The problem asks us to find the lateral area ($L$) and surface area ($S$) of a regular hexagonal pyramid. The side length of the hexagonal base is 8 inches, and the slant height of the pyramid is 20 inches. We are asked to round to the nearest tenth, if necessary.

1. Problem Description

The problem asks us to find the lateral area ($L$) and surface area ($S$) of a regular hexagonal pyramid. The side length of the hexagonal base is 8 inches, and the slant height of the pyramid is 20 inches. We are asked to round to the nearest tenth, if necessary.

2. Solution Steps

The lateral area of a pyramid is given by:

$L = \frac{1}{2}Pl$, where $P$ is the perimeter of the base and $l$ is the slant height.

The perimeter of the hexagonal base is $P = 6 \times \text{side length} = 6 \times 8 = 48$ inches.

The slant height is given as $l = 20$ inches.

Therefore, the lateral area is $L = \frac{1}{2} \times 48 \times 20 = 24 \times 20 = 480$ square inches.

The surface area of a pyramid is given by:

$S = L + B$, where $L$ is the lateral area and $B$ is the area of the base.

We already found $L = 480$ square inches.

The area of a regular hexagon with side length $s$ is given by:

$B = \frac{3\sqrt{3}}{2}s^2$.

In this case, $s = 8$ inches, so $B = \frac{3\sqrt{3}}{2}(8^2) = \frac{3\sqrt{3}}{2}(64) = 3\sqrt{3}(32) = 96\sqrt{3}$ square inches.

$96\sqrt{3} \approx 96 \times 1.732 = 166.272 \approx 166.3$

So, the surface area is $S = 480 + 96\sqrt{3} \approx 480 + 166.3 = 646.3$ square inches.

3. Final Answer

$L = 480.0$ in$^2$

$S = 646.3$ in$^2$