Area · Lesson 6 of 12
Area of Any Polygon
“Find areas of quadrilaterals, pentagons, hexagons, and shaded regions by triangulating, adding, and subtracting simpler areas.”
• Explain why polygons can be decomposed into triangles. • Choose useful diagonals for calculating the area of a quadrilateral or larger polygon. • Find polygon area by adding triangle areas. • Find shaded or cut-out areas by subtraction. • Reason about regular hexagons and other compound figures without depending on a single memorised formula.
Once triangle area is understood, the area of a much larger family of shapes becomes accessible. A polygon can be split into triangles by drawing suitable diagonals. Instead of inventing a new formula for every possible quadrilateral, pentagon, or irregular outline, we reduce the unfamiliar figure to familiar pieces.
The art lies in choosing a useful decomposition. A good diagonal creates triangles whose bases and perpendicular heights are known or easy to find. In shaded-region problems, subtraction may be even shorter than adding many pieces.
Triangulating a Polygon
Dividing a polygon into non-overlapping triangles whose union is the entire polygon.
A quadrilateral can be divided into two triangles by drawing one diagonal. A pentagon can be divided into three triangles from a suitable vertex. More generally, a simple n-sided polygon can be triangulated into n − 2 triangles, though the exact triangle shapes depend on which diagonals are chosen.
Quadrilateral Area from a Diagonal
Suppose diagonal AC of quadrilateral ABCD is 22 cm. If the perpendicular distances from B and D to AC are both 3 cm, the quadrilateral is the sum of triangles ABC and ADC. Both use the same 22 cm diagonal as base.
Problem
A quadrilateral has diagonal AC = 22 cm. The perpendicular distances from the other two vertices to AC are 3 cm and 5 cm. Find its area.
- 1.Triangle ABC area = ½ × 22 × 3 = 33 cm².
- 2.Triangle ADC area = ½ × 22 × 5 = 55 cm².
- 3.Quadrilateral area = 33 + 55 = 88 cm².
Pentagons and Irregular Polygons
For a pentagon, one strategy is to pick a vertex and draw diagonals to the two non-adjacent vertices. This creates three triangles. Another strategy may split the pentagon into a rectangle and triangles. Both methods are valid if the pieces exactly cover the polygon.
Problem
A pentagon is divided into three non-overlapping triangles with areas 18 cm², 25 cm², and 31 cm². Find the pentagon's area.
- 1.Because the three triangles fill the pentagon without overlap, add their areas.
- 2.18 + 25 + 31 = 74.
- 3.The pentagon area is 74 cm².
Different decompositions should produce the same total area. If two correct methods disagree, at least one calculation, measurement, or decomposition contains an error. This gives a useful self-check.
Shaded Regions by Subtraction
Sometimes the simplest route is to find the area of an enclosing rectangle and subtract unshaded triangles or smaller rectangles. This is especially effective when the outer dimensions are simple but the shaded boundary is irregular.
Problem
A rectangle is 18 cm by 10 cm. Two non-overlapping unshaded right triangles inside it have areas 24 cm² and 30 cm². Find the shaded area.
- 1.Rectangle area = 18 × 10 = 180 cm².
- 2.Total unshaded area = 24 + 30 = 54 cm².
- 3.Shaded area = 180 − 54 = 126 cm².
Regular Hexagons
A regular hexagon can be divided into six congruent equilateral triangles by joining its centre to all vertices. If the side length and the perpendicular height of one of those equilateral triangles are known, the total area is six times one triangle's area.
Another problem may divide a regular hexagon into a trapezium, an equilateral triangle, and a rhombus. In that case it is often better to compare the smaller regions by decomposition rather than to calculate the whole hexagon first.
Problem
A regular hexagon is divided from its centre into six congruent triangles. Each triangle has base 8 cm and height 4√3 cm. Find the hexagon's area.
- 1.One triangle area = ½ × 8 × 4√3 = 16√3 cm².
- 2.There are six congruent triangles.
- 3.Hexagon area = 6 × 16√3 = 96√3 cm².
Creating Half-Area Regions
A diagonal does not always divide a quadrilateral into equal areas. To construct a quadrilateral with half the area of a given quadrilateral, one possible route is to triangulate the original, halve the area of each component triangle using a median or midpoint construction, and combine the two half-area pieces consistently.
A decomposition must use non-overlapping parts. If two chosen triangles overlap, adding their areas counts the shared region twice.
Quiz
A quadrilateral can always be split into how many triangles by one diagonal?
A pentagon can be triangulated into how many triangles using a fan from one vertex?
When is subtraction especially useful for a shaded region?
Why can two different correct decompositions give the same polygon area?
What is the main requirement when adding areas of pieces?
Practice Problems
- A quadrilateral has diagonal 16 cm. The perpendicular distances of the other two vertices to the diagonal are 4 cm and 7 cm. Find the area.
- A pentagon is triangulated into areas 12 cm², 19 cm², and 23 cm². Find the total area.
- A 20 cm by 14 cm rectangle contains an unshaded rectangle of area 72 cm². Find the shaded area.
- Explain two different ways to find the area of an irregular polygon.
- A regular hexagon is split into six equal triangles, each of area 15 cm². Find the hexagon area.
- Why can a diagonal of a general quadrilateral fail to split it into equal areas?
- Design a quadrilateral and draw a construction that produces a region with half its area.
Key Takeaways
• Any simple polygon can be reduced to triangles for area calculation. • A quadrilateral split by a diagonal becomes two triangles; a pentagon can become three. • Choose diagonals that make useful bases and heights visible. • Shaded regions are often easiest as whole area minus unshaded area. • Different correct decompositions must produce the same total because they cover the same surface. • Midpoints and medians are useful tools for constructing half-area regions.