Area of a Triangle — All Formulas with Examples

Finding the area of a triangle is really a matter of picking the formula whose condition matches the information you already have. The most direct method uses a base $b$ and the perpendicular height $h$:

$$A = \frac{1}{2}bh$$

But if the height is not given, the same area can come from two sides and an included angle, from all three side lengths, or from coordinates. The whole procedure starts with one question: which formula's condition actually fits this triangle?

When to use each method

The five formulas below each apply under a specific condition. Match the formula to what you are given.

Base and perpendicular height

Use this when a base and its corresponding height are known. It is the most direct formula and usually the fastest.

$$A = \frac{1}{2}bh$$

The condition matters: $h$ must be perpendicular to the base you chose. A slanted side is not a height unless it meets the base at a right angle. The factor $\frac{1}{2}$ appears because a triangle with base $b$ and height $h$ has half the area of a rectangle or parallelogram built on the same base and height.

Two sides and the included angle

Use this when you know sides $a$ and $b$ and the angle $C$ between them.

$$A = \frac{1}{2}ab\sin C$$

This works because the height relative to side $b$ is $a\sin C$.

Heron's formula

Use this when you know all three sides $a$, $b$, and $c$, but not the height.

$$s = \frac{a+b+c}{2}$$ $$A = \sqrt{s(s-a)(s-b)(s-c)}$$

Here, $s$ is the semiperimeter.

Coordinate formula

Use this when the triangle is given by points $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$.

$$A = \frac{1}{2}\left|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)\right|$$

The absolute value is important because area should not be negative.

Equilateral triangle formula

Use this only when all three sides are equal and each side has length $a$.

$$A = \frac{\sqrt{3}}{4}a^2$$

This is a special case, not a general triangle formula. If the triangle is merely isosceles, the shortcut does not apply.

A full worked example: the $3$-$4$-$5$ triangle

Suppose a triangle has side lengths $3$, $4$, and $5$. Since $3^2 + 4^2 = 5^2$, it is a right triangle, so the sides of lengths $3$ and $4$ are perpendicular. That makes them the easiest base and height to use.

Let $b = 4$ and $h = 3$.

$$A = \frac{1}{2}bh = \frac{1}{2}(4)(3) = 6$$

So the area is $6$ square units.

As a check, Heron's formula gives the same result:

$$s = \frac{3+4+5}{2} = 6$$ $$A = \sqrt{6(6-3)(6-4)(6-5)} = \sqrt{36} = 6$$

The point is not to run every formula each time. It is that different formulas agree when their conditions are satisfied, so you can choose the cheapest one and verify with a second.

Where students get stuck, and how to check each step

• Choosing the formula: If you know base and perpendicular height, use $A = \frac{1}{2}bh$; two sides and the included angle, use $A = \frac{1}{2}ab\sin C$; all three sides, Heron's formula; coordinates, the coordinate formula; an equilateral triangle, the special shortcut.
• The height condition: The most common error is using a side length as the height without checking it is perpendicular to the chosen base.
• The included angle: In $A = \frac{1}{2}ab\sin C$, $C$ must be the angle between sides $a$ and $b$. An angle elsewhere gives the wrong area.
• Heron's setup: Compute the semiperimeter $s$ first, and do not confuse it with the full perimeter. Everything sits inside a square root, so small arithmetic slips matter.
• Coordinate sign: Forgetting the absolute value can produce a negative number, which cannot be an area.

To rehearse the decision and the calculation together, take a triangle with sides $5$, $12$, and $13$. First notice what kind of triangle it is, find the area the fastest way, then redo it with Heron's formula and confirm the two answers agree.