Math Core

Contest area problems rarely hand you a shape that fits a formula. Instead you get a strange polygon, a shaded region, or a triangle drawn inside a rectangle. The formulas are the easy part; the real skill is cutting a shape into pieces you know, or surrounding it with a shape you know and subtracting.

The formulas you need

Shape	Area
Rectangle	$\text{length} \times \text{width}$
Triangle	$\dfrac{1}{2} \cdot \text{base} \cdot \text{height}$
Parallelogram	$\text{base} \times \text{height}$
Trapezoid	$\dfrac{1}{2}(b_1 + b_2) \cdot h$ (average of the parallel sides, times the height)
Rhombus (or any kite)	$\dfrac{1}{2} d_1 d_2$ (half the product of the diagonals)
Equilateral triangle, side $s$	$\dfrac{\sqrt{3}}{4}s^2$
Regular hexagon, side $s$	$6 \cdot \dfrac{\sqrt{3}}{4}s^2$ (six equilateral triangles)

The height is always measured perpendicular to the base, not along a slanted side. And any side of a triangle can be the base, as long as you use the height that goes with it.

Three strategies

Add: cut the shape into rectangles and triangles.
Subtract: draw a rectangle around the shape, then subtract the simple pieces outside it. This is often the fastest way to find the area of a tilted triangle.
Compare: use ratios instead of computing every area.

Triangles that share a height

If two triangles have the same height, their areas are in the same ratio as their bases. In particular:

Triangles with the same base and same height have the same area, even if they look completely different.
A segment from a vertex to the opposite side splits a triangle into two triangles whose areas are in the ratio of the two parts of that side. A median (to the midpoint) splits the area exactly in half.

Worked example: A trapezoid

A trapezoid has parallel sides $8$ and $14$ and height $5$ . Find its area.

The average of the parallel sides is $\dfrac{8 + 14}{2} = 11$ , so the area is $11 \cdot 5 = 55$ .

Worked example: Subtracting the corners

Square $ABCD$ has side $12$ , with $A$ at the bottom left and the vertices going counterclockwise. $E$ is the midpoint of $BC$ and $F$ is the midpoint of $CD$ . Find the area of triangle $AEF$ .

Triangle AEF inside square ABCD. The three corner triangles are right triangles.

The square has area $144$ . Outside triangle $AEF$ there are three right triangles:

$ABE$ with legs $12$ and $6$ : area $36$ .
$ECF$ with legs $6$ and $6$ : area $18$ .
$FDA$ with legs $6$ and $12$ : area $36$ .

So triangle $AEF$ has area $144 - 36 - 18 - 36 = 54$ .

Worked example: Splitting a side in a ratio

Triangle $ABC$ has area $60$ . Point $D$ lies on $BC$ with $BD : DC = 2 : 3$ . Find the area of triangle $ABD$ .

Triangles $ABD$ and $ADC$ share the same height from $A$ to line $BC$ , so their areas are in the ratio $2 : 3$ . Triangle $ABD$ gets $\dfrac{2}{5}$ of the total: $\dfrac{2}{5} \cdot 60 = 24$ .

Worked example: A regular hexagon

Find the area of a regular hexagon with side length $4$ .

The three long diagonals cut a regular hexagon into six equilateral triangles.

The long diagonals cut it into six equilateral triangles with side $4$ , each with area $\dfrac{\sqrt{3}}{4} \cdot 16 = 4\sqrt{3}$ . The hexagon's area is $6 \cdot 4\sqrt{3} = 24\sqrt{3}$ .

Common mistake

Don't use a slanted side as the height. In a parallelogram with sides $10$ and $6$ , the area is not $60$ unless the angles are right angles. You need the perpendicular distance between the parallel sides.