Area and perimeter

There are only a few formulas you need to memorize to solve most area problems confidently:

Area of parallelogram (square, rectangle)

$\text{length}\ast \text{width}$ or $\text{base}\ast \text{height}$

Area of triangle

$\frac{1}{2}\ast \text{base}\ast \text{height}$

Area of circle

$\pi \ast {\text{radius}}^2$

It helps to have a simple way to remember these formulas. For example, a triangle is half of a rectangle (or parallelogram):

That picture is a good reminder that the area of a triangle is one-half $\times$ base $\times$ height.

Once you’re comfortable with these formulas, you can move on to the kinds of shape properties that often show up on tests.

Properties of shapes

Let’s start with the different kinds of triangles: equilateral, isosceles, and scalene. Remember: all triangles use the same area formula (base $\times$ height $\times \frac{1}{2}$).

Definitions

Equilateral triangles

Triangles where all three sides are equal to each other.

Isosceles triangles

Triangles where two sides are equal to each other and one is not.

Scalene triangles

Triangles where no sides are equal to each other.

Now for parallelograms. This may feel straightforward, but it’s important to recognize these terms when they appear in a question. Remember: all parallelograms use the same area formula (length $\times$ width or base $\times$ height).

Definitions

Parallelogram

A four-sided shape where opposing sides are equal. Rectangles, squares, and rhombi are all parallelograms.

Rectangle

A four-sided shape where opposing sides are equal and all four angles are right angles.

Square

A four-sided shape where all sides are equal and all four angles are right angles. A square is a rectangle, but a rectangle is not necessarily a square: rectangles don’t require all sides to be equal.

Unfamiliar shapes

You may be shown a strange-looking shape and asked to find its area. These questions can seem difficult, as if there’s a special formula you’re supposed to remember. In most cases, you don’t need a new formula.

Instead, use this strategy: break the shape into shapes you already know (parallelograms/rectangles, triangles, and circles).

When you need the area of an unfamiliar shape, try drawing lines (or imagining removing parts) to create familiar shapes. Trapezoids are a common example.

How do you find the area of a trapezoid? Do you remember the formula? Probably not. So use the strategy: break it into familiar shapes.

Here, you draw two lines to break the trapezoid into a rectangle and two equal triangles. Once you’ve done that, you can use the basic area formulas you already know.

So the strategy is:

• Break down unfamiliar shapes into basic, familiar shapes by drawing or removing lines.

Partial areas

Many problems ask for the area of part of a shape, often in a word problem. A common setup is one shape inside another, like this:

A typical question might describe this as a wall with a window (the shaded rectangle). If you want the area of the wall surface, you do not include the window.

The method is:

• Find the total area.
• Subtract the area of the portion you don’t want.

For this problem specifically, find the area of the whole rectangle (length $\times$ width)

$$5\ast 10=50$$

Then find the area of the window (the part you don’t want included):

$$2\ast 3=6$$

Finally, subtract the unwanted part from the total:

$$50-6=44$$

Area using algebra

Basic algebra questions involving area usually come down to one skill: write the correct area formula and substitute the values you’re given.

Try solving this one using the triangle area formula.

Word problems involving area are often harder than straightforward algebra problems because you have to translate words into equations.

Suppose the word problem is:

The length of a rectangle is twice as large as its width. If the area is $8$, what is the length of the rectangle?

Start with the formula you know:

• Area $=$ length $\times$ width

Now translate the information:

• Area $=8$
• length $=2\times$ width
• width $=W$

Write the equation:

$$8=2W\ast W$$

Now solve for the one variable:

$$\begin{array}{rl}8/2& =W\ast W\\ 4& =W^2\\ \sqrt{4}& =2\\ 2& =W\end{array}$$

Finally, answer what the question asked for: the length.

• length $=2W=2(2)=4$

Perimeter

Perimeter is the total distance around the outside of a shape. The perimeter of a circle has a special name: circumference.

Most perimeter formulas are easy to reason out without memorizing them, but it helps to keep a few in mind:

Circles

The perimeter (circumference) of a circle is $2\pi \times \text{radius}$ (don’t confuse this with the area formula, $\pi \times {\text{radius}}^2$).

Squares

Since each side is equal, the perimeter is $4$ times one side, or $4L$.

Rectangles

Rectangles have $2$ lengths and $2$ widths that are equal to each other, so the perimeter is $(2\times L)+(2\times W)$.

Triangles

To find the hypotenuse of a right triangle, use the Pythagorean Theorem $(a^2+b^2=c^2)$. This is useful when you need the perimeter of a right triangle.

Perimeter is usually more about problem-solving than memorization. You can often build the perimeter by adding the side lengths. The main formula worth memorizing is circumference, $2\pi r$.

Geometry questions often combine area and perimeter in the same problem, which means you may need algebra. Here’s an example:

The area of a rectangle is $8$ and its perimeter is $12$. What are the values of the lengths of the rectangle? Try solving this out, then review your answer. Start by using the area and perimeter formulas for both area and perimeter and trying to relate them to each other.

a. 8 & 4
b. 2 & 4
c. 8 & 1
d. 3 & 4
e. 6 & 2

Be careful! Plugging in answers is a useful shortcut, but not every problem will be set up for it. It’s also worth knowing how to solve this using a system of equations and quadratic equations. Please recap the [Solving a system of equations] and [Factorization of quadratics and cubics] chapters if you are not confident with them!

Here is the proper way to solve this problem using a system of equations:

List the two equations and substitute the given values:

$$\begin{array}{rl}\text{Area}& =\text{base}\ast \text{height}\\ 8& =b\ast h\end{array}$$

$$\begin{array}{rl}\text{Perimeter}& =2\ast \text{base}+2\ast \text{height}\\ 12& =2\ast \text{base}+2\ast \text{height}\\ 12& =2b+2h\end{array}$$

Now solve one equation for one variable.

Using the area equation:

$$\begin{gathered} 8=b\ast h \\ b=\frac{8}{h} \end{gathered}$$

Substitute this expression for $b$ into the perimeter equation:

$$12=2(\frac{8}{h})+2h$$

Multiply into the parentheses:

$$12=\frac{16}{h}+2h$$

Solve for $h$ by clearing the fraction:

$$12-2h=\frac{16}{h}$$

Multiply by $h$:

$$h(12-2h)=16$$

Distribute:

$$12h-2h^2=16$$

Rewrite as a quadratic equation:

$$-2h^2+12h-16=0$$

Solve for $h$ by quadratic formula or by factoring. We’ll factor, but first divide by $-2$ to simplify:

$$h^2-6h+8=0$$

Factor:

$$(h-4)(h-2)$$

Answers:

$$h=4\text{ or }h=2$$

Since we are looking for the base AND the height, and the height could be either of these values, these two values are the answer!

Key points

Area formulas. Memorize the formulas above for circles, parallelograms, and triangles.

Shape properties. Properties of scalene (no similar sides), isosceles (two similar sides), and equilateral triangles (all similar sides) as well as parallelograms (opposing sides equal), rectangles (opposing sides equal, four right angles), and squares (all sides equal, four right angles).

Unfamiliar Shapes. Draw or remove lines to see unfamiliar shapes as more familiar ones, like triangles and rectangles in a trapezoid.

Partial Areas. If you are only looking for part of a whole area, subtract the area of the unwanted portion from the total area.

Area Using Algebra. Plug the information you are given from a word problem into the formulas you know. You may need to replace length and width with a short equation using $x$, then solve for that variable.

Perimeter. Keeping in mind that most of the equations can be thought up without memorizing them, but you need to know the formulas for circumference ($2\pi r$) and the Pythagorean Theorem ($a^2+b^2=c^2$).