The Three Main Methods for Triangle Area

Method 1: Base × Height ÷ 2
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Formula:

\[ \text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} \]

Say you have a triangle with base 10 and height 5:

\[ \text{Area} = \frac{1}{2} \cdot 10 \cdot 5 = 25 \]

The area of a triangle is 150. The height is 5 units less than the base. Find base and height.

Let base = \( b \), then height = \( b - 5 \). Using:

\[ \frac{1}{2} \cdot b \cdot (b - 5) = 150 \]\[ b(b - 5) = 300 \quad \Rightarrow \quad b^2 - 5b - 300 = 0 \]

Solving:

\[ b = 20 \Rightarrow \text{height} = 15 \]

Base = 20, Height = 15

Formula:

\[ \text{Area} = \frac{1}{2} ab \sin(C) \]

Where \( a \) and \( b \) are two sides, and \( C \) is the included angle.

Find the area of a triangle with sides 3 and 4 and an included angle of \( 30^\circ \).

\[ \text{Area} = \frac{1}{2} \cdot 3 \cdot 4 \cdot \sin(30^\circ) = \frac{1}{2} \cdot 12 \cdot \frac{1}{2} = 3 \]

Area = 3

Use when all three sides are known.

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

Triangle with sides 4, 7, and 5

\[ \text{Area} = \sqrt{ 8(8 - 4)(8 - 7)(8 - 5) } = \sqrt{8 \cdot 4 \cdot 1 \cdot 3} \]

\[ \text{Area} = \sqrt{96} = \sqrt{16 \cdot 6} = 4\sqrt{6} \]

Area = \( 4\sqrt{6} \)

All explanations and problems written by Adrian Hernandez Vega, unless otherwise noted.

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