When we try to determine how much space a is covered in a two-dimensional space, we need to use the concept of area. For example, when buying a house we oftentimes will consider the total living space in a house, which is considering how much floor space the house has.
Because area is represented in two-dimensions, we represent quantities of area as squared units. For example, we might say that a square living room has a total area of 625 square feet, which might mean that the length of the room is 25 feet and the width is 25 feet.
Suppose we wanted to paint the outer walls of a house, we may want to estimate the total area of the house’s exterior to get an estimate of the total amount of paint that we would need.
When calculating areas, there are a few common shapes that you will always come across. These shapes and their area formulas are listed below.
Note that the area of a square is a special case of the area of a rectangle formula where the length and width are the same.
Note that the height of a triangle is the distance of the line from one vertex (or corner) of the triangle to the opposite base, such that the line is perpendicular to the base.
Note that the radius of a circle is the distance from the center of a circle to the edge of the circle. Pi is a constant irrational number equal to 3.14159265….
Let’s look at how we can use these different area formulas to find some unknown quantity.
EXAMPLE
![]() |
|
![]() |
![]() |
|
Multiply length by width |
![]() |
|
Units of feet are also multiplied |
![]() |
|
Our Solution |
Notice that when calculating area, we square the units of distance.
EXAMPLE
Here, we are given the diameter of the circle, but we need to know the radius to use the formula. The diameter of a circle is simply the distance of the line passing through the center of a circle and touching the circle’s edge. In other words the diameter, d, is twice as long as the radius, r, (d= 2r).
Since the diameter for this circle was 9 inches, the radius must be 4.5 inches.
![]() |
|
![]() |
![]() |
|
Substitute 4.5 in for r |
![]() |
|
Square the radius |
![]() |
|
Our Solution, rounded to the tenths place |
Sometimes we may be given the area of an object and need to back solve to find the measurements of a given part of an object. Let’s look at some examples.
EXAMPLE
When solving these types of problems, we follow the same process as we did before but then do some algebraic manipulation to solve the problem. Note that the length and width of a square is the same so we can refer to side length using a single variable, s.
![]() |
|
![]() |
![]() |
|
Take the square root of both sides |
![]() |
|
Our Solution |
EXAMPLE
Like in Example 3, we begin by substituting what we know into the appropriate formula. Notice that the area contains pi in it. This is often times the same when we represent area exactly; we leave the pi multiplied to the number.
![]() |
|
![]() |
![]() |
|
Divide both sides by ![]() |
![]() |
|
Take the square root of both sides |
![]() |
|
Our Solution |