Area of a Triangle: ½ab sin C, Heron's Formula and Coordinates

More Than Half Base Times Height

Everyone learns that the area of a triangle is ½ × base × height. But in many problems you are not given the height: you know two sides and the angle between them, or all three sides, or the coordinates of the corners. This tutorial covers every common way to find the area of a triangle: the basic formula, the trigonometric formula ½ab sin C, Heron's formula, and the coordinate (shoelace) formula, with worked examples in the style of WAEC, NECO and JAMB questions and practical uses in land surveying.

Method 1: Half Base Times Height

Area = ½ × base × perpendicular height. The height must be perpendicular to the base you choose. Any side can be the base, as long as you use the matching height.

Example 1. A triangle has base 12 cm and perpendicular height 7 cm. Its area is ½ × 12 × 7 = 42 cm².

For a right-angled triangle, the two shorter sides are perpendicular, so the area is half their product. A triangle with sides 6, 8 and 10 cm has area ½ × 6 × 8 = 24 cm².

Method 2: Two Sides and the Included Angle (½ab sin C)

Label the triangle in the usual way: angles A, B, C at the corners, and sides a, b, c opposite them. If you know two sides and the angle between them (the included angle), then

Area = ½ab sin C (and equally ½bc sin A or ½ac sin B).

Why it works. Take side b as the base. Drop a perpendicular from the opposite corner; in the right-angled triangle formed, the height is h = a sin C. So the area is ½ × b × a sin C. The formula still works when C is obtuse, because sin C = sin(180° − C).

Example 2. Find the area of a triangle with sides 8 cm and 11 cm and an included angle of 40°.

Area = ½ × 8 × 11 × sin 40° = 44 × 0.6428 = 28.3 cm² (to 3 significant figures).

Example 3. In triangle PQR, PQ = 10 cm, QR = 6 cm and angle PQR = 120°. Find its area.

The angle at Q is between PQ and QR, so area = ½ × 10 × 6 × sin 120° = 30 × (√3/2) = 15√3 ≈ 26.0 cm².

Example 4 (working backwards). A triangle has area 30 cm², and two sides of 10 cm and 12 cm. Find the angle between them.

30 = ½ × 10 × 12 × sin C, so sin C = 30/60 = 0.5. So C = 30° or C = 150°, because sin 150° is also 0.5. Both triangles exist and have the same area. Exam questions usually say whether the angle is acute or obtuse.

Method 3: Heron's Formula (Three Sides)

When you know all three sides a, b and c, use Heron's formula. First find the semi-perimeter, s = (a + b + c)/2. Then

Area = √[s(s − a)(s − b)(s − c)].

Example 5. Find the area of a triangle with sides 13 cm, 14 cm and 15 cm.

s = (13 + 14 + 15)/2 = 21.
s − a = 8, s − b = 7, s − c = 6.
Area = √(21 × 8 × 7 × 6) = √7056 = 84 cm².

Example 6. Find the area of a triangle with sides 5 m, 6 m and 9 m.

s = (5 + 6 + 9)/2 = 10.
Area = √(10 × 5 × 4 × 1) = √200 = 10√2 ≈ 14.1 m².

Check for a valid triangle. Each side must be less than the sum of the other two (the triangle inequality). Equivalently, s − a, s − b and s − c must all be positive. Sides 3, 4 and 8 cannot form a triangle: s = 7.5 and s − c = −0.5, so Heron's formula would need the square root of a negative number.

Where Heron's formula comes from. Combine area = ½ab sin C with the cosine rule, cos C = (a² + b² − c²)/(2ab), and the identity sin² C = 1 − cos² C. After factorising the difference of two squares twice, the expression becomes s(s − a)(s − b)(s − c). This is a good algebra exercise for Further Mathematics students.

Method 4: Coordinates (the Shoelace Formula)

If the corners are at (x₁, y₁), (x₂, y₂) and (x₃, y₃), then

Area = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|.

The modulus signs make the answer positive whichever order you list the points in.

Example 7. Find the area of the triangle with vertices A(1, 2), B(5, 3) and C(2, 7).

Area = ½ |1(3 − 7) + 5(7 − 2) + 2(2 − 3)| = ½ |−4 + 25 − 2| = ½ × 19 = 9.5 square units.

Collinear points. If the formula gives 0, the three points lie on a straight line. For example (1, 1), (2, 3) and (4, 7): ½ |1(3 − 7) + 2(7 − 1) + 4(1 − 3)| = ½ |−4 + 12 − 8| = 0, so they are collinear.

Area of a Triangle: Choosing the Right Formula

Base and perpendicular height known: ½ × base × height.
Two sides and the angle between them: ½ab sin C.
All three sides: Heron's formula.
Coordinates of the vertices: the shoelace formula.
Two angles and one side: use the sine rule to find a second side, then ½ab sin C.

Example 8 (two angles and a side). In triangle ABC, A = 50°, B = 60° and c = 10 cm. Find the area.

C = 180° − 50° − 60° = 70°. By the sine rule, a = c sin A / sin C = 10 × 0.7660 / 0.9397 = 8.152 cm. Then area = ½ × a × c × sin B = ½ × 8.152 × 10 × 0.8660 = 35.3 cm².

Applications: Land Surveying

Surveyors divide an irregular plot of land into triangles, measure the sides with a tape or the angles with a theodolite, and add up the triangle areas. Suppose a four-sided plot ABCD has AB = 40 m, BC = 30 m, CD = 35 m, DA = 25 m, and the diagonal AC = 50 m.

Triangle ABC has sides 40, 30 and 50. Since 30² + 40² = 50², it is right-angled at B, with area ½ × 40 × 30 = 600 m².
Triangle ACD has sides 50, 35 and 25: s = 55, area = √(55 × 5 × 20 × 30) = √165000 ≈ 406.2 m².
Total area ≈ 1006 m², a little over a tenth of a hectare.

Practice Questions

Find the area of a triangle with sides 7 cm and 9 cm enclosing an angle of 30°. Answer: ½ × 7 × 9 × 0.5 = 15.75 cm².
Find the area of an equilateral triangle of side 6 cm. Answer: ½ × 6 × 6 × sin 60° = 9√3 ≈ 15.6 cm².
Use Heron's formula for sides 9, 10 and 17. Answer: s = 18, area = √(18 × 9 × 8 × 1) = √1296 = 36.
Find the area of the triangle with vertices (0, 0), (4, 0) and (1, 5). Answer: ½ |0 + 4(5 − 0) + 1(0 − 0)| = 10.

Summary

Area = ½ × base × perpendicular height is the starting point.
Area = ½ab sin C uses two sides and the included angle; an area can come from two different angles, C and 180° − C.
Heron's formula, √[s(s − a)(s − b)(s − c)], uses three sides and the semi-perimeter s.
The shoelace formula gives the area from coordinates; an area of 0 means the points are collinear.