Areas of plane figures

Area measures how much of the plane a figure occupies. It is needed all the time in practical work: a plot of land, a floor, a wall to paint, the amount of material. In area problems knowing the formulas is not enough — you often have to cut a figure into pieces or subtract "unwanted" parts. Two ideas are decisive here: triangles with the same height have areas in the ratio of their bases, and similar figures have areas in the ratio k². This lesson covers the properties of area, the formulas for quadrilaterals, triangles and circles, ratios of areas and composite figures.

§1. Area and its properties

Area is measured in square units: a square of side 1 cm has area 1 cm². When changing units, the length factor is squared: 1 m = 100 cm, so 1 m² = 10 000 cm²; 1 ha = 10 000 m², 1 km² = 100 ha.
The basic properties of area:

• Congruent figures have equal areas.
• Additivity: if a figure is split into parts, its area is the sum of the areas of the parts.
• For similar figures the ratio of areas is the square of the scale factor.

Every formula comes from the rectangle S = ab: a parallelogram can be cut and rearranged into a rectangle, a triangle is half a parallelogram, a trapezium is two triangles.

§2. Areas of quadrilaterals

The parallelogram formula: cut off the triangle at one vertex and move it to the opposite side — you get a rectangle with base a and height h. Since h = b sin α, S = ab sin α. In a rhombus the diagonals are perpendicular, so φ = 90° and S = d₁d₂/2.

Worked example

A parallelogram has sides 8 and 5 and an angle of 30°. Find its area.

1. S = ab sin α.
2. S = 8 · 5 · sin 30° = 40 · ½.
3. S = 20.

Worked example

A rhombus has side 10 and one diagonal 12. Find its area.

1. The diagonals are perpendicular and bisect each other: half-diagonal 6, the other half-diagonal √(100 − 36) = 8.
2. The second diagonal is d₂ = 16.
3. S = 12 · 16/2 = 96.

If only the side and an angle are known, S = a² sin α is handier: a rhombus with side 10 and angle 30° has area 100 · ½ = 50.

§3. Area of a triangle

For a right-angled triangle S = ½ ab (the legs); for an equilateral triangle S = a²√3/4.
For a triangle given by the coordinates of its vertices:

Choose the formula to fit the data: base and height — ½ah; two sides and the included angle — the sine formula; three sides — Heron; coordinates — the last formula.

Worked example

Find the area of the triangle A(1; 1), B(5; 2), C(2; 6).

1. x₁(y₂ − y₃) = 1 · (2 − 6) = −4; x₂(y₃ − y₁) = 5 · (6 − 1) = 25; x₃(y₁ − y₂) = 2 · (1 − 2) = −2.
2. Sum: −4 + 25 − 2 = 19.
3. S = 19/2 = 9.5.

§4. Ratios of areas

• Triangles with a common height have areas in the ratio of their bases (S = ½ah with the same h).
• Triangles sharing an angle: S₁/S₂ = (b₁c₁)/(b₂c₂) (the ratio of the products of the sides forming that angle).
• A median divides a triangle into two parts of equal area.
• Similar figures: S₁/S₂ = k².

These rules often give an area without knowing any lengths. For example, if triangle ABC has area 30, M lies on AB with AM : MB = 1 : 2 and N on AC with AN : NC = 1 : 1, then triangle AMN has area 30 · ⅓ · ½ = 5.

Worked example

Parallelogram ABCD has area 64. F is on AD with AF/FD = 3; K is on AB with AK/KB = 1. Find the area of FKCD.

1. K is the midpoint of AB, S(ABC) = 32 → S(KBC) = 16.
2. S(AKF) = ½ · AK · AF · sin A = ½ · ½AB · ¾AD · sin A = (3/16) · 64 = 12.
3. S(FKCD) = 64 − 16 − 12 = 36.

§5. The circle and its parts

A sector is the fraction α/360° of the disc: the smaller the angle, the proportionally smaller the area. In radians: S = R²α/2. The circumference C = 2πR must not be confused with the area.

Worked example

In a circle of radius 6, find the areas of a 60° sector and the corresponding segment.

1. Sector: 36π · 60°/360° = 36π/6 = 6π.
2. The triangle with central angle 60° is equilateral with side 6: S = 36√3/4 = 9√3.
3. Segment: 6π − 9√3.

§6. Composite figures

Strategy:

• cut the figure into known pieces and add their areas;
• subtract the unwanted part from a larger figure;
• use ratios (common height, similarity).

The diagonals divide a trapezium into 4 triangles: the two at the bases are similar (k = a/b), while the two at the legs have equal areas, each the geometric mean of the areas of the base triangles.

Worked example

A trapezium has bases 6 and 10 and height 5. Find its area and the areas of the triangles formed by the diagonals.

1. S = (6 + 10)/2 · 5 = 40.
2. The base triangles are in the ratio k² = (6/10)² = 9 : 25; each leg triangle is √(9 · 25) = 15 "shares". In total 9 + 25 + 15 + 15 = 64 shares.
3. One share is 40/64 = 0.625: the areas are 5.625, 15.625, 9.375, 9.375.

Check yourself

1. The diagonals of a rhombus are 10 and 12. Find its area.

2. A trapezium has midline 7 and height 4. Find its area.

3. A disc has area 49π. Find the circumference.

4. Triangles with a common height have bases 3 and 5; the first has area 12. Find the area of the second.

5. A square has diagonal 6. Find its area.

6. Find the area of the ring with R = 5, r = 3.

7. Find the area of the triangle with sides 13, 14, 15.

8. How many cm² are there in 2.5 m²?

Common mistakes

• Taking the area of a parallelogram as the product of its sides (you need sin α or the height).
• Forgetting to halve the product of the diagonals of a rhombus.
• Dropping the ½ in the area of a triangle.
• Equating the ratio of areas with the scale factor (it must be k²).
• Converting units as 1 m² = 100 cm² (it is 10 000 cm²).
• Confusing the area of a disc πR² with the circumference 2πR.