Triangle Area Formulas: ½ab sin C and Heron's Formula

You know from geometry that the area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$. But that formula requires a perpendicular height, which you rarely know for oblique triangles. These two trigonometric formulas solve that problem — one for when you know two sides and the included angle (SAS), and one for when you know all three sides (SSS).

Formula 1: Area = ½ab sin C (SAS)

When you know two sides and the angle between them, you can compute the area directly.

The derivation: Drop a perpendicular from vertex A to side $a$. That perpendicular height is $h = b \sin C$ (from basic right-triangle trig). Substituting into the standard area formula:

$\text{Area} = \frac{1}{2} \times a \times h = \frac{1}{2} \times a \times b \sin C$

$\boxed{\text{Area} = \frac{1}{2}ab\sin C}$

The angle $C$ must be the included angle — the angle between sides $a$ and $b$. Using a non-included angle gives a wrong answer.

Example 1: SAS Area

Problem: Find the area of a triangle with sides 8 and 11 and an included angle of 40 degrees.

$\text{Area} = \frac{1}{2}(8)(11)\sin(40°) = \frac{1}{2}(88)(0.6428) = 28.3 \text{ square units}$

Answer: The area is approximately 28.3 square units.

Example 2: Finding an Angle from the Area

Problem: A triangle has sides of 10 and 14 with an area of 50 square units. Find the acute included angle.

$50 = \frac{1}{2}(10)(14)\sin C$

$50 = 70\sin C$

$\sin C = \frac{50}{70} = 0.7143$

$C = \sin^{-1}(0.7143) = 45.6°$

Answer: The included angle is approximately 45.6 degrees.

Note: Since $\sin(180° - \theta) = \sin(\theta)$, the supplementary angle $134.4°$ also has a sine of approximately $0.7143$. So there is also an obtuse solution of 134.4 degrees. The problem specifies the acute angle.

Formula 2: Heron’s Formula (SSS)

When you know all three sides but no angles, Heron’s formula gives the area directly.

Step 1: Compute the semi-perimeter:

$s = \frac{a + b + c}{2}$

Step 2: Plug into Heron’s formula:

$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$

Example 3: Heron’s Formula

Problem: Find the area of a triangle with sides 7, 9, and 12.

Step 1 — Semi-perimeter:

$s = \frac{7 + 9 + 12}{2} = \frac{28}{2} = 14$

Step 2 — Compute each factor:

$s - a = 14 - 7 = 7$ $s - b = 14 - 9 = 5$ $s - c = 14 - 12 = 2$

Step 3 — Apply the formula:

$\text{Area} = \sqrt{14 \times 7 \times 5 \times 2} = \sqrt{980} = 31.3 \text{ square units}$

Answer: The area is approximately 31.3 square units.

When to Use Which Formula

If you know SAS, you could use the Law of Cosines to find the third side, then apply Heron’s formula — but $\frac{1}{2}ab\sin C$ is faster and involves less arithmetic.

Real-World Application: Surveying a Triangular Lot

A surveyor measures a triangular plot of land with sides of 120 ft, 95 ft, and 140 ft. What is the area in square feet and acres?

$s = \frac{120 + 95 + 140}{2} = \frac{355}{2} = 177.5$

$s - a = 177.5 - 120 = 57.5$ $s - b = 177.5 - 95 = 82.5$ $s - c = 177.5 - 140 = 37.5$

$\text{Area} = \sqrt{177.5 \times 57.5 \times 82.5 \times 37.5} = \sqrt{31{,}575{,}585.9} = 5619.2 \text{ sq ft}$

Convert to acres: $5619.2 \div 43{,}560 = 0.129$ acres

Answer: The lot is approximately 5,620 square feet or about 0.13 acres.

Common Mistakes

1. Using a non-included angle with ½ab sin C. The angle must be between the two given sides. Using any other angle produces a wrong answer.
2. Forgetting to compute the semi-perimeter. Heron’s formula uses $s = (a+b+c)/2$, not $a+b+c$ directly. Skipping this step is the most common arithmetic error.
3. Forgetting the final square root. After multiplying $s(s-a)(s-b)(s-c)$, you still need to take the square root. Students who forget end up with an area that is way too large.
4. Rounding intermediate values. In Heron’s formula, keep full precision through the multiplication and only round the final answer. Rounding $s - a$, $s - b$, and $s - c$ individually can compound into significant error.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• Use $\frac{1}{2}ab\sin C$ when you know two sides and the included angle (SAS) — it is faster than any alternative
• Use Heron’s formula when you know all three sides (SSS) and no angles
• The angle in $\frac{1}{2}ab\sin C$ must be the included angle between the two known sides
• Both formulas work for any triangle — right, acute, or obtuse
• For surveying and construction, Heron’s formula is especially practical because field measurements often give three side lengths

Return to Trigonometry for more topics in this section.