We are asked to find the area of a regular hexagon. We are given the apothem, which is the perpendicular distance from the center to one of the sides, as 32 cm. We need to round the answer to the nearest tenth.

1. Problem Description

We are asked to find the area of a regular hexagon. We are given the apothem, which is the perpendicular distance from the center to one of the sides, as 32 cm. We need to round the answer to the nearest tenth.

2. Solution Steps

First, we need to determine the side length of the hexagon. A regular hexagon can be divided into 6 equilateral triangles. The apothem bisects one of these equilateral triangles. Let $s$ be the side length of the hexagon. The apothem, $a$, is related to the side length by the formula $a = \frac{s}{2} \sqrt{3}$.

We are given $a = 32$ cm.

So, we have $32 = \frac{s}{2} \sqrt{3}$.

Multiplying both sides by 2, we get $64 = s \sqrt{3}$.

Dividing both sides by $\sqrt{3}$, we get $s = \frac{64}{\sqrt{3}}$.

Rationalizing the denominator, we have $s = \frac{64 \sqrt{3}}{3}$.

Now, we find the perimeter $P$ of the hexagon. Since it has 6 sides, $P = 6s = 6 \cdot \frac{64 \sqrt{3}}{3} = 2 \cdot 64 \sqrt{3} = 128 \sqrt{3}$.

The area of a regular polygon is given by the formula:

$A = \frac{1}{2} a P$, where $a$ is the apothem and $P$ is the perimeter.

In this case, $A = \frac{1}{2} (32) (128 \sqrt{3}) = 16 \cdot 128 \sqrt{3} = 2048 \sqrt{3}$.

Now, we approximate the value of $2048 \sqrt{3}$:

$A \approx 2048 \times 1.7320508 = 3547.7064$.

Rounding to the nearest tenth, we get $A \approx 3547.7$ cm$^2$.

3. Final Answer

3547.7 cm$^2$