14.3 The Area of Any Polygon

How do we find the area of a four-sided figure that is not a rectangle, a general quadrilateral ABCD? Just draw one diagonal, say BD. That cuts ABCD into two triangles. Work out each triangle’s area and add.

The very same move works for a pentagon, a hexagon, and indeed any polygon:

The Master Idea Every polygon can be cut into triangles. Since the area of a triangle is always within reach, we can find the area of any polygon by chopping it into triangles and adding their areas.

Worked Example

Find the area of quadrilateral ABCD where the diagonal AC = 26\text{ cm}, the perpendicular from B to AC is BM = 4\text{ cm}, and the perpendicular from D to AC is DN = 5\text{ cm}.

The diagonal AC divides ABCD into \triangle ABC (base AC, height BM) and \triangle ACD (base AC, height DN):

\begin{aligned} \text{Area} &= \tfrac12 \times AC \times BM + \tfrac12 \times AC \times DN \\ &= \tfrac12 \times 26 \times 4 + \tfrac12 \times 26 \times 5 \\ &= 52 + 65 = 117 \text{ cm}^2. \end{aligned}

A Bonus: Heron’s Formula What if you know only the three side lengths of a triangle, with no height in sight? An old result attributed to Heron of Alexandria still pins down the area. Let the sides be a, b, c, and let s = \tfrac{a+b+c}{2} be the half-perimeter. Then

\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}.

For a triangle with sides 13, 14, 15: s = 21, and the area is \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = \sqrt{7056} = 84 square units. No altitude required.

Figure it Out: Polygons

Practice

  1. Find the area of quadrilateral ABCD given AC = 26\text{ cm}, BM = 4\text{ cm}, DN = 5\text{ cm}, with BM \perp AC and DN \perp AC.
  2. ABCD is a rectangle of 20\text{ cm} \times 12\text{ cm}. Point E lies on AB with AE = 12\text{ cm} and EB = 8\text{ cm}, and point F lies on AD with AF = 7\text{ cm} and FD = 5\text{ cm}. The segment EF is drawn. Find the area of region EBCDF (the rectangle minus \triangle AEF).
  3. Which measurements would you need to find the area of a regular octagon?
  4. What fraction of a rectangle’s area is the blue triangle whose base is a full side and whose apex lies anywhere on the opposite side?
  5. Describe a method to construct a quadrilateral whose area is half that of a given quadrilateral.
  1. \text{Area} = \tfrac12 \times 26 \times 4 + \tfrac12 \times 26 \times 5 = 52 + 65 = \mathbf{117\text{ cm}^2}.
  2. The whole rectangle has area 20 \times 12 = 240\text{ cm}^2. The cut-off corner \triangle AEF has perpendicular sides AE = 12\text{ cm} and AF = 7\text{ cm}, so its area is \tfrac12 \times 12 \times 7 = 42\text{ cm}^2. Region EBCDF = 240 - 42 = \mathbf{198\text{ cm}^2}.
  3. Split the octagon into 8 triangles from the centre. You need the side length and the apothem (the perpendicular distance from the centre to a side); then area = 8 \times \tfrac12 \times \text{side} \times \text{apothem} = 4 \times \text{side} \times \text{apothem}. (For a regular octagon the side length alone fixes everything, since the apothem follows from it.)
  4. Whichever point on the opposite side serves as the apex, the triangle has the full base of the rectangle and the full height between the two sides, so its area is \tfrac12 \times \text{base} \times \text{height}, exactly half the rectangle.
  5. Draw a diagonal to split the quadrilateral into two triangles. In one of those triangles, draw a median to the midpoint of a side; the median halves that triangle’s area. Combining the median piece with the untouched second triangle in the right way produces a quadrilateral of half the original area. (Equivalently: a diagonal gives two triangles, and a median of either triangle cuts off a region of half that triangle.)