Area of Any Triangle [H]

Area of Any Triangle

The standard area formula \( \frac{1}{2}bh \) requires a perpendicular height. When this is not directly available but two sides and the included angle are known, the following formula gives the area directly.

The Formula

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

where \( a \) and \( b \) are two sides and \( C \) is the angle between them (the included angle). The formula works for any triangle — acute, obtuse or right-angled.

Why It Works

The perpendicular height from \( B \) to \( AC \) has length \( h=a\sin C \). Substituting into \( \frac{1}{2}bh \):

\[ \text{Area} = \frac{1}{2}\times b \times a\sin C = \frac{1}{2}ab\sin C \]

Applications

  • Finding a side or angle: rearrange \( \text{Area}=\frac{1}{2}ab\sin C \) for any unknown.
  • Segment area: sector area \( - \frac{1}{2}r^2\sin\theta \) uses this formula with \( a=b=r \).
  • Polygon area: divide into triangles, apply the formula to each.

Worked Examples

Triangle with \( a=8 \) cm, \( b=11 \) cm, included angle \( C=65° \). Find the area.
\[ \text{Area} = \frac{1}{2}\times8\times11\times\sin65° \approx \frac{1}{2}\times88\times0.906 \approx 39.9 \text{ cm}^2 \]
Find the exact area of an equilateral triangle with side 6 cm.
\[ \text{Area} = \frac{1}{2}\times6\times6\times\sin60° = 18\times\frac{\sqrt{3}}{2} = 9\sqrt{3} \text{ cm}^2 \]
The area of triangle \( ABC \) is 40 cm². \( AB=10 \) cm, \( BC=12 \) cm. Find \( \angle ABC \).
\[ 40 = \frac{1}{2}\times10\times12\times\sin B \implies \sin B = \frac{80}{120} = \frac{2}{3} \implies B = \sin^{-1}\!\left(\tfrac{2}{3}\right) \approx 41.8° \]

(or \( 138.2° \) — check context).

Find the area of the minor segment of a circle with radius 9 cm and central angle 80°.
\[ \text{Sector area} = \frac{80}{360}\pi\times81 = 18\pi \] \[ \text{Triangle area} = \frac{1}{2}\times9\times9\times\sin80° = \frac{81\sin80°}{2} \approx 39.83 \text{ cm}^2 \] \[ \text{Segment} = 18\pi - 39.83 \approx 56.55 - 39.83 = 16.72 \text{ cm}^2 \]

 Key Takeaways

  • \( \text{Area}=\frac{1}{2}ab\sin C \). The angle \( C \) must be between sides \( a \) and \( b \).
  • Works for any triangle — no perpendicular height needed.
  • To find an angle: \( \sin C = \frac{2\times\text{Area}}{ab} \). To find a side: rearrange.
  • Equilateral triangle with side \( s \): area \( =\frac{s^2\sqrt{3}}{4}=\frac{1}{2}s^2\sin60° \).
  • Segment area: \( \frac{\theta}{360}\pi r^2 - \frac{1}{2}r^2\sin\theta \).