How to Find the Area of a Triangle: Six Formulas, Worked Examples, and the Cases That Break Each One

The area of a triangle is half the base times the height: A = ½bh. That single formula covers most everyday needs, but it fails the moment you don't know a perpendicular height, and that's where the other five methods earn their keep. Pick the formula that matches what you actually have: three sides, two sides and the included angle, three coordinates, or one side and two angles all work without a height. The most common mistake is treating a slanted side as the height, which overstates the answer by the inverse of the sine of the angle between them.

Why ½bh Works and What b and h Actually Measure

The factor of one-half comes from a simple geometric fact: two congruent triangles tile a parallelogram exactly. A parallelogram's area is base times perpendicular height, so one of the two triangles filling it gets half. You can see this yourself by duplicating any triangle, rotating the copy 180 degrees, and matching the slanted edges. The result is always a parallelogram with the same base and the same perpendicular height.

The height must be perpendicular to the chosen base. It can fall inside the triangle, land exactly on a vertex, or drop outside the shape entirely if the triangle is obtuse. All three cases give the same area when you use the right h. A triangle with base 10 cm and perpendicular height 6 cm has an area of 30 cm² regardless of how slanted the other two sides look.

Sides that aren't heights are the most frequent source of wrong answers. If you plug a side length into h, your answer will be too large by a factor of 1/sin of the angle between the base and that side. For an equilateral triangle with side 10, using the side as the height gives 50, but the correct area is about 43.3 because the real height is 10 × sin(60°) ≈ 8.66.

The unit works the same way every time: base in meters and height in meters gives square meters, base in feet and height in feet gives square feet. Convert both measurements to the same unit first. Mixing inches and centimeters is silent and produces nonsense — 6 inches and 10 centimeters gives 30, which has no meaning in either system.

Three Sides Only: Heron's Formula With a Worked Triangle

Heron's formula needs no height at all. Label the sides a, b, c, compute the semi-perimeter s = (a + b + c) / 2, then apply A = √[s(s−a)(s−b)(s−c)]. For a triangle with sides 7, 8, and 9, s = 12, so the area is √(12 × 5 × 4 × 3) = √720 ≈ 26.83 square units.

The formula has ancient roots — it's named after Hero of Alexandria, who recorded it in his Metrica around the first century CE, though the earliest known statement appears in Chinese mathematical text Mathematical Treatise in Nine Sections, compiled by Qin Jiushao in 1247. Both traditions arrived at an equivalent statement, which is one reason the formula is so trusted.

Heron's formula fails quietly on impossible triangles. If a + b ≤ c, no triangle exists, but the formula will often produce either a negative radicand (which stops you with a domain error in most calculators) or — worse — a real answer if you've mislabeled sides. Always check the triangle inequality a + b > c for all three pairings before trusting the output. A triangle with sides 2, 3, and 6 does not exist.

A faster variant for programming or repeated work is Brahmagupta's identity rearrangement: A = ¼√[(a+b+c)(−a+b+c)(a−b+c)(a+b−c)]. It avoids computing s and produces the same result. For large numbers with many decimal places, Heron's formula can lose precision in floating-point arithmetic when the triangle is very thin, so computer implementations sometimes switch to Kahan's stable variant — worth knowing if you're writing code rather than solving a homework problem.

Two Sides and the Included Angle: What Sine Buys You

When you know sides a and b and the angle C between them, the area formula is A = ½ab sin(C). Take a triangle with a = 5, b = 7, and C = 40°. The area is ½ × 5 × 7 × sin(40°) ≈ ½ × 35 × 0.6428 ≈ 11.25 square units.

The included angle is critical. If C is obtuse, sin(C) is still positive, so the area still comes out right — sin(140°) ≈ 0.6428 gives the same result. That's a friendly feature of the sine function and one of the reasons this formula is more forgiving than it looks. Angles above 90° don't produce negative areas, unlike cosine-based formulas you might use for side lengths.

For a right triangle, C = 90°, so sin(C) = 1, and the formula collapses to A = ½ab — exactly the rule for legs. That gives you an easy check: if the sine formula gives a different answer than ½ab on a right triangle, one of your angles is wrong. It also means you never need to remember the right-triangle area rule as separate knowledge; it's a special case.

Sine only fails when the units of the two sides disagree. sin returns a pure ratio, so half of a × b carries the product of the units. Feet times feet gives square feet; feet times inches gives a hybrid that requires conversion before reporting. The angle unit also matters: calculator sine functions accept degrees or radians, and entering 40 radians instead of 40 degrees shifts the answer wildly — sin(40°) is about 0.643, while sin(40 rad) is about 0.745. Check your calculator's mode before computing.

Three Vertices: Coordinates on a Grid and the Shoelace Shortcut

For triangles drawn on a coordinate plane, the area is half the absolute value of a determinant: A = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|. With vertices at (1, 2), (4, 6), and (7, 2), the computation is ½ |1(6 − 2) + 4(2 − 2) + 7(2 − 6)| = ½ |4 + 0 − 28| = ½ × 24 = 12 square units.

The shoelace formula generalizes this to any polygon. List the vertices in order, write each x-coordinate above the y below, cross-multiply diagonally, sum one direction, subtract the sum of the other, and halve. For the same triangle: (1×6 + 4×2 + 7×2) − (2×4 + 6×7 + 2×1) = (6 + 8 + 14) − (8 + 42 + 2) = 28 − 52 = −24, and halving the absolute value gives 12. Same answer, and the method scales to pentagons and hexagons without modification.

Vertex order matters for the sign, not the size. Listing the points counterclockwise produces a positive doubled area; clockwise produces negative. Taking absolute value erases the difference. If you list the vertices out of order — say, (1, 2), (7, 2), (4, 6) — you get a self-intersecting path, and the shoelace formula returns the signed area of the polygon traced, which won't be your triangle. Plot the points first or sort them by angle around the centroid if you aren't certain.

A useful cross-check: the shoelace area equals what you'd get from computing side lengths and running Heron's formula. For the same triangle, sides are √25 = 5, √25 = 5, and 6, so Heron gives s = 8 and area = √(8 × 3 × 3 × 2) = √144 = 12. Two formulas, one answer. If they disagree, your coordinates are wrong somewhere.

One Side and Two Angles: Turning Angles Into a Ratio

If you know one side and have enough angle information, the law of sines supplies the missing side, and the sine formula finishes the job. Suppose side a = 8 with angle A = 35° and angle B = 75°. The third angle is 180° − 35° − 75° = 70°. The law of sines gives b = a × sin(B) / sin(A) = 8 × 0.9659 / 0.5736 ≈ 13.47. Then A = ½ × a × b × sin(C) = ½ × 8 × 13.47 × sin(70°) ≈ ½ × 8 × 13.47 × 0.9397 ≈ 50.65 square units.

The angle-side assignment rule trips most people. Side a is opposite angle A, side b opposite angle B, and so on. Mixing them up gives a plausible-looking but wrong answer. A quick sanity check is to compare side ratios: the longer side should be opposite the larger angle. If side b is longer than a, then angle B must be larger than A — if it isn't, revisit your labels.

This case comes up more in surveying and astronomy than in textbooks. Surveyors using triangulation measure one baseline on the ground and then measure angles to two distant points. The distance to each point follows from law of sines, and the area of the resulting plotting triangle follows from the sine formula. GPS triangulation works differently — it uses ranges from known satellites rather than measured angles — but the geometric logic of "know one length, measure angles, get the rest" is the same.

Angle units carry the same trap as before. In the example above, sin(70°) is about 0.9397, but if the calculator is in radian mode, sin(70) is about 0.7739, producing 41.71 instead of 50.65. Always confirm degrees before running the law of sines or the sine area formula. If you're given radians, convert: one radian is about 57.2958 degrees.

Match the Formula to What You Actually Know

The six formulas overlap, but each has one clear best use. The table below is the fastest way to pick when you're staring at a problem and don't want to derive anything.

An equilateral triangle with side s has its own shortcut: A = (√3/4)s², which for s = 10 gives about 43.30. That result matches ½bh with h = 8.66, and if your h comes out to something else for an equilateral triangle, one of your measurements is off. Similarly, an isosceles triangle with equal sides a and base b uses A = (b/4) × √(4a² − b²), which is just Heron's formula after substituting c = a.

For a right triangle, remember that the legs are the base and height when you pick either as b and the other as h. The hypotenuse never plays a role in ½bh unless you draw the perpendicular altitude from the right angle to the hypotenuse — and then the height is not a side. This alternative form gives the same area and comes up in problems about the altitude: if the legs are 6 and 8, the hypotenuse is 10, and the altitude to it is (6 × 8) / 10 = 4.8. Two area formulas, one triangle.

Real Numbers and the Cases That Fail Quietly

An acre equals 43,560 square feet. A triangular field with base 220 feet and perpendicular height 180 feet has an area of ½ × 220 × 180 = 19,800 square feet, or about 0.4545 acres. Surveying applications frequently report fields in acres, so halving and then dividing by 43,560 is worth committing to memory. The same triangle measured in hectares — one hectare is 10,000 square meters — gives about 0.184 hectares, computed from 1,839.5 square meters.

The triangle inequality is the most common silent failure. If you measure three sides as 4, 6, and 11, Heron's formula requires s = 10.5, then s − 11 = -0.5, and the product under the radical turns negative. The formula returns an error, which is correct. But if you mislabel and the error disappears, you'll get a real answer for a shape that doesn't exist. Check that the two shorter sides sum to more than the longest before computing anything.

Recomputing the answer with a second formula is cheap insurance. Heron's and the sine formula give the same result within rounding error on any real triangle. If Heron's says 26.83 and the sine formula says 27.4, either the triangle is very nearly degenerate (thin enough that rounding matters) or your numbers are wrong. The other low-cost audit: the area cannot exceed ½ × (product of the two longest sides). If the two longest sides are 9 and 8, no answer above 36 is possible for that triangle, so an answer of 45 tells you something went wrong.

Degenerate "triangles" — three collinear points — have area zero. Heron's formula returns zero when the three points lie on a line because the product inside the radical hits zero. The coordinate formula also returns zero. If your answer is zero, you haven't found a tiny triangle; you've found three points that form a line. That's a genuine edge case worth naming, because it comes up in checking collinearity.

Angle-Degree Traps and Calculator Mode

Sine and cosine functions in every scientific calculator and every programming language accept either degrees or radians, and the default varies. Google's built-in calculator historically defaulted to radians; iOS's calculator app uses degrees when you tap the "Rad" button to switch to "Deg" but has caught users out in radians mode. Excel's SIN function always expects radians, so for degrees you must wrap the angle: SIN(RADIANS(40)). Python's math.sin also expects radians, so math.sin(math.radians(40)) is the correct form.

The error factor is substantial. sin(40°) ≈ 0.6428, but sin(40 rad) ≈ 0.7451, a 16% overshoot. On a triangle with ½ab = 35, that's the difference between 22.50 and 26.08 — enough to fail a mechanical tolerance check. For a 90° angle, sin(90°) = 1 but sin(90 rad) ≈ 0.8940, and the error gets worse as the angle grows because radians wrap around every 2π.

The reverse trap exists too: if you're given radians and enter them as degrees. sin(1 rad) ≈ 0.8415, but sin(1°) ≈ 0.01745 — a factor of 48 error. In code, the safest habit is to standardize on one unit inside your function and name the unit in every variable: angle_deg or angle_rad, not just angle. In Excel, the same discipline applies: an unambiguous column header stops weeks of misreported triangles later.

Area problems reported in the wrong units compound the angle mistake. If the inputs are in meters and the reporting system expects square feet, multiplying by 10.7639 converts square meters to square feet, but only after the area is correctly computed. Converting the sides first and then applying the formula gives the same answer, but only if every side and every angle is converted consistently. The safest workflow is to convert all linear measurements to a single unit, compute the area once, then convert the area at the end.

FAQ

What is the area of a triangle with sides 5, 5, and 6?

Twelve square units, since s = 8 and √(8 × 3 × 3 × 2) = √144 = 12. This is an isosceles triangle with base 6 and equal sides 5, and the height to the base is 4 (from the 3-4-5 Pythagorean triple split in half), so ½ × 6 × 4 = 12 confirms the answer. If your answer doesn't match, check whether you've used a slanted side as the height. The formula that always works when three sides are known is Heron's.

Can I use the triangle area formula for any polygon?

No, but the shoelace formula extends it cleanly. For any polygon with known vertex coordinates, list them in order, apply the diagonal cross-multiplication, and halve the absolute result. A regular hexagon with side s can also be split into six equilateral triangles, giving 6 × (√3/4)s² = (3√3/2)s². Splitting into triangles and summing is the general approach, though the shoelace formula is shorter once you have coordinates.

Does it matter which side I call the base?

No, as long as you pair it with the correct perpendicular height. All three of a triangle's base-height pairings give the same area because the product b × h is invariant under side choice, by similar-triangle reasoning. A triangle with sides 6, 8, and 10 gives area 24 whichever side you label base, provided you compute or look up the matching height. The half-and-multiply structure ensures the answer is consistent.

What happens if the three sides don't form a triangle?

Heron's formula fails, usually with a negative number under the square root, when the triangle inequality a + b > c is violated. For sides 2, 3, and 6, s = 5.5, so s − 6 = -0.5 and the product inside the radical is negative, giving no real answer. Geometry software typically returns "null" or "invalid." Always check the inequality before computing, and treat a partial-success output as a red flag rather than a result.

Is Heron's formula or ½ab sin(C) more accurate?

They're algebraically equivalent and give the same exact answer, so the question is numerical precision. Modern calculators handle both fine except for very thin triangles, where Heron's formula can suffer catastrophic cancellation in the subtraction step. For high-precision computing, Kahan's stable variant of Heron's formula avoids this, or use ½ab sin(C) if you have an angle available. On paper, they are interchangeable.

How do I find a triangle's area if I only know the perimeter and one angle?

Not enough information on its own — perimeter plus one angle leaves the triangle underdetermined. Many triangles share a perimeter and have the same one angle but different areas. Among all triangles with a fixed perimeter, the equilateral triangle maximizes the area, which follows from the isoperimetric inequality. If you know the perimeter is 30, the largest possible area is 43.30 (equilateral with side 10); actual triangles could be smaller, including nearly zero if they collapse.

What's the area of a triangle on a coordinate grid with vertices at the origin, (4, 0), and (0, 3)?

Six square units, from the coordinate formula: ½ |0(0−3) + 4(3−0) + 0(0−0)| = ½ × 12 = 6. The same answer appears from base 4 and height 3 using ½bh, since the origin−(4,0) side lies along the x-axis and the perpendicular from (0,3) is exactly 3 units. If you got 12, you forgot the leading ½. This triangle is a right triangle with legs 4 and 3, so this is a special case of ½ab.

Why is the area of an equilateral triangle with side 10 about 43.3 rather than 50?

Because the height is not 10. An equilateral triangle with side 10 has height 10 × sin(60°) ≈ 8.66, so ½ × 10 × 8.66 ≈ 43.3. Using 10 as the height produces 50, which is the area of a parallelogram with those sides and a right angle — not the actual triangle. The (√3/4)s² formula is the direct route, and the 43.3 result is the correct one.