10.6 Area

Area of Triangles

The formula for the area of a triangle is given as follows.

For example, consider the triangle in Figure 10.103.

Figure 10.103 Triangle 1

The base measures 4 cm and the height measures 5 cm. Using the formula, we can calculate the area:

In Figure 10.104, the triangle has a base equal to 7 cm and a height equal to 3.5 cm. Notice that we can only find the height by dropping a perpendicular to the base. The area is then

Figure 10.104 Triangle 2

Example 10.41

Finding the Area of a Triangle

Find the area of this triangle that has a base of 4 cm and the height is 6 cm (Figure 10.105).

Figure 10.105

Solution

Using the formula, we have

$$A=\frac{1}{2}(4)(6)=\frac{24}{2}=12{\text{cm}}^2\text{.}$$

Your Turn 10.41

1.

Find the area of the triangle with a base equal to 4 cm and the height equal to 4 cm.

Area of Quadrilaterals

A quadrilateral is a four-sided polygon with four vertices. Some quadrilaterals have either one or two sets of parallel sides. The set of quadrilaterals include the square, the rectangle, the parallelogram, the trapezoid, and the rhombus. The most common quadrilaterals are the square and the rectangle.

Square

In Figure 10.106, a $12\text{in}\times 12\text{in}$ grid is represented with twelve $1\text{in}\times 1\text{in}$ squares across each row, and twelve $1\text{in}\times 1\text{in}$ squares down each column. If you count the little squares, the sum equals 144 squares. Of course, you do not have to count little squares to find area—we have a formula. Thus, the formula for the area of a square, where $s=\text{length of a side}$, is $A=s\cdot s$. The area of the square in Figure 10.106 is $A=12\text{in}\times 12\text{in}=144{\text{in}}^2.$

Figure 10.106 Area of a Square

FORMULA

The formula for the area of a square is $A=s\cdot s$ or $A=s^2.$

Rectangle

Similarly, the area for a rectangle is found by multiplying length times width. The rectangle in Figure 10.107 has width equal to 5 in and length equal to 12 in. The area is $A=5(12)=60{\text{in}}^2.$

Figure 10.107 Area of a Rectangle

Many everyday applications require the use of the perimeter and area formulas. Suppose you are remodeling your home and you want to replace all the flooring. You need to know how to calculate the area of the floor to purchase the correct amount of tile, or hardwood, or carpet. If you want to paint the rooms, you need to calculate the area of the walls and the ceiling to know how much paint to buy, and the list goes on. Can you think of other situations where you might need to calculate area?

Parallelogram

The area of a parallelogram can be found using the formula for the area of a triangle. Notice in Figure 10.108, if we cut a diagonal across the parallelogram from one vertex to the opposite vertex, we have two triangles. If we multiply the area of a triangle by 2, we have the area of a parallelogram:

Figure 10.108 Area of a Parallelogram

For example, if we have a parallelogram with the base be equal to 10 inches and the height equal to 5 inches, the area will be $A=(10)(5)=50{\text{in}}^2.$

Example 10.44

Finding the Area of a Parallelogram

In the parallelogram (Figure 10.109), if $FB=10,AD=15,$ find the exact area of the parallelogram.

Figure 10.109

Solution

Using the formula of $A=bh,$ we have

$$A=10(15)=150{\text{cm}}^2.$$

Your Turn 10.44

1.

Find the area of the parallelogram.

Example 10.45

Finding the Area of a Parallelogram Park

The boundaries of a city park form a parallelogram (Figure 10.110). The park takes up one city block, which is contained by two sets of parallel streets. Each street measures 55 yd long. The perpendicular distance between streets is 39 yd. How much sod, sold by the square foot, should the city purchase to cover the entire park and how much will it cost? The sod is sold for $0.50 per square foot, installation is $1.50 per square foot, and the cost of the equipment for the day is $100.

Figure 10.110

Solution

Step 1: As sod is sold by the square foot, the first thing we have to do is translate the measurements of the park from yards to feet. There are 3 ft to a yard, so 55 yd is equal to 165 ft, and 39 yd is equal to 117 ft.

Step 2: The park has the shape of a parallelogram, and the formula for the area is $A=bh$:

$$A=165(117)=\mathrm{19,305}{\text{ft}}^2.$$

Step 3: The city needs to purchase $19,305{\text{ft}}^2$ of sod. The cost will be $0.50 per square foot for the sod and $1.50 per square foot for installation, plus $100 for equipment:

$$\begin{array}{ccc}\hfill \mathrm{19,305}(\text{\$}0.50)+\mathrm{19,305}(\text{\$}1.50)& \hfill =\hfill & \text{\$}\mathrm{9,652.5}+\text{\$}\mathrm{28,957.5}+\text{\$}100.00\hfill \\ \hfill & \hfill =\hfill & \text{\$}\mathrm{38,710.00}\hfill \end{array}$$

Trapezoid

Another quadrilateral is the trapezoid. A trapezoid has one set of parallel sides or bases. The formula for the area of a trapezoid with parallel bases $a$ and $b$ and height $h$ is given here.

For example, find the area of the trapezoid in Figure 10.111 that has base $a$ equal to 8 cm, base $b$ equal to 6 cm, and height equal to 6 cm.

Figure 10.111 Area of a Trapezoid

The area is $A=\frac{1}{2}(6)\left(6+8\right)=42{\text{cm}}^2$.

Example 10.46

Finding the Area of a Trapezoid

$ABCD$ (Figure 10.112) is a regular trapezoid with $\overline{AB}\Vert \overline{CD}.$ Find the exact perimeter of $ABCD$, and then find the area.

Figure 10.112

Solution

The perimeter is the measure of the boundary of the shape, so we just add up the lengths of the sides. We have $P=31+13.5+11+13.5=69\text{in}.$ Then, the area of the trapezoid using the formula is $A=\frac{1}{2}(10)(11+31)=210{\text{in}}^2$.

Your Turn 10.46

1.

Find the area of the trapezoid shown.

Rhombus

The rhombus has two sets of parallel sides. To find the area of a rhombus, there are two formulas we can use. One involves determining the measurement of the diagonals.

FORMULA

The area of a rhombus is found using one of these formulas:

• $A=\frac{d_1{d}_2}{2},$ where $d_1$ and $d_2$ are the diagonals.
• $A=\frac{1}{2}bh,$ where $b$ is the base and $h$ is the height.

For our purposes here, we will use the formula that uses diagonals. For example, if the area of a rhombus is $220{\text{cm}}^2,$ and the measure of $d_2=11,$ find the measure of $d_1.$ To solve this problem, we input the known values into the formula and solve for the unknown. See Figure 10.113.

Figure 10.113 Area of a Rhombus

We have that

$$\begin{array}{ccc}\hfill 220& \hfill =\hfill & \frac{11d_1}{2}\hfill \\ \hfill 220(2)& \hfill =\hfill & 11d_1\hfill \\ \hfill \frac{440}{11}& \hfill =\hfill & d_1=40\hfill \end{array}$$

Example 10.47

Finding the Area of a Rhombus

Find the measurement of the diagonal $d_1$ if the area of the rhombus is $240{\text{cm}}^2,$ and the measure of $d_2=24\text{cm}.$

Solution

Use the formula with the known values:

$$\begin{array}{ccc}\hfill 240& \hfill =\hfill & \frac{d_1(24)}{2}\hfill \\ \hfill 240(2)& \hfill =\hfill & d_1(24)\hfill \\ \hfill \frac{480}{24}& \hfill =\hfill & d_1=20\hfill \end{array}$$

Your Turn 10.47

1.

A rhombus has an area of $40{\text{ in}}^2$, the measure of $d_1=8$. Find the measure of $d_2$.

Example 10.48

Finding the Area of a Rhombus

You notice a child flying a rhombus-shaped kite on the beach. When it falls to the ground, it falls on a beach towel measuring $36\text{in}$ by $72\text{in}.$ You notice that one of the diagonals of the kite is the same length as the $36\text{in}$ width of the towel. The second diagonal appears to be 2 in longer. What is the area of the kite (Figure 10.114)?

Figure 10.114

Solution

Using the formula, we have:

$$\begin{array}{ccc}\hfill A& \hfill =\hfill & \frac{d_1{d}_2}{2}\hfill \\ \hfill & \hfill =\hfill & \frac{36(38)}{2}=684{\text{in}}^2\hfill \end{array}$$

Your Turn 10.48

1.

You have a kite that measures $500\text{i}{\text{n}}^2.$ If one of the diagonals measures 25 in, what is the length of the other diagonal?

Area of Polygons

To find the area of a regular polygon, we need to learn about a few more elements. First, the apothem $a$ of a regular polygon is a line segment that starts at the center and is perpendicular to a side. The radius $r$ of a regular polygon is also a line segment that starts at the center but extends to a vertex. See Figure 10.115.

Figure 10.115 Apothem and Radius of a Polygon

For example, consider the regular hexagon shown in Figure 10.116 with a side length of 4 cm, and the apothem measures $a=2\sqrt{3}.$

Figure 10.116 Area of a Hexagon

We have the perimeter, $p=6(4)=24\text{cm}.$ We have the apothem as $a=2\sqrt{3}.$ Then, the area is:

Example 10.49

Finding the Area of a Regular Octagon

Find the area of a regular octagon with the apothem equal to 18 cm and a side length equal to 13 cm (Figure 10.117).

Figure 10.117

Solution

Using the formula, we have the perimeter $p=8(13)=104\text{cm}.$ Then, the area is $A=\frac{1}{2}(10)(104)=936{\text{cm}}^2.$

Your Turn 10.49

1.

Find the area of the regular pentagon with the apothem equal to 5.5 cm and the side length equal to 7 cm.

Area of Circles

Just as the circumference of a circle includes the number $\pi ,$ so does the formula for the area of a circle. Recall that $\pi$ is a non-terminating, non-repeating decimal number: $\pi =3.14159\dots$. It represents the ratio of the circumference to the diameter, so it is a critical number in the calculation of circumference and area.

For example, to find the of the circle with radius equal to 3 cm, as shown in Figure 10.118, is found using the formula $A=\pi r^2.$

Figure 10.118 Circle with Radius 3

We have

Example 10.53

Applying Area to the Real World

You want to purchase a tinted film, sold by the square foot, for the window in Figure 10.119. (This problem should look familiar as we saw it earlier when calculating circumference.) The bottom part of the window is a rectangle, and the top part is a semicircle. Find the area and calculate the amount of film to purchase.

Figure 10.119

Solution

First, the rectangular portion has $A=5(10)=50{\text{ft}}^2$. For the top part, we have a semicircle with a diameter of 5 ft, so the radius is 2.5 ft. We want one half of the area of a circle with radius 2.5 ft, so the area of the top semicircle part is $A=\frac{1}{2}\pi {(2.5)}^2=9.8{\text{ft}}^2.$ Add the area of the rectangle to the area of the semicircle. Then, the total area to be covered with the window film is $A=50+9.8=59.8{\text{ft}}^2.$

Your Turn 10.53

1.

You decide to install a new front door with a semicircle top as shown in the figure. How much area will the new door occupy?

Area within Area

Suppose you want to install a round hot tub on your backyard patio. How would you calculate the space needed for the hot tub? Or, let’s say that you want to purchase a new dining room table, but you are not sure if you have enough space for it. These are common issues people face every day. So, let’s take a look at how we solve these problems.

Example 10.54

Finding the Area within an Area

The patio in your backyard measures 20 ft by 10 ft (Figure 10.120). On one-half of the patio, you have a 4-foot diameter table with six chairs taking up an area of approximately 36 sq feet. On the other half of the patio, you want to install a hot tub measuring 6 ft in diameter. How much room will the table with six chairs and the hot tub take up? How much area is left over?

Figure 10.120

Solution

The hot tub has a radius of 3 ft. That area is then $A=\pi {(3)}^2=9\pi =28.27{\text{ft}}^2.$ The total square feet taken up with the table and chairs and the hot tub is $36+28.27=64.27{\text{ft}}^2.$ The area left over is equal to the total area of the patio, $200{\text{ft}}^2$ minus the area for the table and chairs and the hot tub. Thus, the area left over is $200-64.27=135.7{\text{ft}}^2.$

Your Turn 10.54

1.

Find the area of the shaded region in the given figure.

Your Turn 10.55

1.

You want to install sod on one-half of your parallelogram-shaped backyard as shown. The patio covers the other half. Sod costs $50 a bag and covers $25\text{f}{\text{t}}^2.$ How much will it cost to buy the sod?

People in Mathematics

Heron of Alexandria

Figure 10.121 Heron of Alexandria (credit: "Heron of Alexandria" from 1688 German translation of Hero's Pneumatics/Wikimedia Commons, Public Domain)

Heron of Alexandria, born around 20 A.D., was an inventor, a scientist, and an engineer. He is credited with the invention of the Aeolipile, one of the first steam engines centuries before the industrial revolution. Heron was the father of the vending machine. He talked about the idea of inserting a coin into a machine for it to perform a specific action in his book, Mechanics and Optics. His contribution to the field of mathematics was enormous. Metrica, a series of three books, was devoted to methods of finding the area and volume of three-dimensional objects like pyramids, cylinders, cones, and prisms. He also discovered and developed the procedures for finding square roots and cubic roots. However, he is probably best known for Heron’s formula, which is used for finding the area of a triangle based on the lengths of its sides. Given a triangle $ABC$ (Figure 10.122),

Figure 10.122

Heron’s formula is $A=\sqrt{s(s-a)(s-b)(s-c)},$ where $s$ is the semi-perimeter calculated as $s=\frac{a+b+c}{2}.$