DWG NO. 6.3 — Lesson 3 of 6

Area of a Triangle

Unit 6: Additional Topics in Trigonometry · ~20 min

The familiar formula Area = ½·base·height runs into trouble when you don't have the height handed to you — and for an oblique triangle, you usually don't. Trigonometry supplies the height instead, and if you know only the three side lengths, a second formula skips height entirely.

The trig area formula

Take two sides, a and b, and the angle C between them. Dropping an altitude from the vertex between sides a and c down to side b gives a height of h = a·sinC. Substituting into Area = ½·base·height with base b:

• Area = ½ab·sinC   (equivalently ½bc·sinA or ½ac·sinB)

This needs exactly the SAS information — two sides and their included angle — and never requires solving the rest of the triangle first.

Heron's Formula

When all three sides are known but no angle is, first find the semi-perimeter s, then:

• s = \(\frac{a + b + c}{2}\)
• Area = \(\sqrt{}\)[s(s−a)(s−b)(s−c)]

Guided practice

• 1.Find the area with a = 8, b = 12, C = 60°.
  Show answer

  Area = ½(8)(12)sin60° = 48(0.8660) ≈ 41.6 square units.
• 2.Use Heron's Formula with a = 9, b = 10, c = 11.
  Show answer

  s = 15. Area = \(\sqrt{}\)[15·6·5·4] = \(\sqrt{1800}\) ≈ 42.4 square units.
• 3.A triangle with sides a = 6, b = 10 has area 24. Find the included angle C (acute solution).
  Show answer

  24 = ½(6)(10)sinC = 30sinC, so sinC = 0.8 and C ≈ 53.1° (126.9° is also geometrically possible from the same sine value).
• 4.Use Heron's Formula to find the area of an equilateral triangle with side length 6.
  Show answer

  s = 9. Area = \(\sqrt{}\)[9·3·3·3] = \(\sqrt{243}\) = 9\(\sqrt{3}\) ≈ 15.6 square units.
• 5.Explain why the trig area formula doesn't require ever finding the triangle's height explicitly.
  Show answer

  The height is already built into the formula — a·sinC is the height relative to base b, so multiplying by sinC folds the height calculation directly into the area formula instead of requiring it as a separate step.