CEOMETRY What is the area of a regular heptagon with a side length of 8 in . and an apothem of 8.3 in.? Round to the nearest whole number. $$ 2232 \mathrm{in.}^{2} $$

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To find the area of a regular heptagon (a seven-sided polygon) with a given side length and apothem, you can use the following formula:

$$
\text{Area} = \frac{1}{2} \times \text{Perimeter} \times \text{Apothem}
$$

Let's break down the steps:

1. **Calculate the Perimeter:**
   
   The perimeter $$ P $$ of a regular heptagon with side length $$ s $$ is:
   
   $$
   P = 7 \times s = 7 \times 8\ \text{in} = 56\ \text{in}
   $$
   
2. **Use the Area Formula:**
   
   With the apothem $$ a = 8.3\ \text{in} $$, plug the values into the area formula:
   
   $$
   \text{Area} = \frac{1}{2} \times 56\ \text{in} \times 8.3\ \text{in} = 28 \times 8.3\ \text{in}^2 = 232.4\ \text{in}^2
   $$
   
3. **Round to the Nearest Whole Number:**
   
   $$
   \text{Area} \approx 232\ \text{in}^2
   $$
   
**Conclusion:**

The area of the regular heptagon is approximately **232 square inches**.

It seems there might have been a miscalculation in your initial answer of $$ 2232\ \text{in}^2 $$. The correct area, based on the given measurements, is **232 square inches**.
The area of the regular heptagon is approximately 232 square inches.

CEOMETRY What is the area of a regular heptagon with a side length of 8 in . and an apothem of 8.3 in.? Round to the nearest whole number. \( 2232 \mathrm{in.}^{2} \)

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