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# 8.1: Area of Polygons

Created by: CK-12

## Introduction

The School Olympics

Montgomery Middle School is going to host a school wide Olympics for the first time. The students have been studying the Greek Olympics. There has been so much enthusiasm over the content that the administration has decided to promote this event. Each class is taking on one piece of the project. The initial preparations have already started and a fantastic two day Olympic event will take place in six weeks.

Mrs. Hamilton’s class is going to prepare an area for students to be awarded prizes after the competition. The area is in the shape of a trapezoid. Mr. Samuels the custodian dug out a great spot for a platform and the students will need to use cement to fill in the area for the award area.

“How are we going to do this?” Kelly asked Mrs. Hamilton.

“Well, let’s think it through. We know that the area will be level and Mr. Samuels already helped us with that piece. Now we need to figure out the area of the trapezoid and how much cement we will need.”

“We can start with the area,” Casey said smiling.

“Probably a good idea,” Mrs. Hamilton agreed.

If the bases of the trapezoid are 35 feet and 41 feet and the height of the trapezoid is 7.5 feet, what is the area of the trapezoid?

If one bucket of cement covers 25 square feet, how many buckets will the students need?

This is the work that you will learn about in this lesson. Pay close attention and you will be able to solve this dilemma very soon.

What You Will Learn

In this lesson, you will learn how to complete the following skills.

• Find areas and linear dimensions of any triangle.
• Find areas and linear dimensions of any quadrilateral.
• Find areas and linear dimensions of any polygon.
• Solve real – world problems involving area of polygons, including metric and customary units of length and area.

Teaching Time

I. Find Areas and Linear Dimensions of any Triangle

In this lesson we will explore the area of polygons. Polygons can have any number of angles and sides, but their sides are always straight. Triangles are polygons that have three angles and sides. Quadrilaterals (including squares, rectangles, and parallelograms) have four angles and sides. Other polygons, such as pentagons, hexagons, and octagons, have even more angles and sides.

Area is the amount of two-dimensional space a figure covers. In this lesson we will learn how to find the area of polygons using formulas and other concepts. We will see how we can break a polygon into triangles to find its area. Let’s see how this works by first understanding the formula for the area of triangles.

How do we find the area of a triangle?

Area, as we have said, is the amount of space a figure covers. To find area, we multiply the dimensions, or sides, of the figure. In a triangle, those dimensions are its height, $h$, and its base, $b$. The area formula for triangles is

$A =\frac{1}{2} bh$

Write this formula for area of a triangle down in your notebook. Be sure to write “Triangle Area” with it.

The base is the area at the bottom of the triangle opposite the vertex or top point. When finding the area of triangles, remember that the height of a triangle is always perpendicular to the base. The height is not necessarily a side of the triangle; this happens only in right triangles, because the two sides joined by a right angle are perpendicular.

You can see in the right triangle that the left side is also the height of the triangle. It is perpendicular to the base. The equilateral triangle has a dotted line to show you the measurement for the height of the triangle.

Example

Find the area of the triangle below.

We can see that the base is 11 centimeters and the height is 16 centimeters. We simply put these numbers into the appropriate places in the formula.

$A &=\frac{1}{2} bh\\A &=\frac{1}{2}11(16)\\A &=\frac{1}{2}(176)\\A &=88 \ cm^2$

Remember that we always measure area in square units because we are combining two dimensions.

The area of this triangle is 88 square centimeters.

Finding the area of a triangle is just that simple. We can also use this same formula to figure out a missing dimension of the triangle. This is possible if we are given the area and one dimension to start with. Then we can use the formula, substitute the values and solve for the missing dimension. Let’s take a look.

Example

A triangle has an area of $44 \ {m^2}$. The base of the triangle is 8 m. What is its height?

In this problem, we know the area and the base of the triangle. We put these numbers into the formula and solve for the height, $h$.

$A &=\frac{1}{2} bh\\44 &=\frac{1}{2} 8h\\44 \div \frac{1}{2} &=8h\\44 \left (\frac{2}{1} \right) &=8h\\88 &=8h\\11 \ m &=h$

Remember, when you divide both sides by a fraction, you need to multiply by its reciprocal. To divide by $\frac{1}{2}$, then, we multiply by 2. Keep this in mind when you use the area formula.

By solving for $h$, we have found that the height of the triangle is 11 meters.

Now that we understand how area works for triangles, let’s see how we can use this information to find the area of quadrilaterals.

II. Find Areas and Linear Dimensions of any Quadrilateral

Quadrilaterals are four-sided polygons. Rectangles, squares, parallelograms, rhombi, and trapezoids are all quadrilaterals. Each of these has its own area formula. We can also find the area of any of these quadrilaterals by using what we know about triangles, because we can divide them all into two triangles. Take a look.

But even though we know that we can find the area of these quadrilaterals by dividing them into triangles, this is still a bit tricky. For example, think about figuring out the height of a trapezoid? This is why it is very helpful to have area formulas for each of the different types of quadrilaterals.

Let’s start by looking at rectangles.

To find the area of a rectangle, we multiply the length times the width. Let’s apply it with an example.

$A=lw$

Example

What is the area of a rectangle with a length of 6 inches and a width of 4 inches?

To figure this out, we substitute the values for length and width into the formula.

$A &=lw\\A &=(6)(4)\\A &=24 \ sq.in.$

This is the answer. Notice that we write the units for measurement as square inches because we are solving for area.

What about parallelograms?

Parallelograms are very similar to rectangles. In fact, a rectangle is a type of parallelogram. However, some parallelograms do not have four right angles. Because of this, we have to use a different formula for them. To find the area of a parallelogram, we multiply the base times the height.

Now we can substitute these values into the formula.

$A &=bh\\A &=(5)(4)\\A &=20 \ sq.meters$

This is the answer.

What about trapezoids?

A trapezoid is an interesting figure because it has two bases and a height. We have to consider the lengths of both of the bases and the height of the trapezoid to figure out its area. Here is the formula that we can use for finding the area of a trapezoid.

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

If we substitute the given values into this formula, then we can find the area of any trapezoid.

Now let’s look at applying some of this formula

Example

Find the area of this trapezoid.

$A &=\frac{1}{2}(5+8)(4)\\A &=\frac{1}{2}(13)(4)\\A &=\frac{1}{2}(52)\\A &=26 \ sq.inches$

This is the answer.

In the same way that we did for triangles, we can use the area formulas of quadrilaterals to solve for unknown dimensions. We simply fill the information we have been given into the appropriate formula and solve for the unknown variable. Let’s give it a try.

Example

A parallelogram has an area of $105 \ m^2$ The height of the parallelogram is 7 m. What is its base?

First of all, what kind of quadrilateral are we dealing with in this problem? It tells us that the figure is a parallelogram, so we will need to use the formula $A = bh$. We know the area and the height of the parallelogram, so we can put these numbers into the formula and solve for the base, $b$.

$A &= bh\\105 &= b (7)\\105 \div 7 &= b\\15 \ m &= b$

By solving for $b$, we have found that the base of the parallelogram is 15 meters.

Let’s check our calculation to be sure. We can check by putting the base and height into the formula and solving for area:

$A &= bh\\A &= 15 (7)\\A &= 105 \ m^2$

We know the area is $105 \ m^2$, so our calculation is correct.

We can follow this same process whenever we are given the area of a quadrilateral and need to solve for an unknown dimension. Just be sure you know what kind of quadrilateral you’re dealing with so that you know which area formula to use.

III. Find Areas and Linear Dimensions of Any Polygon

We can use this same method for even larger polygons. Every regular polygon can be divided into congruent triangles. A regular polygon has congruent sides and congruent angles. Take a look below.

We can divide any regular polygon into smaller triangles by drawing diagonals. A diagonal is a line drawn from one vertex, or angle, to any other vertex in the polygon. The number of diagonals depends on the number of angles and sides in the polygon. The dotted lines in the polygons above represent diagonals.

As you can see, the diagonals divide each polygon into congruent triangles. Therefore if we can find the area of one triangle, we can multiply it to find the area of the whole figure. Let’s see how this works with an example.

Example

What is the area of the figure below?

To find the area of this pentagon, we divide it into triangles whose base and height we know (because we will need the base and height to calculate the area of the triangle). The line inside the pentagon is called an apothem. It is perpendicular to one side of the pentagon and connects that side to the center. We can use this as the height of our triangle.

Now let’s work with only one of the triangles. We have a triangle with a base of 6 inches and a height of 4 inches. We can use the formula to calculate its area.

$A & =\frac{1}{2} bh\\A &= \frac{1}{2} (4) (5.81)\\ A &= 2 (5.81)\\A &= 11.62 \ in.^2$

The area of the triangle is 11.62 square inches.

Next, we can think about the five congruent triangles that make up the pentagon, so we can find its area by multiplying the area of one triangle by 5: $11.62 \times 5 = 58.1$.

The area of the whole pentagon is 58.1 square inches.

You can work with any polygon in this way. Think about a hexagon or an octagon. The number of triangles will change, but as long as you have the apothem and the length of one side, you can calculate the area of the polygon.

We can work the other way around too. If you know the area of a figure and one other dimension, either the apothem or the base you can figure out the missing dimension by working backwards. First, you will have to divide the area by the number of triangles in the figure, and then you will have to use the formula to work backwards.

Take a look at this example.

Example

The area of a hexagon is 105 sq.inches. The length of the base is 7 inches, what is the height or apothem of the figure?

To work on this problem, we first have to figure out the area of each individual triangle in the hexagon. There are six triangles in this hexagon, so we can divide the total area by six.

$105 \div 6=17.5 \ sq.inches$

Now we can use the formula for area of a triangle to figure out the missing height.

$17.5 &=\frac{1}{2}(7h)\\17.5 &=3.5h\\5 \ inches &=h$

The height of the figure is 5 inches.

IV. Solve Real – World Problems involving Area of Polygons, including Metric and Customary Units of Length and Area

We have seen that we can find the area of different polygons and use area formulas in different kinds of situations. Sometimes we need to solve for the area, but other times we may need to find the height or the base of a figure. We can also use area formulas when we are given measurements of real objects.

Let’s try a few problems involving finding area in the real world.

Example

Mr. Okada is buying new carpet to redecorate his parallelogram-shaped shop. To determine how much carpet he needs, he measured the length of the room and found it was 36 meters. Then he measured across the room with a line perpendicular to the first and found it was 16 meters across. How many square meters of carpet does Mr. Okada need to buy?

First, let’s decide which kind of polygon is involved in the problem. It tells us that Mr. Okada’s shop is in the shape of a parallelogram, so now we know what we are dealing with. Now we need to figure out what the problem is asking us to find. We need to find how much carpet Mr. Okada needs to cover the floor of his shop, so we have to find the area of the shop. That means we will use the area formula for parallelograms to solve for $A$. In order to use the formula, we need to know the base and height of the shop.

We know that one side of the shop is 36 meters. Let’s call this the base. We also know that Mr. Okada made a perpendicular line in order to measure the height, or the distance across the room. The height given in the problem is 16 meters. Let’s put this information into the formula and solve for area.

$A &= bh\\A &= 36(16)\\A &= 576 \ m^2$

Mr. Okada will need to buy 576 square meters of carpet.

Now let’s look at how we can apply what we have learned about area to the problem from the introduction.

## Real-Life Example Completed

The School Olympics

Here is the original problem once again. Reread it and then solve both parts of the problem.

Montgomery Middle School is going to host a school wide Olympics for the first time. The students have been studying the Greek Olympics. There has been so much enthusiasm over the content that the administration has decided to promote this event. Each class is taking on one piece of the project. The initial preparations have already started and a fantastic two day Olympic event will take place in six weeks.

Mrs. Hamilton’s class is going to prepare an area for students to be awarded prizes after the competition. The area is in the shape of a trapezoid. Mr. Samuels the custodian dug out a great spot for a platform and the students will need to use cement to fill in the area for the award area.

“How are we going to do this?” Kelly asked Mrs. Hamilton.

“Well, let’s think it through. We know that the area will be level and Mr. Samuels already helped us with that piece. Now we need to figure out the area of the trapezoid and how much cement we will need.”

“We can start with the area,” Casey said smiling.

“Probably a good idea,” Mrs. Hamilton agreed.

If the bases of the trapezoid are 35 feet and 41 feet and the height of the trapezoid is 7.5 feet, what is the area of the trapezoid?

If one bucket of cement covers 25 square feet, how many buckets will the students need?

Remember there are two parts of your answer.

Solution to Real – Life Example

The first thing we need to do is decide which kind of polygon we are dealing with. The driveway in the picture has two parallel sides, but one is shorter than the other. This is a trapezoid, so we know we’ll need to use the area formula for trapezoids.

What is the problem asking us to find? We need to find the number of buckets of cement necessary to cover the award area. In order to figure this out, though, we first need to find the area of the trapezoid so we know how much space needs to be covered. We will use the formula to solve for the area of the trapezoid.

What information have we been given? We know that one base, the long side of the trapezoid, is 41 feet. The shorter base is 35 feet. In this case the height of the trapezoid is the distance across the platform, and this is 7.5 feet. Let’s put this information into the formula and solve for $A$:

$A &= \frac{1}{2}(b_{1}+b_{2})h\\A &= \frac{1}{2}(41 + 35)(7.5)\\A &= \frac{1}{2}(76) (7.5)\\A &= 38 (7.5)\\A &= 285 \ ft^2$

The area of the platform is 285 square feet.

But we’re not done yet! Remember, we need to find a number of buckets of cement. What information have we been given about the cement? We know that one bucket of cement covers 25 square feet.

To find the number of buckets of cement necessary to cover the whole driveway, we need to divide the area by 25.

$285 \div 25= 11.4$

The students therefore must buy 12 buckets of cement to pave the award area.

## Vocabulary

Here are the vocabulary words that are found in this lesson.

Polygon
a simple closed figure made up of at least three line segments.
Triangle
a three-sided polygon.
Quadrilateral
a four-sided polygon.
Pentagon
a five-sided polygon.
Hexagon
a six-sided polygon.
Octagon
an eight sided polygon.
Area
the two-dimensional space that a figure occupies.
Height of the triangle
the line perpendicular to the base.
Base of a triangle
the line perpendicular to the height.
Diagonal
a line that connects one vertex to another.
Apothem
the height of a triangle in a polygon-it is a perpendicular to the base that is connected to the center of the figure.

## Time to Practice

Directions: Find the area of each triangle described below.

1. $b=10 \ inches, \ h=5 \ inches$
2. $b=7 \ inches, \ h=5.5 \ inches$
3. $b=8 \ feet, \ height=6 \ feet$
4. $b=9 \ feet, \ height=7.5 \ feet$
5. $b=12 \ meters, \ h=9 \ meters$
6. $b=15 \ feet, \ h=12 \ feet$

Directions: Find the missing dimension for each triangle given the area and one other dimension.

1. $A=4.5 \ sq.in, \ b=4.5 \ in, \ h= ?$
2. $A=21 \ sq.ft, \ b=7 \ ft, \ h= ?$
3. $A=60 \ sq.in, \ h=10 \ in, \ b= ?$
4. $A=97.5 \ sq.ft, \ h=13 \ ft, \ b= ?$
5. $A=187 \ sq.ft, \ b=22 \ ft, \ h= ?$
6. $A=405 \ sq.ft, \ b=30 \ ft, \ h= ?$

Directions: Find the area of the following rectangles.

1. $l=10 \ in, \ w=7.5 \ in$
2. $l=12 \ ft, \ w=9 \ ft$
3. $l=14 \ ft, \ w=11 \ ft$
4. $l=21 \ ft, \ w=19 \ ft$

Directions: Find the area of each parallelogram.

1. $b=11 \ ft, \ h=9 \ ft$
2. $b=13 \ in, \ h=11 \ in$
3. $b=22 \ ft, \ h=19 \ ft$
4. $b=31 \ meters, \ h=27 \ meters$

Directions: Find the area of each trapezoid.

1. $Bases=5 \ in \ and \ 8 \ in, \ height=4 \ inches$
2. $Bases=6 \ in \ and \ 8 \ in, \ height=5 \ inches$
3. $Bases=10 \ feet \ and \ 12 \ feet, \ height=9 \ feet$

Jul 25, 2013

## Last Modified:

Jun 04, 2014
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CK.MAT.ENG.SE.1.Middle-School-Math-Grade-8.8.1