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Pre-Calculus Triangle trigonometry

Area formula A = ½ ab sin C

20 practice questions 0 video lessons Theory + worked examples

Area of a Triangle

Texas Precalculus (TEKS) • Standard P.4(E) • Triangle Trigonometry

Area of a Triangle is a topic in Triangle Trigonometry in the Texas Essential Knowledge and Skills (Precalculus, §111.42). It is aligned to Standard P.4(E), which requires students to determine the area of a triangle using trigonometric ratios.

The area of a triangle is \(\dfrac12 ab\sin C\) from two sides and the included angle, or Heron's formula from all three sides.

Texas Precalculus (TEKS) › Triangle Trigonometry › Area of a Triangle  —  Standard P.4(E)

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Theory

For a right triangle, area is \(\dfrac12(\text{base})(\text{height})\), but for a general triangle the height may be unknown. Two trig-based formulas fix that:

  • SAS area: \(\text{Area}=\dfrac12 ab\sin C\), from two sides and the included angle.
  • Heron's formula: from all three sides, using the semi-perimeter \(s=\dfrac{a+b+c}{2}\).
The SAS formula is just base times height in disguise: \(b\sin C\) is the height dropped onto side \(a\).
Area from two sides and the included angle The area of a triangle equals one half a b sine C, using two sides and the angle between them. a b C
\(\text{Area}=\dfrac12 ab\sin C\) uses two sides and the angle between them.
Heron's formula for three sides Heron's formula gives the area from the three side lengths using the semi-perimeter s. s = (a+b+c)/2Area = √( s(s−a)(s−b)(s−c) )
Heron's formula gives the area from the three side lengths.

The two formulas:

\[\text{Area}=\dfrac12 ab\sin C,\qquad \text{Area}=\sqrt{s(s-a)(s-b)(s-c)},\ \ s=\dfrac{a+b+c}{2}\]
area equals one half a b sine C; Heron's area is the square root of s times s minus a times s minus b times s minus c
Choose by what you're given: SAS \(\to\) \(\dfrac12 ab\sin C\); SSS \(\to\) Heron.

How to find a triangle's area

  1. Two sides + included angle: apply \(\dfrac12 ab\sin C\).
  2. Three sides: compute \(s\), then Heron's formula.
  3. Solving for a side: substitute the known area and solve.
  4. Include units squared in the answer.
Example 1 — Two sides and the included angle
Find the area of a triangle with \(a=8\), \(b=5\), and included angle \(C=30^\circ\).
Solution

Use \(\text{Area}=\dfrac12 ab\sin C\).

\(\text{Area}\)\(=\)\(\dfrac12(8)(5)\sin 30^\circ\)
\(=\)\(20\cdot\dfrac12=10\)
area is 10 square units
Example 2 — Heron's formula
Find the area of a triangle with sides \(a=6\), \(b=8\), \(c=10\).
Solution

First the semi-perimeter, then Heron's formula.

\(s\)\(=\)\(\dfrac{6+8+10}{2}=12\)
\(\text{Area}\)\(=\)\(\sqrt{12(6)(4)(2)}\)
\(=\)\(\sqrt{576}=24\)
area is 24 square units
Example 3 — Solve for a missing side
A triangle with area \(30\) has \(b=12\) and included angle \(C=90^\circ\). Find \(a\).
Solution

Set up \(\dfrac12 ab\sin C=30\) with \(\sin 90^\circ=1\).

\(\dfrac12\cdot a\cdot 12\cdot 1\)\(=\)\(30\)
\(6a\)\(=\)\(30\)
\(a\)\(=\)\(5\)
side a is 5
Example 4 — A real-world area
A triangular plot has two sides \(40\ \text{ft}\) and \(55\ \text{ft}\) meeting at \(105^\circ\). Find its area.
Solution

Two sides and the included angle \(\Rightarrow\) \(\dfrac12 ab\sin C\).

\(\text{Area}\)\(=\)\(\dfrac12(40)(55)\sin 105^\circ\)
\(\approx\)\(1062\ \text{ft}^2\)
area is about 1062 square feet

Common pitfalls

The angle must be included. \(\dfrac12 ab\sin C\) needs \(C\) between \(a\) and \(b\).
Use the semi-perimeter in Heron's. \(s\) is half the perimeter, not the perimeter.
Area is in square units. Don't drop the \(^2\).

Frequently asked questions

How do you find the area with two sides and an angle?

Use \(\text{Area}=\dfrac12 ab\sin C\), where \(C\) is the angle included between sides \(a\) and \(b\).

What is Heron's formula?

\(\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}\), with \(s=\dfrac{a+b+c}{2}\). It gives the area from the three side lengths.

When do you use each area formula?

Use \(\dfrac12 ab\sin C\) for two sides and the included angle (SAS); use Heron's formula when you know all three sides (SSS).

Why does 1/2 a b sin C work?

Because \(b\sin C\) is the triangle's height on base \(a\), so \(\dfrac12 ab\sin C\) is just \(\dfrac12\,\text{base}\times\text{height}\).