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Lesson 3: Area of Triangles and Heron's Formula

Estimated time: 25-30 minutes

Learning Objectives

SAS Area Formula

Area = (1/2) a b sin C

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

Example 1

a=8, b=12, C=30°. Find the area.

Area = (1/2)(8)(12)sin30 = (1/2)(96)(1/2) = 24 square units

Example 2

b=15, c=20, A=110°. Find the area.

Area = (1/2)(15)(20)sin110 = 150 sin110 ≈ 150(0.9397) ≈ 140.95 sq units

Heron's Formula

Heron's Formula: Given sides a, b, c, let s = (a+b+c)/2 (semi-perimeter). Then:

Area = √[s(s−a)(s−b)(s−c)]

Example 3

a=7, b=8, c=9.

s = (7+8+9)/2 = 12

Area = √[12(12−7)(12−8)(12−9)] = √[12·5·4·3] = √720 = 12√5 ≈ 26.83

Example 4

a=3, b=4, c=5 (right triangle).

s=6. Area = √[6·3·2·1] = √36 = 6 (matches (1/2)(3)(4)=6).

Choosing the Right Formula

Check Your Understanding

1. a=10, b=14, C=65°. Find the area.

A=(1/2)(10)(14)sin65 = 70sin65 ≈ 63.44.

2. Sides 5, 12, 13. Find area using Heron's.

s=15. √[15·10·3·2]=√900=30.

3. When is (1/2)ab sinC = 0?

When C = 0° or 180° (degenerate triangle — the vertices are collinear).

Key Takeaways

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