Perimeter and Area of Polygons

Start with perimeter. Because perimeter is the total distance around, one rule works for every polygon: add the lengths of all its sides. When the side lengths are positive and in the same unit, the perimeter $P$ is

$P = \text{sum of all the side lengths}$,

and $P$ is in that same linear unit.

Equal sides give you shortcuts, because adding the same number several times is the same as multiplying:

• A rectangle (four sides and four right angles) with length $l$ and width $w$ has two sides of each measure, so $P = 2l + 2w$.
• A square is a rectangle with four equal sides, which also makes it a rhombus (a four-sided polygon with four equal sides). If each side is $s$, then $P = 4s$.
• An equilateral triangle has three equal sides. If each side is $s$, then $P = 3s$. It is also isosceles, which means it has at least two equal sides.

Use a shortcut only when the problem states that the sides are equal. Never judge that from how a picture looks.

Now for area. Think of a rectangle that is 6 cm long and 4 cm tall. It holds 4 rows with 6 unit squares in each row, so it covers $6 \times 4 = 24$ unit squares. Its area is $24\text{ cm}^2$.

Call one side of a rectangle the base, $b$, and call the side that meets it at a right angle the height, $h$. Counting rows of unit squares gives the area $A$:

$A = b \times h$.

This rule holds for any positive $b$ and $h$ in the same unit, including fractions and decimals, and $A$ comes out in that unit squared. A square is a rectangle, so a square with side $s$ has $A = s \times s$.

A parallelogram is a four-sided polygon with two pairs of parallel sides, which are sides that always stay the same distance apart. It can lean, so unit squares do not line up neatly inside it. The figure below shows how to straighten one without changing its area. The horizontal axis is the $x$-axis, the vertical axis is the $y$-axis, grid lines are 1 unit apart, and 1 unit stands for 1 cm.

• The parallelogram has vertices (corner points) $(1, 1)$, $(7, 1)$, $(10, 5)$ and $(4, 5)$.
• The right triangle inside its left end has vertices $(1, 1)$, $(4, 1)$ and $(4, 5)$.
• Slide that triangle 6 units to the right. It lands exactly on the triangle with vertices $(7, 1)$, $(10, 1)$ and $(10, 5)$.

No area is added or lost, and the result is the rectangle with vertices $(4, 1)$, $(10, 1)$, $(10, 5)$ and $(4, 5)$. So a parallelogram covers the same area as a rectangle with the same base and the same perpendicular height:

$A = b \times h$.

Here $h$ is the distance from the base straight across to the opposite side, measured at a right angle to the base. Use positive lengths in one unit.

A triangle is a polygon with three sides. The next figure uses the same axes and the same scale, with 1 unit standing for 1 cm. The triangle has vertices $(1, 1)$, $(7, 1)$ and $(4, 5)$. Next to it is an exact copy, turned halfway around, with vertices $(7, 1)$, $(10, 5)$ and $(4, 5)$.

Together, the two triangles make the parallelogram you just measured, with vertices $(1, 1)$, $(7, 1)$, $(10, 5)$ and $(4, 5)$. The triangles are exactly the same shape and size, so each one is half of the parallelogram.

This works for any triangle: a copy turned halfway around fits against it to make a parallelogram with the same base and the same perpendicular height. So a triangle with base $b$ and height $h$ has area

$A = \tfrac{1}{2} \times b \times h$.

The base $b$ can be any side you choose. The height $h$ is the perpendicular distance from the opposite vertex to the line the base lies on. Both must be positive and in the same unit.

A trapezoid is a four-sided polygon with at least one pair of parallel sides. Call the two parallel sides the bases, $b_1$ and $b_2$. The height $h$ is the perpendicular distance between them.

In the next figure (same axes, 1 unit stands for 1 cm), a trapezoid with vertices $(1, 1)$, $(9, 1)$, $(5, 4)$ and $(1, 4)$ is cut along a diagonal, a segment joining two opposite corners, from $(1, 1)$ to $(5, 4)$. The cut makes two triangles:

• one with vertices $(1, 1)$, $(9, 1)$ and $(5, 4)$, whose base is the trapezoid's bottom side;
• one with vertices $(1, 1)$, $(5, 4)$ and $(1, 4)$, whose base is the trapezoid's top side.

The bases are parallel, so they are the same distance apart everywhere, and that distance $h$ is the height of both triangles. Add the two triangle areas:

$A = \tfrac{1}{2} b_1 h + \tfrac{1}{2} b_2 h = \tfrac{1}{2}(b_1 + b_2)h$.

The last step uses the distributive property, because both parts share the factor $\tfrac{1}{2}h$. A diagonal splits every trapezoid this way, so the rule works for all of them when the lengths are positive and in one unit.

You now have two tools for measuring polygons. For perimeter, add every side length and answer in linear units. For area, start from a base and its perpendicular height, and answer in square units: use the whole product for a rectangle or parallelogram, half of it for a triangle, and half of the height times the sum of the two bases for a trapezoid. Keep every length positive and in one unit.

Next, you will run these rules backward to find a missing dimension when you already know the area or the perimeter, and you will use them in word problems.