Precalculus / Unit 6: Applications of Trigonometry
6.3 Area of a Triangle
Notes
- With two sides and the included angle: $\displaystyle \text{Area} = \frac{1}{2}ab\sin C$.
- Heron's formula: with semi-perimeter $\displaystyle s = \frac{a + b + c}{2}$, $\displaystyle \text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$.
- Example: $a = 6$, $b = 8$, $C = 30^\circ$ gives $\displaystyle \frac{1}{2}(6)(8)\cdot\frac{1}{2} = 12$.
Practice
A triangle has sides 6 and 8 with an included angle of $30^\circ$. What is its area?
$\displaystyle \frac{1}{2}(6)(8)\sin 30^\circ = 24 \cdot \frac{1}{2} = 12$.
A triangle has two sides of length 10 with an included angle of $60^\circ$. What is its area?
$\displaystyle \frac{1}{2}(10)(10)\cdot\frac{\sqrt{3}}{2} = 25\sqrt{3}$.
Use Heron's formula to find the area of a triangle with sides 5, 6 and 7.
$s = 9$, so area $\displaystyle = \sqrt{9 \cdot 4 \cdot 3 \cdot 2} = \sqrt{216} = 6\sqrt{6}$.
Ready for more? Take the Unit 6 test.