Triangle Area Formulas
The area of a triangle can be found in several ways: from a base and the height, from three sides with Heron’s formula, from two sides and the included angle, or from coordinates. In brief: A = (1/2) · base · height is the standard formula. For sides 13, 14 and 15, Heron’s formula gives A = √(21 · 8 · 7 · 6) = 84. For two sides a and b with included angle C, A = (1/2)ab sin C.
Which formula to use depends on the information given: a drawing with a marked height, three side lengths, an angle between sides or the coordinates of the vertices. This guide presents each formula, explains where it comes from, shows worked examples and gives a table that matches data to formulas. It goes beyond the basic base-and-height rule.
How do you use base and height?
A = (1/2) · b · h, where b is any side chosen as the base and h is the perpendicular distance from the opposite vertex to the line containing that base. The height may fall outside the triangle in an obtuse triangle. Every side can serve as a base, with its own height, and all choices give the same area: b1h1 = b2h2 = b3h3.
Example: a triangle with base 12 cm and height 7 cm has area (1/2) · 12 · 7 = 42 cm2. If the side 10 cm has the height h, then (1/2) · 10 · h = 42, so h = 8.4 cm.
What is Heron’s formula?
With side lengths a, b, c and semi-perimeter s = (a + b + c)/2, the area is A = √(s(s − a)(s − b)(s − c)).
Example 1: sides 13, 14, 15
s = 21. A = √(21 · 8 · 7 · 6) = √7056 = 84. The height onto the side 14 is 2 · 84/14 = 12.
Example 2: sides 7, 8, 9
s = 12. A = √(12 · 5 · 4 · 3) = √720 = 12√5 ≈ 26.83.
Example 3: equilateral triangle with side a
s = 3a/2. A = √((3a/2)(a/2)(a/2)(a/2)) = √(3a4/16) = (√3/4)a2. For a = 6: A = 9√3 ≈ 15.59.
If the three lengths do not satisfy the triangle inequality, one of the factors s − a is zero or negative, and no triangle exists.
How do you use two sides and the included angle?
The height onto side b is h = a sin C, where C is the angle between a and b. So A = (1/2)ab sin C. This formula can be used with any pair of sides and the angle between them.
Example 4
Sides 8 and 11 with included angle 30°: A = (1/2) · 8 · 11 · sin 30° = 44 · 0.5 = 22. With included angle 150°, sin 150° = 0.5 as well, so the area is again 22. For 90°, the formula reduces to (1/2)ab.
Example 5: a parallelogram
Two copies of the triangle make a parallelogram, so a parallelogram with sides a, b and angle C has area ab sin C.
How do you find the area from coordinates?
For vertices A(x1, y1), B(x2, y2), C(x3, y3), the shoelace formula is A = (1/2)|x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)|. It is the half of the absolute value of a 2 × 2 determinant.
Example 6
A(1, 1), B(7, 2), C(4, 6). The sum is 1(2 − 6) + 7(6 − 1) + 4(1 − 2) = −4 + 35 − 4 = 27. So A = 13.5. If the area is zero, the three points lie on one line.
What are the formulas for special triangles?
- Right triangle: A = (1/2)ab with the legs a and b.
- Equilateral triangle: A = (√3/4)a2.
- Isosceles triangle with legs l and base b: A = (b/4)√(4l2 − b2).
- With the inradius r and the semi-perimeter s: A = rs.
- With the circumradius R: A = abc/(4R).
Check with the 3-4-5 triangle: s = 6 and A = 6, so r = 1, and R = 5/2 gives A = 60/10 = 6 ✓.
Which formula fits which data?
| Given | Formula | Example | Area |
|---|---|---|---|
| Base and height | (1/2)bh | b = 12, h = 7 | 42 |
| Three sides | Heron | 13, 14, 15 | 84 |
| Two sides and angle between | (1/2)ab sin C | 8, 11, 30° | 22 |
| Coordinates | shoelace | (1,1), (7,2), (4,6) | 13.5 |
| Equilateral | (√3/4)a2 | a = 6 | 9√3 ≈ 15.59 |
| Right triangle | (1/2) · leg · leg | 5, 12 | 30 |
Where do these formulas come from?
The formula with base and height was known in ancient Egypt and Babylon. Heron’s formula is named after Heron of Alexandria, who gave a proof in his work Metrica in the first century CE; some sources attribute the result to Archimedes, but this is uncertain. The formula with the sine of the included angle came with the development of trigonometry, and the coordinate formula with analytic geometry.
Where are they used today?
Land surveyors compute the area of a plot from side measurements with Heron’s formula or from coordinates with the shoelace formula, which works for any polygon. Computer graphics use the cross product to find the area of triangle meshes, and engineers compute the cross-section areas of beams.
Which mistakes are common?
Using a slant side as the height
The height is perpendicular to the base. In a right triangle, a leg can serve as the height.
Forgetting the factor 1/2
Base times height gives a parallelogram or rectangle. The triangle is half of it.
Using the wrong angle
In (1/2)ab sin C, the angle C is between the sides a and b, not any angle.
Mixing s and the perimeter
In Heron’s formula s is half the perimeter. Using the full perimeter gives a wrong result.