Question

An architect's sketch of plans for the front of a garage in the shape of pentagon is shown below. What is the approximate perimeter of thefront of the garage?-8-6 -4 -2A. about 36 ftB. about 21 ftC. about 10 ftD. about 77 ft

Answer 1

Using the given pentagon, let's find the perimeter.

From the graph, we can deduce the vertices of the pentagon below:

(0, 9.5), (5.5, 7), (5.5, 0), (-5.5, 0), (-5.5, 7)

Let's find the perimeter.

To find the perimeter, let's first find the length of each side using the distance formula:

`d=√((x_2-x_1)^2+(y_2+y_1)^2)`

Now, let's label the figure:

Thus, we have the following:

• Length of AB:

Where:

(x1, y1) ==> (0, 9.5)

(x2, y2) ==> (5.5, 7)

We have:

`\begin{gathered} AB=√((5.5-0)^2+(7-9.5)^2) \\ \\ AB=√((5.5)^2+(-2.5)^2) \\ \\ AB=√(30.25+6.25)=√(36.50) \\ \\ AB=6.04\text{ ft} \end{gathered}`

The length of AB = 6 ft

Also the length of AE will be 6 ft.

• Length of BC:

Where:

(x1, y1) ==> (5.5, 7)

(x2, y2) ==> (5.5, 0)

Thus, we have:

`\begin{gathered} BC=√((5.5-5.5)^2+(0-7)^2) \\ \\ BC=√(0+(-7)^2) \\ \\ BC=7\text{ ft} \end{gathered}`

The length of BC = 7 ft

The length of DE will also be 7 ft.

• Length of CD:

Where:

(x1, y1) ==> (5.5, 0)

(x2, y2) ==> (-5.5, 0)

Thus, we have:

`\begin{gathered} CD=√((5.5-(-5.5))^2+(0-0)^2) \\ \\ CD=√((5.5+5.5)^2) \\ \\ CD=√(11^2) \\ \\ CD=11\text{ ft} \end{gathered}`

Therefore, we have the following side lengths.

• AB = 6 ft

,

• BC = 7 ft

,

• CD = 11 ft

,

• DE = 7 ft

,

• AE = 6 ft

To find the perimeter, let's sum up the side lengths:

Perimeter = AB + BC + CD + DE + AE

Perimeter = 6 + 7 + 11 + 7 + 6

Perimeter = 37 ft

Therefore, the perimeter of the front garage is about 36 ft.

• ANSWER:

A. about 36 ft.