To find the area of a triangle, multiply half the base by the height drawn perpendicular to that base: area = ½ × base × height. For a 3-4-5 right triangle that gives (3 × 4)/2 = 6 square units. When you do not know the height, use Heron’s formula from the three sides, ½ab·sin C from two sides and the included angle, or the coordinate formula from the vertices.
The area of a triangle is the amount of flat space it covers, in square units. A triangle with base 10 and height 4 covers 20 square units: exactly half the 10 × 4 rectangle that encloses it. That “half a rectangle” picture is the key to the whole topic. But the formula has one condition students keep missing: the “height” must be measured at right angles to the base.
Check any value instantly with a free triangle calculator, which shows the working step by step.
Formula 1: Base × Height (The One to Learn First)
Area = ½ × base × height, where the height is measured at right angles to the base. Pick any side as the base; the height is the perpendicular line from the opposite vertex down to that base (or to the base extended, for obtuse triangles).
Worked example: the 3-4-5 right triangle. Take base 3. The side at right angles to it is 4, so the height is 4: area = ½ × 3 × 4 = 6. Take base 4 instead: the height is 3, and area = ½ × 4 × 3 = 6 again. Two bases, one answer — that agreement is itself a check. Take base 5 (the hypotenuse): the height to it is 12/5 = 2.4, and area = ½ × 5 × 2.4 = 6 a third time.
More examples: base 10, height 4 → 20; base 8, height 5 → 20; base 12, height 7 → 42.
The most common error is using a slanted side as the height. In a 3-4-5 triangle the tempting wrong answer is ½ × 4 × 5 = 10 — wrong, because 5 is slanted relative to the base 4, not perpendicular to it. The height must meet the base at 90°.
Formula 2: Heron’s Formula (When You Know Three Sides)
s = (a + b + c)/2, then area = √(s(s − a)(s − b)(s − c)).
Heron’s formula answers how to find the area of a triangle without the height. Add the three sides, halve the total to get the semi-perimeter s, then take the square root of s(s − a)(s − b)(s − c).
Worked example: 3-4-5. s = (3 + 4 + 5)/2 = 6. Area = √(6 × 3 × 2 × 1) = √36 = 6. It matches the base × height answer exactly — and that match is the point. Whenever two different formulas give the same number, both are almost certainly right.
Worked example: 5-5-6 (isosceles). s = (5 + 5 + 6)/2 = 8. Area = √(8 × 3 × 3 × 2) = √144 = 12. Cross-check by base × height: the height to the base 6 is √(5² − 3²) = 4, so area = ½ × 6 × 4 = 12. Match again.
Worked example: 7-8-9 (scalene). s = (7 + 8 + 9)/2 = 12. Area = √(12 × 5 × 4 × 3) = √720 = 26.8328. There is no neat height here, which is exactly when Heron earns its keep. Worked example: 13-14-15. s = 21, area = √(21 × 8 × 7 × 6) = √7056 = 84 — a whole number, which is why textbooks love this triangle.
Formula 3: ½ab·sin C (Two Sides and the Included Angle)
Area = ½ × a × b × sin C, where C is the angle between sides a and b. This is the SAS formula: two sides and the angle sandwiched between them. It is really the base × height formula in disguise — the height is b·sin C — but it saves you from finding the height first.
Worked example: sides 5 and 7 with a 60° angle between them. Area = ½ × 5 × 7 × sin 60° = ½ × 5 × 7 × 0.8660 = 15.1554. The third side, if you need it, comes from the law of cosines: c² = 25 + 49 − 35 = 39, so c = 6.2450 — and Heron on 5, 7, 6.2450 gives 15.1554 as well.
Worked example: sides 6 and 6 with a 60° angle. Area = ½ × 6 × 6 × sin 60° = 15.5885 — an equilateral triangle of side 6, matching the base × height result from Formula 1. Worked example: sides 8 and 15 with a 90° angle. Area = ½ × 8 × 15 × sin 90° = ½ × 8 × 15 × 1 = 60. With a 90° angle, sin C = 1 and the formula collapses to ½ × base × height — which is why the 8-15-17 right triangle has area 60, and 5-12-13 has area ½ × 5 × 12 = 30.
The word that matters is “included”: C must be the angle between the two sides you multiplied. Any other angle gives a wrong area — and if the given angle is not between the sides, that is SSA, a different case this formula does not cover.
Formula 4: Coordinates (When You Know the Vertices)
Area = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)| for vertices (x₁, y₁), (x₂, y₂), (x₃, y₃). This is the shoelace formula for three points, and it is the fastest method when a triangle is given on a coordinate grid — no side lengths, no angles, no heights needed.
Worked example: (0, 0), (4, 0), (0, 3). Area = ½ |0×(0 − 3) + 4×(3 − 0) + 0×(0 − 0)| = ½ |0 + 12 + 0| = 6. It is the 3-4-5 triangle again, sitting on the axes — and the answer is 6, as always.
Worked example: (1, 2), (7, 2), (4, 6). Area = ½ |1×(2 − 6) + 7×(6 − 2) + 4×(2 − 2)| = ½ |−4 + 28 + 0| = ½ × 24 = 12. By base × height: the base from (1, 2) to (7, 2) is 6, the height is 6 − 2 = 4, so area = ½ × 6 × 4 = 12. Match. The absolute value matters: depending on the order you list the vertices, the inside of the formula can come out negative, and area is never negative.
Which Formula Should You Use?
| What you know | Use | Why |
|---|---|---|
| Base and perpendicular height | ½ × base × height | Fastest; the definition of area |
| Three sides | Heron’s formula | No height needed; works for any triangle |
| Two sides + included angle | ½ab·sin C | No height needed; one multiplication |
| Three vertices | Coordinate / shoelace formula | No lengths or angles needed |
| Right triangle (two legs) | ½ × leg × leg | The legs are base and height already |
Pick the row that matches what the problem gives you. A right triangle needs no formula-hunting at all — the two legs are already a base-height pair.
The 3-4-5 Trap, in Full
The single most-tested area mistake in school mathematics is the 3-4-5 triangle, because it punishes exactly one misunderstanding: what “height” means. The wrong working is ½ × 4 × 5 = 10. It uses the two biggest numbers, it follows the shape of the formula, and it fails the only test that matters — 5 does not meet 4 at a right angle. The height to base 4 is 3; the height to base 3 is 4; the height to base 5 is 2.4. Every correct pairing gives 6.
The size check catches it even without the geometry: a 3-4-5 triangle fits inside a 3 × 4 rectangle of area 12 and fills exactly half of it — so the area must be 6. An answer of 10 would mean the triangle fills more than three-quarters of its own bounding rectangle, which a right triangle never does.
4 Mistakes That Cost Marks
1. Using a slanted side as the height. Covered above: ½ × 4 × 5 = 10, not 6. The height must be perpendicular to the base.
2. Forgetting to halve. Area = ½ × base × height, not base × height. Forgetting the ½ doubles the answer — the 3-4-5 triangle becomes 12 instead of 6.
3. Mixing up the semi-perimeter. In Heron’s formula, s is half the perimeter: (a + b + c)/2. Using the full perimeter gives nonsense — for 3-4-5, the wrong s = 12 gives √6048 ≈ 77.77, wildly too big, which is the clue.
4. Using the wrong angle in ½ab·sin C. The angle must be the one between the two sides. If the problem gives a different angle, find the included one first (angles sum to 180°) or use a different formula.
Frequently Asked Questions
How do you find the area of a triangle?
Multiply half the base by the perpendicular height: area = ½ × base × height. For a 3-4-5 right triangle that is (3 × 4)/2 = 6. Without a height, use Heron’s formula from three sides, ½ab·sin C from two sides and the included angle, or the coordinate formula from the vertices.
What is the area of a 3-4-5 triangle?
6 square units. Area = (3 × 4)/2 = 6. The answer is not 10 — 5 is the hypotenuse, never the height. Heron’s formula agrees: s = 6, area = √(6 × 3 × 2 × 1) = 6.
How do you find the area of a triangle with three sides?
Use Heron’s formula: s = (a + b + c)/2, then area = √(s(s − a)(s − b)(s − c)). For 5-5-6, s = 8 and area = √(8 × 3 × 3 × 2) = 12. For 13-14-15, s = 21 and area = 84.
How do you find the area of a triangle without the height?
Use Heron’s formula if you know all three sides, or ½ab·sin C if you know two sides and the angle between them. Both give the area with no height to find: 5, 7, 60° gives ½ × 5 × 7 × sin 60° = 15.1554.
What is the area of an equilateral triangle of side 6?
15.5885 square units: height 6 × √3/2 = 5.1962, area = ½ × 6 × 5.1962. In general, an equilateral triangle of side s has area (√3/4)s².
Why is the area half base times height?
Because any triangle is exactly half of a parallelogram with the same base and height, and a parallelogram has area base × height.
Keep Practicing
Work these by hand, then verify each with the free triangle calculator: base 10, height 4 → 20; sides 5, 5, 6 → 12; sides 7, 8, 9 → 26.8328; sides 13, 14, 15 → 84; sides 5 and 7 with 60° between them → 15.1554; vertices (0, 0), (4, 0), (0, 3) → 6. If two formulas disagree, one of them used a slanted side as a height.