Area of a Quadrilateral

Area of a quadrilateral

A quadrilateral is a figure, or shape, with four sides. Rectangles and squares are both quadrilaterals.

The formula of area of rectangles and squares are rather straightforward--it is width times length. However, for squares, the width equals the length.

$$A=\text{width}\times \text{length}$$

What about the area of a parallelogram?

Can we still use the $\text{length} \times \text{width}$ formula?

The answer is yes!

In fact, you can cut up any quadrilateral and rearrange the pieces to make a rectangle. Since you are using the pieces from the old quadrilateral, the new rectangle will has the same area. This way, the area formula of a rectangle applies to all quadrilaterals!

Try it!

Notice that if we have two of these quadrilaterals, we can make a rectangle:

The area of this large rectangle is $(4+7)\times 4=44 \text{ units}^2$.

The original quadrilateral is half the size of this large rectangle, therefore its area is half of $44 \text{ units}^2$, which is $22\text{ units}^2$.

$$\textbf{Areas of Different Quadrilaterals}$$

Here are some graphics that show you the formulas to find the areas of different quadrilaterals: squares, rectangles, parallelograms, rhombi, trapezoids, and kites.

(Video) Area of Quadrilaterals

by SmithMathAcademy

In this video, SmithMathAcademy goes over how to find the area of squares, rectangles, general parallelograms, and irregular polygons.

$\large{\text{Summary}}$

Finding the Area of a Square

A square has four sides of equal lengths. Let's say each side has length $s.$

Then $A,$ the area, is given by the following equation: $$\begin{align} A&=s \cdot s\\ \text{or } A&=s^2 \end{align}$$

Finding the Area of a Rectangle

Let's say a rectangle has a base of length $b$ and a height of length $h.$ (Note: you can also say that the rectangle has length $b$ and width $h.$)

Then $A,$ the area, is given by the following equation: $$A=bh$$

Finding the Area of a Parallelogram

Say you're given a parallelogram with a base of length $b$ and a height of length $h.$

Then $A,$ the area, is given by the following equation: $$A=bh$$

Finding the Area of Irregular Polygons

To find the area of an irregular polygon, simply break down the shape into smaller polygons, and find the area of those. Let's look at an example where we find the area of the following polygon from the video:

There are multiple ways of breaking this irregular polygon down into smaller quadrilaterals. Here's one way from the video:

We have broken the irregular polygon into two rectangles: a blue one and a purple one. To find the area of the big irregular polygon, we simply add the areas of these smaller rectangles.

The blue rectangle has a base of $30\text{ft}$ and a height of $20\text{ft}.$ Using our area formula for rectangles, we can find the area of the blue rectangle: $$\begin{align} A&=bh\\ &=30\text{ ft} \times 20\text{ ft}\\ &=600 \text{ ft}^2 \end{align}$$ Notice how the top side of the big irregular polygon is $40\text{ ft}$ and the bottom side is $30\text{ ft}.$ What happened to the remaining $10\text{ ft}?$

The remaining $10\text{ ft}$ is actually the length of the base of the purple rectangle! This rectangle also has a height of $12\text{ ft}.$ Using our area formula for rectangles, we can find the area of the purple rectangle:

$$\begin{align} A&=bh\\ &=10\text{ ft} \times 12\text{ ft}\\ &=120 \text{ ft}^2 \end{align}$$

The area of the big irregular polygon, $A,$ is equal to the sum of the areas of the blue and purple rectangles. Thus:

$$\begin{align} A&=600 \text{ ft}^2+120 \text{ ft}^2\\ &=720 \text{ ft}^2 \end{align}$$

The irregular polygon has an area of $720 \text{ ft}^2.$

The area of any quadrilateral is its length times its height. $$A=lh$$

The definition for the area of a shape can be found here. The formula for finding the area of a quadrilateral is: $$\color{blue}{A}=b\cdot h$$ Where $b=\text{base}$ and $h=\text{height}$. $\\$ This formula can be used to find the area of the rectangle below:

$$\color{blue}{A}=7\text{ units}\times 5\text{ units} \\ \color{blue}{A}=35\text{ units}^2$$

Area of quadrilaterals including irregular shapes

It is important to know the types of quadrilaterals to understand when you can use the following formulas to find the areas of these shapes. It is also important to know that all quadrilaterals are polygons. A polygon is a closed figure formed using only straight lines. A quadrilateral is a polygon with exactly four sides.

This is a video by SmithMathAcademy that reviews the definitions of quadrilateral and polygon. It then discusses how to take the areas of the types of quadrilaterals.

Lets start with the area of a square. In order to find the area of a square, you need to know the measurement of only side. This is because all the sides of a square are the same, so the formula for the area of a square is $A$=$s^2$.

Now that we know how to find the area of a square, we will learn how to find the area of other quadrilaterals that may seem more complicated. These shapes may look more complicated, but it is still pretty easy to find their areas!

The area of a rectangle is $A=l\cdot w$ or $A=b\cdot h$. $l$ stands for length, which is the same as the base, $b$. This is typically the long side of the rectangle, but it does not actually matter. $w$ stands for width, which is the same as $h$. This is typically the shorter side of the rectangle, but again, it does not actually matter. The equation $A=b\cdot h$ can also be used for parallelograms.

Now to find the area of the trapezoid and isosceles trapezoid, we use the formula $A=\frac{1}{2}h\left(B+b\right)$. In this formula, $h$ is the hight just like in a parallelogram. The $B$ is for the bottom base or the longer side of the trapezoid, and the $b$ is for the shorter base or side of the trapezoid that is opposite of the $B$ bigger base.

Now, we get to the areas for the rhombus and kite. The formula for these shapes is $A=\frac{1}{2}d_1\cdot d_2$ where $d_1$ and $d_2$ are the diagonals in the shape as seen in the image below.

Now for the most fun part! How do we find the area of irregular shapes? This is discussed at the end of the video at the beginning of this explanation. I will also give an explanation here. To find the area of irregular shapes, we simply "chop up" the irregular shapes into shapes that we know. We then find the areas of those known shapes and add them together to get the area of our irregular shape! Lets try one!

If that question was confusing for you, I recommend you looking at the last 3 minutes of the video at the top of this explanation. Now we know how to find the area of quadrilaterals and even irregularly shaped polygons!

Area: An Origin Story

So we know that the area of a parallelogram is the product of its base and height: $$A_{\square}=bh$$ But where does this formula come from?

Rectangles

Let's start with thinking about rectangles on a coordinate grid. We can visualize the formula for a rectangle area ($bh$) by looking at how many units are in the length of each side:

In the picture above, the rectangle has a base of $3$ units and a height of $2$ units; by our formula, the area is $6$, which we can confirm by counting how many units are contained within the perimeter of the rectangle.

But what else does $bh$ tell us? We can think of the rectangle above as two rows of three units or three columns of two units. This means that the base and height encode even more information: how many units per row and column, and how many rows and columns total!

Parallelograms

Now, let's take a look at parallelograms. Parallelograms are quadrilaterals with two pairs of parallel sides; this category actually includes all rectangles (and squares as well)!

Parallelograms also have area $bh$ - the same as rectangles. But parallelograms don't look like rectangles most of the time, so what's going on?

In the picture above, we see a parallelogram with base $2$ and height $3$, so we know it has area $2\cdot 3=6$, the same as a rectangle with the same dimensions! Here's an illustration of this relationship:

This is exactly why the $bh$ formula works! We can "chop off" a triangle from the end of the parallelogram and stick it on the other end to create a rectangle with the same base and height!

Trapezoids

Trapezoids are a little weird, but we can take advantage of their weirdness to derive their area. Recall that a trapezoid must have a pair of parallel lines. What else has parallel lines? A parallelogram!

In fact, any trapezoid can be made into a parallelogram by taking a copy of that trapezoid and sticking the two trapezoids together in a way that preserves the original height of that trapezoid.

This means we have a parallelogram with the same height as our trapezoid and a base equal to the sum of the trapezoid's bases. Moreover, the parallelogram is exactly twice as large as our trapezoid. Since we know the area of a parallelogram ($bh$), we can simply plug in the new base and divide this area by two to get the area of our trapezoid!

Thus, the area of a trapezoid is $\frac{h(b_1+b_2)}{2}$.