A city pays a manufacturing company to produce more stop signs for the city due to an increase in traffic incidents at intersections. After doing some research, the company manager learns that a standard stop sign is a regular octagon that is 30 inches wide and 30 inches tall.

The manager must calculate the area of a stop sign so that he can order enough metal and paint to get the job done.

Which of these choices best shows how the manager can compute the area of each sign?

A.
[tex]\[
\begin{array}{l}
A=(B)(H) \\
A \approx(30)(30) \\
A \approx 900 \text { square inches }
\end{array}
\][/tex]

B.
[tex]\[
A=\frac{1}{2}(B)(H) \\
A \approx \frac{1}{2}(15)(30) \\
A \approx 225 \text { square inches }
\][/tex]

C.
[tex]\[
A=\frac{1}{2}(a)(P) \\
A \approx \frac{1}{2}(30)(99.4) \\
A \approx 1491 \text { square inches }
\][/tex]

D.
[tex]\[
\begin{array}{l}
A=\frac{1}{2}(a)(P) \\
A \approx \frac{1}{2}(15)(99.4) \\
A \approx 746 \text { square inches }
\end{array}
\][/tex]



Answer :

To determine the correct way to calculate the area of the stop sign, let's walk through the process step-by-step.

### Step 1: Understand the Shape and Dimensions

Given:
- The stop sign is a regular octagon.
- The width and height of the octagon are both 30 inches.

### Step 2: Calculate the Side Length of the Regular Octagon

The width (or height) of a regular octagon relates to the side length and a geometric constant. The formula relating the width to the side length [tex]\( a \)[/tex] is:

[tex]\[ \text{width} = 2a(1 + \sqrt{2}) \][/tex]

Given the width is 30 inches, we can solve for the side length [tex]\( a \)[/tex]:

[tex]\[ 30 = 2a(1 + \sqrt{2}) \implies a = \frac{30}{2(1 + \sqrt{2})} \][/tex]

Using the calculated result, we get:
[tex]\[ a \approx 6.213 \text{ inches} \][/tex]

### Step 3: Calculate the Perimeter of the Octagon

The perimeter [tex]\( P \)[/tex] of a regular octagon is given by multiplying the side length [tex]\( a \)[/tex] by 8:

[tex]\[ P = 8a \][/tex]

Using the calculated side length:

[tex]\[ P = 8 \times 6.213 \approx 49.706 \text{ inches} \][/tex]

### Step 4: Calculate the Apothem of the Octagon

The apothem [tex]\( a_p \)[/tex] of a regular octagon can be calculated using the side length [tex]\( s \)[/tex]:

[tex]\[ a_p = \frac{s}{2 \tan(\pi/8)} \][/tex]

Using the side length:

[tex]\[ a_p \approx 7.5 \text{ inches} \][/tex]

### Step 5: Calculate the Area of the Octagon

The area [tex]\( A \)[/tex] of a regular octagon is given by:

[tex]\[ A = \frac{1}{2} \times \text{perimeter} \times \text{apothem} \][/tex]

Using the calculated values:

[tex]\[ A = \frac{1}{2} \times 49.706 \times 7.5 \approx 186.396 \text{ square inches} \][/tex]

### Conclusion

Looking at the options provided:

1. [tex]\( A = (B)(H) \Rightarrow A \approx 900 \text{ square inches} \)[/tex]
2. [tex]\( A = \frac{1}{2}(B)(H) \Rightarrow A \approx 225 \text{ square inches} \)[/tex]
3. [tex]\( A = \frac{1}{2}(a)(P) \Rightarrow A \approx 1491 \text{ square inches} \)[/tex]
4. [tex]\( A = \frac{1}{2}(a)(P) \cos A \approx 746 \text{ square inches} \)[/tex]

None of these options perfectly match [tex]\( 186.396 \)[/tex] square inches exactly, which was determined through detailed calculation. However, the method for the closest more elegant approximation involves using [tex]\( A = \frac{1}{2} \times \text{perimeter} \times \text{apothem} \)[/tex], so none of the options provided exactly align with the step-by-step solution done here, but this should be the ideal method to compute the area of the octagon.