Practice Maths

Area of a Triangle

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Key Terms

Area formula
A = ½ab sin C — works for ANY triangle; a and b are two sides, C is the included angle.
Included angle
The angle between the two sides a and b; the angle must be between the two sides used.
Right triangle special case
When C = 90°: sin 90° = 1, so A = ½ab, which is the familiar formula.
Heron’s formula
When all three sides are known: A = √[s(s−a)(s−b)(s−c)] where s = (a+b+c)/2.
When to use
Use A = ½ab sin C when two sides and their included angle are given (or findable).
Units
Area is always in square units (cm², m², etc.); never just cm or m.

Beyond ½bh

The standard formula Area = ½bh only works when the perpendicular height h is known. In real applications, we often know two sides and the included angle — not the height. Trigonometry lets us find the area directly.

A B C h a b C h = a sin C Area = ½bh = ½ab sin C
Trig Area Formula (QCAA Formula Sheet): \[\text{Area} = \frac{1}{2}ab\sin C\]

where a and b are two sides and C is the included angle between them.

When to Use

  • Given SAS (two sides + included angle): apply formula directly.
  • Given AAS/ASA: find all sides first using sine rule, then apply formula.
  • Given SSS: find an angle using cosine rule, then apply formula (or use Heron’s formula).

Heron’s Formula (SSS — Bonus)

When all three sides a, b, c are known but no angle:

\[s = \frac{a+b+c}{2} \quad \text{(semi-perimeter)}\] \[\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}\]

Worked Example 1

Problem: A triangle has sides a = 9, b = 13 and included angle C = 55°. Find the area.

\[\text{Area} = \frac{1}{2}ab\sin C = \frac{1}{2}(9)(13)\sin 55° = 58.5 \times 0.8192 \approx \mathbf{47.92 \text{ cm}^2}\]

Worked Example 2 — Using Cosine Rule First (SSS)

Problem: A triangle has sides 7 cm, 9 cm and 11 cm. Find the area.

Step 1: Find angle C (between sides a = 7 and b = 9, opposite c = 11):

\[\cos C = \frac{49 + 81 - 121}{2(7)(9)} = \frac{9}{126} = 0.07143 \implies C \approx 85.9°\]

Step 2: Apply area formula:

\[\text{Area} = \frac{1}{2}(7)(9)\sin 85.9° = 31.5 \times 0.9975 \approx \mathbf{31.42 \text{ cm}^2}\]

Verify with Heron’s: s = (7+9+11)/2 = 13.5

\[\text{Area} = \sqrt{13.5 \times 6.5 \times 4.5 \times 2.5} = \sqrt{988.59} \approx 31.44 \text{ cm}^2 \checkmark\]
Hot Tip: The formula Area = ½ab sin C gives the same result no matter which angle you choose, as long as a and b are the two sides adjacent to that angle. Choosing the wrong side (one that is not adjacent) will give the wrong answer — always check that C is the angle between sides a and b.

Deriving the Formula

In any triangle with base b and height h:

Area = ½bh

The perpendicular height from C to side b (the base AB) can be expressed using trigonometry. If we drop a perpendicular from C to AB and call its length h, then in the right triangle on the right:

sin C = h / a  ⇒  h = a sin C

Substituting into the base formula:

Area = ½ × b × a sin C = ½ab sin C

This works even when C is obtuse, because sin C = sin(180° − C) > 0 for all angles 0° < C < 180°.

Guide Example

Problem: A triangular sail has two sides of 4.5 m and 6.2 m with an included angle of 84°. Find the area of fabric needed.

\[\text{Area} = \frac{1}{2}(4.5)(6.2)\sin 84° = 13.95 \times 0.9945 \approx 13.88 \text{ m}^2\]
Tip: When the angle is exactly 90°, sin 90° = 1, and the formula gives Area = ½ab — which is just the standard right-triangle area formula. This confirms the trig formula is a generalisation of the familiar rule.

Finding a Missing Angle from Area

If you are given Area, a and b and need to find C:

\[\sin C = \frac{2 \times \text{Area}}{ab} \implies C = \sin^{-1}\!\left(\frac{2\,\text{Area}}{ab}\right)\]

Note: there may be two solutions (C and 180° − C) since sin is positive for both acute and obtuse angles. Consider the context to decide which is appropriate.

Mastery Practice

See Answers ➔
  1. Fluency Find the area of each triangle:
    1. a = 8, b = 12, C = 40°
    2. a = 15, b = 10, C = 75°
    3. a = 6, b = 6, C = 120°
    View Solution
  2. Fluency Two sides of a triangle are 14 cm and 20 cm, enclosing an angle of 62°. Find:
    1. The area of the triangle.
    2. The perimeter (find the third side using the cosine rule first).
    View Solution
  3. Fluency Two sides of a triangle are 9 m and 14 m. The area is 42 m². Find the included angle between these two sides. View Solution
  4. Understanding A triangular garden has sides 18 m, 22 m and 30 m. Find:
    1. The area using the cosine rule to find an angle, then the trig area formula.
    2. The area using Heron’s formula. Verify both methods give the same answer.
    View Solution
  5. Understanding A rhombus has all sides of length 12 cm. One interior angle is 65°. Find:
    1. The area of the rhombus using the trig area formula.
    2. The lengths of the two diagonals using the cosine rule.
    3. Verify your area result using Area = ½d&sub1;d&sub2;.
    View Solution
  6. Understanding A triangular plot of land has angle A = 70° at corner A, with sides AB = 42 m and AC = 58 m. Find:
    1. The area of the plot.
    2. The length of side BC.
    3. A fence is to be built around all three sides. If fencing costs $85 per metre, find the total cost.
    View Solution
  7. Understanding A parallelogram has sides 15 cm and 22 cm with an angle of 48° between them. Find:
    1. The area of the parallelogram.
    2. The area of each triangle formed by one diagonal.
    3. The lengths of the two diagonals.
    View Solution
  8. Understanding A hiking track forms a triangular route. From base camp A, a hiker goes 4.5 km to checkpoint B on a bearing of 050°, then 6 km to checkpoint C on a bearing of 140°. Find:
    1. Angle ABC (the angle at B in the triangle).
    2. The distance AC.
    3. The area of triangle ABC.
    View Solution
  9. Problem Solving A quadrilateral land parcel ABCD is divided by diagonal BD. Triangle ABD has AB = 85 m, AD = 110 m, angle A = 68°. Triangle BCD has BC = 95 m, CD = 75 m, angle C = 54°. Find:
    1. The area of triangle ABD.
    2. The area of triangle BCD.
    3. The total area of the parcel ABCD.
    4. The length of diagonal BD using the cosine rule on triangle ABD.
    5. Verify BD using the cosine rule on triangle BCD.
    View Solution
  10. Problem Solving Three points A, B, C form a triangle. A is at the top of a hill 120 m above sea level. From A, point B is at a bearing of 080° with an angle of depression of 12°. Point C is at a bearing of 160° from A with an angle of depression of 8°. Find:
    1. The horizontal distance from A to B.
    2. The horizontal distance from A to C.
    3. The angle BAC (at A in the base triangle, using bearings 080° and 160°).
    4. The area of the base triangle (the horizontal projection of triangle ABC at sea level).
    View Solution