or its extension). Below we prove this by fitting the triangle inside a rectangle and halving, and show that the area of any triangle can be found this way.

2. Proof: fit it in a rectangle and halve

Fit triangle $ABC$ exactly inside a rectangle so that the base $BC$ is the bottom edge and the vertex $A$ lies on the top edge. Drop the altitude $AH$ from $A$ to the base $BC$; the foot $H$ divides the base into two parts of length $p$ ($=BH$) and $q$ ($=HC$), and the length of $AH$ equals the height $h$ (the height of the rectangle itself).

Proof that a triangle's area = base × height ÷ 2 (congruent triangles shown in the same color) A B C H p q h congruent (each ph/2) congruent (each qh/2)
Figure 1. Fit triangle $ABC$ in a rectangle and drop the altitude $AH$ from $A$ to the base. The left right triangle $ABH$ (blue ○) is half of a $p\times h$ rectangle cut by the diagonal $AB$; the right right triangle $ACH$ (orange □) is half of a $q\times h$ rectangle cut by the diagonal $AC$. Triangles of the same color are congruent, and each is exactly half of the rectangle enclosing it.

The altitude $AH$ splits triangle $ABC$ into two right triangles $ABH$ and $ACH$.

  • Left (blue): the right triangle $ABH$ is half of a rectangle with sides $p$ ($=BH$) and $h$, cut by the diagonal $AB$. Since a rectangle's diagonal halves its area, the two blue triangles are congruent, and its area is $$\triangle ABH = \frac{1}{2}\,p\,h = \frac{ph}{2}$$
  • Right (orange): the right triangle $ACH$ is half of a rectangle with sides $q$ ($=HC$) and $h$, cut by the diagonal $AC$. The two orange triangles are also congruent, with $$\triangle ACH = \frac{1}{2}\,q\,h = \frac{qh}{2}$$

Triangle $ABC$ is the sum of these two, so

$$S = \frac{ph}{2} + \frac{qh}{2} = \frac{1}{2}(p+q)\,h = \frac{1}{2}\times\underbrace{(p+q)}_{\text{base}}\times\underbrace{h}_{\text{height}}$$

The base is $p+q$ and the height is $h$, so we indeed obtain "base $\times$ height $\div 2$". $\blacksquare$

3. The case of an obtuse triangle

The proof above uses the fact that "the foot $H$ of the altitude lands inside the base". For an obtuse triangle, choosing a short side as the base makes the foot stick out beyond the base, so the straightforward addition fails (Figure 2).

An obtuse triangle: re-taking the longest side as the base brings the altitude foot inside A B C H longest side foot outside rotate: longest → base A C B H foot inside longest side = base
Figure 2. (Left) For an obtuse triangle, taking the non-longest side $BC$ as the base makes the altitude foot $H$ fall outside the base (on its extension). (Right) Rotating the same triangle to re-take the longest side $AC$ (green) as the base always brings the foot inside, so the addition proof of Figure 1 applies directly.

So, choose the longest side as the base. Let the longest side be $BC$ and let $H$ be the foot of the altitude dropped from $A$ to $BC$; then $H$ always lies inside the segment $BC$.

Reason: suppose the foot $H$ stuck out beyond $BC$. If, say, it is beyond $B$, the order along the line is $H,\,B,\,C$, so $HC > BC$. But $AC$ is the hypotenuse of the right triangle $AHC$, so it is longer than the horizontal side $HC$ (a slanted segment is longer than its horizontal shadow). Hence

$$AC > HC > BC$$

which contradicts $BC$ being the longest side. If it stuck out beyond $C$, then similarly $AB > BC$, again a contradiction. Therefore the foot $H$ always lies on $BC$ (inside). There is no exception for obtuse or right triangles, as long as the longest side is taken as the base (for an acute triangle, any side works as the base).

If the foot falls outside, AC > HC > BC, contradicting the longest side A B C H foot outside AC (hypotenuse) HC (horizontal) BC
Figure 3. Suppose the foot $H$ falls outside the base $BC$ (beyond $B$). The hypotenuse $AC$ (purple) of right triangle $AHC$ is longer than the horizontal side $HC$ (green), and $HC$ contains $BC$ (blue) so is longer than $BC$. Hence $AC > HC > BC$, which contradicts $BC$ being the longest side.

If necessary, rotate the triangle so the longest side is at the bottom (rotation changes neither the area, nor the side lengths, nor the heights). Therefore, for any triangle, taking the longest side as the base lets us find the area $S = \dfrac{\text{base} \times \text{height}}{2}$ using the proof of Figure 1 alone, with no case analysis. $\blacksquare$

Frequently asked questions

Q1. Why is the area of a triangle base × height ÷ 2?

Dropping the altitude from the apex to the base splits the triangle into two right triangles. Each right triangle is half of a rectangle whose two sides are part of the base and the height, so its area is half of that rectangle. Adding them gives $\dfrac{ph}{2}+\dfrac{qh}{2}=\dfrac{1}{2}(p+q)h=\dfrac{1}{2}\times$ base $\times$ height.

Q2. Does the same proof work for an obtuse triangle?

Yes. If you take the longest side as the base, the foot of the altitude from the apex always lands inside the base. If the foot were outside, the side from that apex to the far endpoint would be longer than the longest side (because the hypotenuse is longer than its horizontal projection), contradicting that it is the longest side. So even for an obtuse triangle, taking the longest side as the base lets the addition proof of Figure 1 apply directly.

Related pages and references

References

  • Euclid, Elements, Book I, Proposition 41 (a triangle is half of a parallelogram on the same base and of the same height)
  • Wikipedia: Triangle
  • Wikipedia: Area of a triangle