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Area of a triangle: ratios and the sine formula

Triangles with the same height have areas in the ratio of their bases; with two sides and the angle between them, Area = ½ · a · b · sin C.

High school Lesson 8 of 9 Interactive figure Practice

First, the rules

Two ways to think about area

Base times height
Example 2
Area = 12 · base · height
Any side can be the base; the height is the perpendicular distance from the opposite vertex to that side (or its extension).
★ Same height → ratio of the bases
Examples 1, 2
Triangles with the same apex and bases on one line have the same height, so their areas are in the ratio of their bases. A point D on BC splits triangle ABC into areas in the ratio |BD| : |DC|.
Parallel lines keep the area
Example 2
If a vertex slides along a line parallel to the base, the height does not change — neither does the area.
★ The sine formula
Example 3
Area = 12 · a · b · sin C
Two sides and the angle between them. For a right angle sin 90° = 1 and it becomes ½ · leg · leg.
Vertical angles
Example 3
Two lines crossing make equal opposite angles — and equal angles have equal sines.
Supplementary angles have the same sine
sin(180° − x) = sin x
So the sine formula gives the same area whether you use the angle or its supplement.

See it

Slide the point, turn the angle

Left: drag D along the base and watch the two areas keep the ratio |BD| : |DC|. Right: change the angle between two fixed sides and watch ½ · a · b · sin C — largest at 90°, and the same at C and 180° − C.

Step by step

Worked examples

1

Area of ABC from two pieces

area ratio = base ratio

D is on AB with |AD| = 3|DB|, E is on AC. A(ADE) = 12 and A(DEC) = 6. Find A(ABC).

ABCDE126

Ask yourself: ADE and DEC share the apex D and their bases lie on AC. What does that say about their heights?

Show the solution
  1. ADE and DEC: apex D, bases on AC, same height, so |AE| : |EC| = 12 : 6 = 2 : 1. Together A(ADC) = 18.
  2. ADC and DBC: apex C, bases on AB. |AD| = 3|DB|, so the areas are 3 : 1: A(DBC) = 18 / 3 = 6.
  3. A(ABC) = 18 + 6

Answer A(ABC) = 24

2

Area of a pool

parallel line · equal area

AB ⊥ BC, EF ∥ AC, |AE| = 12 m, |BF| = 9 m. E is on AB and F on BC; D is a point of AC. Find the area of triangle EFD.

BACEFD129

Ask yourself: Slide D along AC. Does the area of EFD change? (No — so move D to A.)

Show the solution
  1. EF ∥ AC, so every point of AC is the same distance from EF: A(EFD) = A(EFA) (common base EF, same height).
  2. In triangle AEF take base AE = 12. The height from F to line AB is BF = 9, because AB ⊥ BC.
  3. A(AEF) = 12 · 92 = 54

Answer 54 m²

3

Area with the sine formula

vertical angles · sine

Segments AE and BD cross at C, and DE ⊥ AE. |AC| = 8, |BC| = 26, |CE| = 5, |DE| = 12. Find A(ACB).

ACEDB826512

Ask yourself: How are angle ACB and angle DCE related? (Vertical angles — equal.)

Show the solution
  1. Right triangle CED has legs 5 and 12: a 5-12-13 triple, so |CD| = 13.
  2. sin(DCE) = oppositehypotenuse = 1213.
  3. ACB and DCE are vertical angles, so sin(ACB) = 1213.
  4. A(ACB) = 12 · 8 · 26 · 1213 = 4 · 2 · 12 (26 and 13 cancel).

Answer A(ACB) = 96

Practice: areas

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Split a triangle with a point on a side, or compute ½ · a · b · sin C with a special angle — answers are exact. Your best score is kept in this browser only.

About this lesson

Two ways to compare and compute triangle areas used constantly in exams: triangles that share a height have areas in the ratio of their bases (so a point on a side splits the area in the same ratio as the side), and the sine formula Area = ½ · a · b · sin C for two sides and the included angle. Move the point, move the angle, then practise.

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