Triangle Area: Perpendicular Height and Missing Dimensions

The figure shows an acute triangle with vertices $(1, 1)$, $(9, 1)$ and $(4, 7)$. It is drawn in two parts.

Base. Choose the side from $(1, 1)$ to $(9, 1)$. Both ends are on the same horizontal gridline, so subtract the $x$-coordinates: $9 - 1 = 8$ units.

Height. The side shared by the two parts goes from $(4, 1)$ on the base up to the vertex $(4, 7)$. It lies on a vertical gridline, so it is perpendicular to the base. Subtract the $y$-coordinates: $7 - 1 = 6$ units.

The shared side meets the base at a right angle, so each part is a right triangle with its right angle at $(4, 1)$.

• Left part: the sides forming the right angle are $4 - 1 = 3$ units and $6$ units long. Area: $\frac{1}{2} \times 3 \times 6 = 9$ square units.
• Right part: the sides forming the right angle are $9 - 4 = 5$ units and $6$ units long. Area: $\frac{1}{2} \times 5 \times 6 = 15$ square units.

The whole triangle is the two parts together: $9 + 15 = 24$ square units.

Look at that sum again. Both parts use the same factor $\frac{1}{2}$ and the same height, $6$. The two products differ in only one number: $3$ in the first and $5$ in the second. So you can add those two numbers first and then multiply once. This is the distributive property:

$\frac{1}{2} \times 3 \times 6 + \frac{1}{2} \times 5 \times 6 = \frac{1}{2} \times (3 + 5) \times 6 = \frac{1}{2} \times 8 \times 6 = 24$

The pieces $3$ and $5$ make up the whole base, $8$. So the area of this triangle is half of its base times its height. The same reasoning works whenever the height meets the base between its two ends, because the two base pieces then add up to the whole base. You do not need to cut the triangle every time. You need only a base and the height that is perpendicular to it.

Why does the formula still work when the height is outside the triangle? The next figure shows an obtuse triangle with vertices $(4, 1)$, $(10, 1)$ and $(1, 5)$. Take the side from $(4, 1)$ to $(10, 1)$ as the base.

The base goes to the right from $(4, 1)$. A side going straight up from $(4, 1)$ would make a right angle with it. The side to $(1, 5)$ tilts past that, up and to the left. So the angle at $(4, 1)$ is larger than a right angle, and no point of the base is straight below $(1, 5)$.

The perpendicular path from $(1, 5)$ meets the line that the base lies on at $(1, 1)$, outside the triangle. This happens whenever the base is one of the two sides that form the obtuse angle.

The figure also shows a right triangle with vertices $(1, 1)$, $(4, 1)$ and $(1, 5)$. It fills the gap beside the obtuse triangle. Its side from $(1, 1)$ to $(1, 5)$ is the height you need.

Sometimes you know the area and either the base or the height, and the other length is missing. The formula $A = \frac{1}{2} \times b \times h$ ties the three numbers together, so you can work backward.

The formula says the area is half of the product $b \times h$. Doubling undoes halving, so doubling the area gives the whole product: $2 \times A = b \times h$. Now you know a product and one of its two factors. Dividing undoes multiplying, so divide the doubled area by the length you know. That length is greater than zero, so you can always divide by it.

Each step does the same thing to both sides of the equation, so the two sides stay equal. One condition matters: the base and the height must be a matching pair. The height is the one perpendicular to that base, and both are in the same unit.

Takeaway. The area of any triangle is half of a base times the height that is perpendicular to that base. The height can fall inside the triangle, as it did for the acute triangle, or outside it, as it did for the obtuse triangle. When you know the area and one of those two lengths, double the area and divide by the length you know to find the other. Next, you will use this formula in word problems and learn to spot mistakes in triangle area work.