Area of Polygons
Apply formulas for the area of triangles, parallelograms, trapezoids, rhombuses, and regular polygons (A = ½·apothem·perimeter). Decompose composite figures and solve problems involving shaded regions.
A lesson of Geometry, in Area and Volume, about 10 minutes.
By the end of this lesson, you will be able to find the areas of triangles, quadrilaterals, and regular polygons using the appropriate formulas, apply the area addition and area subtraction techniques, and solve problems requiring multiple area computations.
The Big Idea
Area is the measure of a two-dimensional region in square units. Every polygon's area can be computed from a set of base formulas, with complex shapes decomposed into simpler ones. Regular polygons require one additional concept — the apothem — but their formula follows directly from decomposing into triangles.
Fundamental Area Formulas
Rectangle: $A = bh$ (base × height)
Parallelogram: $A = bh$ (base × perpendicular height — not the slanted side)
Triangle: $A = \dfrac{1}{2}bh$
Trapezoid: $A = \dfrac{1}{2}(b_1 + b_2)h$ (half the sum of bases times height)
Rhombus/Kite: $A = \dfrac{1}{2}d_1 d_2$ (half the product of the diagonals)
Regular Polygons
A regular polygon can be divided into $n$ congruent isosceles triangles from the center.
The apothem ($a$) is the perpendicular distance from the center to the midpoint of a side. It is the height of each triangle.
$$A = \frac{1}{2} \cdot \text{perimeter} \cdot \text{apothem} = \frac{1}{2} P a$$
Derivation: Each triangle has area $\dfrac{1}{2} \cdot s \cdot a$ (where $s$ = side length). With $n$ triangles: $A = n \cdot \dfrac{1}{2} \cdot s \cdot a = \dfrac{1}{2}(ns)(a) = \dfrac{1}{2}Pa$.
Example 1 — Parallelogram
Parallelogram with base 14 and a slanted side of 10. The height is 8. Find the area.
$$A = 14 \times 8 = 112 \text{ sq units}$$
The slanted side (10) is not used in the area formula — only the perpendicular height.
Example 2 — Triangle on the Coordinate Plane
Triangle with vertices $A(0,0)$, $B(6,0)$, $C(3,8)$.
Base: $AB = 6$ (horizontal). Height: perpendicular distance from $C$ to $\overline{AB}$ = $8$ (just the $y$-coordinate since $\overline{AB}$ is on the $x$-axis).
$$A = \frac{1}{2}(6)(8) = 24 \text{ sq units}$$
Example 3 — Trapezoid
Trapezoid with parallel bases 7 and 13, height 6.
$$A = \frac{1}{2}(7 + 13)(6) = \frac{1}{2}(20)(6) = 60$$
Example 4 — Regular Hexagon
Regular hexagon with side length 8.
A regular hexagon can be divided into 6 equilateral triangles. The apothem of a regular hexagon with side $s$ is $a = \dfrac{s\sqrt{3}}{2}$.