Question
What is the area of the regular pentagon? Round your answer to the nearest whole number. (1 point)
in2
in2
Answers
Answered by
GPT-4o mini
To calculate the area of a regular pentagon, you can use the formula:
\[
\text{Area} = \frac{1}{4} \sqrt{5(5 + 2\sqrt{5})} s^2
\]
where \( s \) is the length of a side of the pentagon.
However, since the length of the side \( s \) is not given in your question, I won’t be able to provide an exact area value. If you provide the length of a side, I can calculate the area for you.
Without that information, if we assume a side length of \( s = 1 \) for demonstration purposes:
\[
\text{Area} \approx \frac{1}{4} \sqrt{5(5 + 2\sqrt{5})} \cdot 1^2 \approx \frac{1}{4} \sqrt{5(5 + 4.472)} \approx \frac{1}{4} \sqrt{5 \times 9.472} \approx \frac{1}{4} \sqrt{47.36} \approx \frac{1}{4} \times 6.88 \approx 1.72
\]
Rounding this to the nearest whole number gives an area of approximately 2 square units for a pentagon with \( s = 1 \).
Please provide the side length if you need a specific calculation.
\[
\text{Area} = \frac{1}{4} \sqrt{5(5 + 2\sqrt{5})} s^2
\]
where \( s \) is the length of a side of the pentagon.
However, since the length of the side \( s \) is not given in your question, I won’t be able to provide an exact area value. If you provide the length of a side, I can calculate the area for you.
Without that information, if we assume a side length of \( s = 1 \) for demonstration purposes:
\[
\text{Area} \approx \frac{1}{4} \sqrt{5(5 + 2\sqrt{5})} \cdot 1^2 \approx \frac{1}{4} \sqrt{5(5 + 4.472)} \approx \frac{1}{4} \sqrt{5 \times 9.472} \approx \frac{1}{4} \sqrt{47.36} \approx \frac{1}{4} \times 6.88 \approx 1.72
\]
Rounding this to the nearest whole number gives an area of approximately 2 square units for a pentagon with \( s = 1 \).
Please provide the side length if you need a specific calculation.