Question

A clock in the shape of a regular dodecagon has side measures of 5.5 inches. What is the clock's area to the nearest tenth?

The clock's area is approximately ____ square inches

The answer isn't 169.3 and 338.2

Answer 1

To find the area of a regular dodecagon (a 12-sided polygon), we can use the formula for the area \( A \) of a regular polygon:

\[
A = \frac{1}{4} n s^2 \cot \left( \frac{\pi}{n} \right)
\]

where \( n \) is the number of sides, and \( s \) is the length of a side.

For a dodecagon:
- \( n = 12 \)
- \( s = 5.5 \) inches

Now we can substitute these values into the formula:

\[
A = \frac{1}{4} \times 12 \times (5.5)^2 \cot \left( \frac{\pi}{12} \right)
\]

Calculating \( (5.5)^2 \):

\[
(5.5)^2 = 30.25
\]

Now compute the area:

\[
A = \frac{1}{4} \times 12 \times 30.25 \times \cot \left( \frac{\pi}{12} \right)
\]

First, we can simplify \( \frac{1}{4} \times 12 = 3 \):

\[
A = 3 \times 30.25 \times \cot \left( \frac{\pi}{12} \right)
\]

Next, we calculate \( \cot \left( \frac{\pi}{12} \right) \). Using the identity \( \cot(x) = \frac{\cos(x)}{\sin(x)} \) and knowing \( \frac{\pi}{12} \approx 15^\circ \):

\[
\cot \left( \frac{\pi}{12} \right) = \cot(15^\circ) \approx 3.732
\]

Now, substitute this value into the area calculation:

\[
A \approx 3 \times 30.25 \times 3.732
\]

Calculating \( 3 \times 30.25 \):

\[
3 \times 30.25 = 90.75
\]

Now multiply:

\[
A \approx 90.75 \times 3.732 \approx 338.20
\]

Rounding to the nearest tenth:

\[
A \approx 338.2 \text{ square inches}
\]

Thus, the area of the clock is approximately **338.2 square inches**.