Math Core

Area measures how much flat surface a figure covers, in square units. You have used area formulas for years. In geometry, the goal is to see why each formula works, so you can rebuild any of them from the rectangle and handle figures that don't come with a formula attached.

Start with the rectangle

A rectangle with base $b$ and height $h$ can be tiled by $b$ rows of $h$ unit squares, so its area is $A = bh$ . Every other polygon formula in this lesson comes from cutting, rearranging or doubling a shape until it becomes a rectangle.

Two facts make this legal:

Congruent figures have equal areas. Moving a piece doesn't change its size.
Area is additive. If a figure is split into pieces that don't overlap, its area is the sum of the areas of the pieces.

Parallelograms and triangles

Cut a right triangle off one end of a parallelogram and slide it to the other end. The result is a rectangle with the same base and the same height, so a parallelogram also has area $A = bh$ .

In a parallelogram, the height is the perpendicular distance between the bases, not the slanted side.

The catch is the word height. The height is the perpendicular distance between the two bases, not the length of the slanted side. The slanted side is always at least as long as the height, so using it makes your answer too big.

Now take any triangle and glue a congruent copy to it, rotated $180^\circ$ . The two copies form a parallelogram with the same base and height. The triangle is half of it:

A_{\text{triangle}} = \frac{1}{2}bh

The height h is perpendicular to the base b and meets it at a right angle.

Any side of a triangle can serve as the base. The height that goes with it is the perpendicular segment from the opposite vertex to the line containing that base. For an obtuse triangle, that segment may land outside the triangle, on an extension of the base.

Trapezoids

Two congruent copies of a trapezoid, one flipped upside down, fit together into a parallelogram whose base is $b_1 + b_2$ and whose height is $h$ . One trapezoid is half of that parallelogram.

A trapezoid with parallel bases b₁ and b₂ and height h.

Rhombuses and kites

The diagonals of a rhombus or a kite are perpendicular. Draw a rectangle around the figure with sides parallel to the diagonals. The rectangle measures $d_1$ by $d_2$ , and the kite fills exactly half of it (each of the four small rectangles is cut in half by a side of the kite). So $A = \frac{1}{2}d_1 d_2$ .

Area formulas for polygons

Figure	Area
Rectangle or parallelogram	$A = bh$
Triangle	$A = \frac{1}{2}bh$
Trapezoid	$A = \frac{1}{2}(b_1 + b_2)h$
Rhombus or kite (diagonals $d_1$ , $d_2$ )	$A = \frac{1}{2}d_1 d_2$
Regular polygon (apothem $a$ , perimeter $P$ )	$A = \frac{1}{2}aP$

In every formula, the height is measured perpendicular to the base.

Worked example: A parallelogram with a slanted side

A parallelogram has sides of 12 cm and 8 cm. The height to the 12 cm side is 6 cm. Find its area.

The base is 12 cm and the height to that base is 6 cm. The 8 cm side is slanted, so it plays no part in the area:

A = bh = 12 \cdot 6 = 72 \text{ cm}^2

Common mistake

Using the slanted side as the height is the most common area mistake. In the last example, $12 \cdot 8 = 96$ is wrong. Before you multiply, check that the height you chose makes a right angle with the base.

Worked example: A trapezoid with a missing height

An isosceles trapezoid has bases of 7 in and 19 in, and each leg is 10 in. Find its area.

Drop a height from each end of the short base. The two heights cut off a $7$ in rectangle in the middle, so the long base has $19 - 7 = 12$ in left over, or 6 in on each side. Each end is a right triangle with hypotenuse 10 and leg 6:

h = \sqrt{10^2 - 6^2} = \sqrt{64} = 8 \text{ in}

Now use the trapezoid formula:

A = \tfrac{1}{2}(7 + 19)(8) = \tfrac{1}{2}(26)(8) = 104 \text{ in}^2

Regular polygons

A regular polygon has all sides congruent and all angles congruent. Its center is the same distance from every vertex, and its apothem $a$ is the perpendicular segment from the center to a side.

A regular hexagon. The apothem a runs from the center to the midpoint of a side, at a right angle to it.

Segments from the center to each vertex split a regular $n$ -gon into $n$ congruent isosceles triangles. Each triangle has base $s$ (one side) and height $a$ (the apothem). Adding them up:

A = n \cdot \tfrac{1}{2}as = \tfrac{1}{2}a(ns) = \tfrac{1}{2}aP

since $ns$ is the perimeter $P$ .

Worked example: A regular hexagon

Find the exact area of a regular hexagon with side length 6.

The six triangles in a regular hexagon are equilateral, because each central angle is $360^\circ \div 6 = 60^\circ$ . The apothem is the height of an equilateral triangle with side 6, which splits it into two $30^\circ$ - $60^\circ$ - $90^\circ$ triangles with short leg 3:

a = 3\sqrt{3}

The perimeter is $6 \cdot 6 = 36$ , so

A = \tfrac{1}{2}aP = \tfrac{1}{2}(3\sqrt{3})(36) = 54\sqrt{3} \approx 93.5

Tip

For a regular polygon that isn't a hexagon or a square, find the apothem with trigonometry. The triangle from the center to the midpoint of a side has a central angle of $\dfrac{180^\circ}{n}$ , and $\tan\left(\dfrac{180^\circ}{n}\right) = \dfrac{s/2}{a}$ .

Polygons on the coordinate plane

When a polygon's vertices are points on a grid, its sides are often slanted, which makes heights awkward. A reliable method is to box it in: draw the smallest rectangle around the polygon with horizontal and vertical sides, then subtract the right triangles (and rectangles) in the corners.

Worked example: Boxing in a triangle

Find the area of the triangle with vertices $A(0, 0)$ , $B(6, 2)$ and $C(2, 5)$ .

The box runs from $x = 0$ to $x = 6$ and from $y = 0$ to $y = 5$ , so its area is $6 \cdot 5 = 30$ . Three right triangles sit outside $\triangle ABC$ :

Below $\overline{AB}$ : legs 6 and 2, area $\tfrac{1}{2}(6)(2) = 6$ .
Right of $\overline{BC}$ : legs 3 and 4, area $\tfrac{1}{2}(3)(4) = 6$ .
Left of $\overline{CA}$ : legs 2 and 5, area $\tfrac{1}{2}(2)(5) = 5$ .

A = 30 - (6 + 6 + 5) = 13 \text{ square units}