Have you ever been to a school Olympics? Take a look at this dilemma.

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 Concept. Pay close attention and you will be able to solve this dilemma very soon.**

### Guidance

*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.

\begin{align*}A=lw\end{align*}

Now let's apply that formula.

**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.

\begin{align*}A &=lw\\ A &=(6)(4)\\ A &=24 \ sq.in.\end{align*}

**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.**

\begin{align*}A &=bh\\ A &=(5)(4)\\ A &=20 \ sq.meters \end{align*}

**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.**

\begin{align*}A=\frac{1}{2}({b_1}+{b_2})h\end{align*}

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

Take a look at this one.

Find the area of this trapezoid.

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

**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. Just be sure that you know which formula to use.

Find the area of each quadrilateral.

#### Example A

A square with a side length of 4.5 inches.

**Solution: \begin{align*}20.25\end{align*} sq. inches**

#### Example B

A rectangle with a length of 8 feet and a width of 6.25 feet.

**Solution: \begin{align*}50\end{align*} sq. feet**

#### Example C

A parallelogram with a base of 10 meters and a height of 7.5 meters.

**Solution: \begin{align*}75\end{align*} square meters**

Now let's go back to the dilemma from the beginning of the Concept.

**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 \begin{align*}A\end{align*}:**

\begin{align*}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\end{align*}

**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.**

\begin{align*}285 \div 25= 11.4\end{align*}

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

### Vocabulary

- Polygon
- a simple closed figure made up of at least three line segments.

- Quadrilateral
- a four-sided polygon.

- Area
- the two-dimensional space that a figure occupies.

### Guided Practice

Here is one for you to try on your own.

**A parallelogram has an area of \begin{align*}105 \ m^2\end{align*} The height of the parallelogram is 7 m. What is its base?**

**Solution**

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 \begin{align*}A = bh\end{align*}. We know the area and the height of the parallelogram, so we can put these numbers into the formula and solve for the base, \begin{align*}b\end{align*}.

\begin{align*}A &= bh\\ 105 &= b (7)\\ 105 \div 7 &= b\\ 15 \ m &= b\end{align*}

**By solving for \begin{align*}b\end{align*}, 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:

\begin{align*}A &= bh\\ A &= 15 (7)\\ A &= 105 \ m^2\end{align*}

**We know the area is \begin{align*}105 \ m^2\end{align*}, so our calculation is correct.**

### Video Review

### Practice

Directions: Find the area of the following rectangles.

- \begin{align*}l=10 \ in, \ w=7.5 \ in\end{align*}
- \begin{align*}l=12 \ ft, \ w=9 \ ft\end{align*}
- \begin{align*}l=14 \ ft, \ w=11 \ ft\end{align*}
- \begin{align*}l=21 \ ft, \ w=19 \ ft\end{align*}

Directions: Find the area of each parallelogram.

- \begin{align*}b=11 \ ft, \ h=9 \ ft\end{align*}
- \begin{align*}b=13 \ in, \ h=11 \ in\end{align*}
- \begin{align*}b=22 \ ft, \ h=19 \ ft\end{align*}
- \begin{align*}b=31 \ meters, \ h=27 \ meters\end{align*}

Directions: Find the area of each trapezoid.

- \begin{align*}Bases=5 \ in \ and \ 8 \ in, \ height=4 \ inches\end{align*}
- \begin{align*}Bases=6 \ in \ and \ 8 \ in, \ height=5 \ inches\end{align*}
- \begin{align*}Bases=10 \ feet \ and \ 12 \ feet, \ height=9 \ feet\end{align*}

Directions: Find the area of each square.

- side length of 8 inches
- side length of 15 feet
- side length of 22.5 mm
- side length of 18.25 cm