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?

Answer 1

To find the area of a regular dodecagon (a polygon with 12 equal sides), we can use the formula for the area 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 one side.

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

Now we substitute these values into the area formula:

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

Calculating the square of the side length:

\[
(5.5)^2 = 30.25
\]

Now plug this into the formula:

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

This simplifies to:

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

Now we need to find \( \cot\left(\frac{\pi}{12}\right) \). The cotangent is the reciprocal of the tangent:

\[
\cot\left(\frac{\pi}{12}\right) = \frac{1}{\tan\left(\frac{\pi}{12}\right)}
\]

Using a calculator, we find:

\[
\tan\left(\frac{\pi}{12}\right) \approx 0.2679
\]

Thus,

\[
\cot\left(\frac{\pi}{12}\right) \approx \frac{1}{0.2679} \approx 3.728
\]

Now substituting this back into the area formula:

\[
A \approx 3 \times 30.25 \times 3.728
\]

Calculating that:

\[
3 \times 30.25 \approx 90.75
\]

\[
A \approx 90.75 \times 3.728 \approx 338.6
\]

Therefore, the area of the dodecagon to the nearest tenth is:

\[
\boxed{338.6} \text{ square inches}
\]