The problem asks to find the lateral area ($L$) and surface area ($S$) of a regular triangular pyramid, given the side length of the base ($b = 14$ ft) and the slant height ($l = 12$ ft). We need to round the answers to the nearest tenth, if necessary.

1. Problem Description

The problem asks to find the lateral area ($L$) and surface area ($S$) of a regular triangular pyramid, given the side length of the base ($b = 14$ ft) and the slant height ($l = 12$ ft). We need to round the answers to the nearest tenth, if necessary.

2. Solution Steps

The formula for the lateral area of a regular pyramid is:

$L = \frac{1}{2}Pl$

where $P$ is the perimeter of the base and $l$ is the slant height.

The base is a triangle with side length $b = 14$ ft. The perimeter of the triangular base is:

$P = 3b = 3(14) = 42$ ft.

The slant height is given as $l = 12$ ft.

Substitute these values into the lateral area formula:

$L = \frac{1}{2}(42)(12) = \frac{1}{2}(504) = 252$ square feet.

The formula for the surface area of a regular pyramid is:

$S = L + B$

where $L$ is the lateral area and $B$ is the area of the base.

We already found $L = 252$ square feet. Now we need to find the area of the triangular base $B$. Since the problem stated "assume a base that appears to be a regular polygon is a regular polygon", we assume the base is an equilateral triangle.

The area of an equilateral triangle with side $b$ is:

$B = \frac{\sqrt{3}}{4}b^2$

$B = \frac{\sqrt{3}}{4}(14)^2 = \frac{\sqrt{3}}{4}(196) = 49\sqrt{3} \approx 49(1.732) \approx 84.868 \approx 84.9$ square feet.

Now we can find the surface area:

$S = L + B = 252 + 49\sqrt{3} \approx 252 + 84.868 \approx 336.868 \approx 336.9$ square feet.

3. Final Answer

L ≈ 252.0 $ft^2$

S ≈ 336.9 $ft^2$