How to Find the Area of a Triangle: Every Formula Explained

Finding the area of a triangle is one of those topics that follows you from elementary school into the SAT, the GED, and even into engineering. The good news: there’s a formula for every situation, and they’re all friendlier than they look.

In this guide, we’ll cover five different formulas, when to use each one, and how to avoid the classic mistakes that drop test scores. By the end, you’ll be able to look at any triangle — by sides, by coordinates, or by sketch — and find its area in under a minute.

The classic formula

$$A = \tfrac{1}{2} \cdot b \cdot h$$

where $b$ is the base and $h$ is the perpendicular height (not the slanted side).

Example: a triangle with base 8 and height 5 has area $\tfrac{1}{2}(8)(5) = 20$ square units.

The factor of $\tfrac{1}{2}$ exists because every triangle is exactly half of a parallelogram with the same base and height. If you draw a duplicate triangle and rotate it 180°, the two together form a parallelogram — that’s the geometric reason behind the $\tfrac{1}{2}$.

Right triangles are the easiest

For a right triangle, the two legs are the base and the height. No measuring needed — just multiply and halve.

A right triangle with legs 6 and 9 has area $\tfrac{1}{2}(6)(9) = 27$ square units.

Heron’s formula (when you only know the sides)

If you know all three sides $a$, $b$, $c$:

$s = \tfrac{a+b+c}{2}$ (the semi-perimeter)

$A = \sqrt{s(s-a)(s-b)(s-c)}$.

Example: sides 5, 6, 7. $s = 9$. $A = \sqrt{9 \cdot 4 \cdot 3 \cdot 2} = \sqrt{216} \approx 14.7$.

Another worked example. Sides 13, 14, 15. $s = 21$. $A = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = \sqrt{7056} = 84$ square units.

SAS formula (two sides + included angle)

If you know two sides and the angle between them:

$$A = \tfrac{1}{2} \cdot a \cdot b \cdot \sin(C)$$

This shows up on the ACT and in trigonometry.

Example: sides 10 and 12 with a 30° angle between them. $A = \tfrac{1}{2}(10)(12)\sin(30°) = 60 \cdot 0.5 = 30$ square units.

Equilateral triangle formula

For an equilateral triangle with side $s$:

$$A = \tfrac{\sqrt{3}}{4} s^2$$

A nice memorize-once shortcut.

Example: equilateral triangle with side 6. $A = \tfrac{\sqrt{3}}{4} \cdot 36 = 9\sqrt{3} \approx 15.59$.

Coordinate-geometry shoelace formula

If you know the vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$:

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

Example: vertices $(1, 1)$, $(5, 1)$, $(3, 6)$. $A = \tfrac{1}{2}|1(1 – 6) + 5(6 – 1) + 3(1 – 1)| = \tfrac{1}{2}|-5 + 25 + 0| = 10$.

Use this when the SAT gives you triangle vertices in the coordinate plane.

Common mistakes

• Using the slant side as the height (you must use the perpendicular height).
• Forgetting the $\tfrac{1}{2}$.
• Mixing up area (square units) and perimeter (linear units).
• Forgetting absolute value bars in the shoelace formula — you can’t have negative area.
• Using $\sin$ with sides that don’t form the included angle.

Quick practice

1. Find the area of a triangle with base 10 and height 7. Answer: 35 square units.
2. Sides 9, 12, 15. Find the area. Answer: This is a 3-4-5 multiple — right triangle. $A = \tfrac{1}{2}(9)(12) = 54$.
3. Find the area of an equilateral triangle with side 10. Answer: $25\sqrt{3} \approx 43.30$.
4. Sides 8 and 6 with an included angle of 60°. Answer: $\tfrac{1}{2}(8)(6)\sin(60°) = 24 \cdot \tfrac{\sqrt{3}}{2} = 12\sqrt{3} \approx 20.78$.
5. Triangle with vertices $(0, 0), (4, 0), (4, 3)$. Answer: 6 square units.
6. Isosceles triangle with legs 13 and base 10. Answer: drop the altitude (which bisects the base) → height $= \sqrt{13^2 – 5^2} = 12$. Area $= \tfrac{1}{2}(10)(12) = 60$.
7. Triangle with sides 7, 8, 9. Answer: $s = 12$, $A = \sqrt{12 \cdot 5 \cdot 4 \cdot 3} = \sqrt{720} \approx 26.83$.
8. Two sides of 14 and 9 with a 45° included angle. Answer: $\tfrac{1}{2}(14)(9)\sin(45°) \approx 44.55$.

Special triangles you should know

• 30-60-90 triangle. Sides in ratio $1 : \sqrt{3} : 2$. If the short leg is $x$, the longer leg is $x\sqrt{3}$ and the hypotenuse is $2x$. Area: $\tfrac{1}{2} \cdot x \cdot x\sqrt{3} = \tfrac{x^2\sqrt{3}}{2}$.
• 45-45-90 (isosceles right) triangle. Sides in ratio $1 : 1 : \sqrt{2}$. If a leg is $x$, the area is $\tfrac{x^2}{2}$.

These two patterns show up on virtually every SAT and ACT geometry section. Memorize them — they save real time.

Finding height from area (going backward)

Sometimes you’re given the area and one side, and you need the height. Just rearrange:

$$h = \frac{2A}{b}$$

Example: a triangle has area 48 and base 12. What’s the height? $h = \tfrac{2 \cdot 48}{12} = 8$.

Connecting to similar triangles

If two triangles are similar with a scale factor $k$, their areas relate by a factor of $k^2$. Scale a triangle by 3 and the area grows by 9. This is one of the most-tested ideas on the SAT — and the source of many trap answer choices.

Real-world applications

• Architecture. Roof pitches, gables, and trusses are all triangular calculations.
• Land surveying. Plots of land are often partitioned into triangles — Heron’s formula is the surveyor’s friend.
• Computer graphics. Every 3D-rendered scene is built from tiny triangles. The shoelace formula is at the heart of rendering.
• Sailing and aviation. Triangulation locates a vessel by measuring angles to known landmarks.

Test-prep checklist

Before your geometry test or SAT:

• Memorize the five formulas above.
• Practice each one twice with different numbers.
• Drill the 3-4-5, 5-12-13, 30-60-90, and 45-45-90 ratios.
• Practice 5 coordinate-plane area problems using the shoelace formula.
• Review the area-to-scale-factor relationship ($k^2$).

FAQ

What’s the basic formula for the area of a triangle?

$A = \tfrac{1}{2} \cdot b \cdot h$.

What if I only know the three sides?

Use Heron’s formula.

Does the formula change for obtuse or acute triangles?

No — as long as you use the perpendicular height.

How do I find the area of an equilateral triangle?

Use $A = \tfrac{\sqrt{3}}{4} s^2$ where $s$ is the side length.

Are triangle area problems on the SAT?

Yes, often in coordinate-geometry questions. Practice spotting the right base/height pair on a graph.

What if my height falls outside the triangle (obtuse case)?

Use the extension of the base. The perpendicular height may live outside the triangle, but the formula still works.

How is the SAS formula derived?

It comes from $\tfrac{1}{2} \cdot \text{base} \cdot \text{height}$, where the height is one side times $\sin$ of the angle between the two sides.

Can I find the area of a triangle from its perimeter alone?

No — perimeter alone doesn’t fix the shape. You also need at least one angle or one altitude.

What is the centroid of a triangle?

The centroid is where the three medians (lines from each vertex to the midpoint of the opposite side) meet. It splits the triangle into three smaller triangles of equal area, so it’s also the triangle’s “center of mass.”

How can a triangle be both right and isosceles?

A right isosceles triangle has a 90° angle and two equal legs. Its third angle is 45°. The 45-45-90 triangle is the most famous example.

Does the formula change if the triangle is obtuse?

No — the formula $A = \tfrac{1}{2}bh$ still works, but the altitude might fall outside the triangle. You have to extend the base to meet the height. The number you get is exactly the same.

What’s the area of an equilateral triangle in terms of its side length?

$A = \dfrac{s^2 \sqrt{3}}{4}$. Memorize this — it shows up on standardized tests often.

Can two triangles have the same area but look totally different?

Yes. A tall, thin triangle and a short, wide one can have identical areas. Area depends only on the product of base and height, not on the shape itself.

What’s the relationship between the area of a triangle and the area of a parallelogram with the same base and height?

The triangle is exactly half. That’s why the formula has a $\tfrac{1}{2}$ in front — a parallelogram is $bh$, and a triangle is half of one cut along a diagonal.

Why do we need so many different area formulas?

Because different problems give you different starting information. If you have base and height, use $\tfrac{1}{2}bh$. If you have all three sides, Heron’s formula. If you have two sides and the angle between them, the sine formula. Each formula matches a real-world data situation.

One last reminder

Progress in math compounds. A 1% improvement every day for 100 days yields nearly a 3x improvement overall, because each new concept builds on the last. The students who pull ahead aren’t the ones who study the longest — they’re the ones who study consistently, review their mistakes, and refuse to skip the foundations. Show up tomorrow. Then show up the day after. The results take care of themselves.

If you found something useful here, save this article and revisit it after your next practice session. You’ll catch nuances on the second read that you missed on the first, because by then you’ll have the experience to recognize them. Happy practicing.

For more practice, jump into our 6th-grade math worksheets or grab the Middle School Math Bundle for a deep dive.