Question

a rectangular pyramid has a square base with an area of 324 square meters. what is the surface area, in square meters, of one of the triangular faces if the rectangular pyramid has a volume of 4,320 cubic meters?

Explanation:

Step1: Find the side - length of the base

Since the area of the square base $A = s^{2}=324$ square meters, then $s=\sqrt{324}=18$ meters.

Step2: Find the height of the pyramid

The volume formula of a rectangular pyramid is $V=\frac{1}{3}Bh$, where $B$ is the base - area and $h$ is the height. Given $V = 4320$ cubic meters and $B = 324$ square meters. We can solve for $h$ from $4320=\frac{1}{3}\times324\times h$. First, simplify the right - hand side: $\frac{1}{3}\times324\times h = 108h$. Then, $h=\frac{4320}{108}=40$ meters.

Step3: Find the slant height of the triangular face

For a triangular face of the pyramid, consider the right - triangle formed by half of the base side and the height of the pyramid. The base of this right - triangle is $\frac{s}{2}=\frac{18}{2}=9$ meters and the height is $h = 40$ meters. Using the Pythagorean theorem $l=\sqrt{(\frac{s}{2})^{2}+h^{2}}$, where $l$ is the slant height. So $l=\sqrt{9^{2}+40^{2}}=\sqrt{81 + 1600}=\sqrt{1681}=41$ meters.

Step4: Find the area of one triangular face

The area of a triangular face of the pyramid is $A_{triangle}=\frac{1}{2}sl$, where $s$ is the base side of the triangle (equal to the side of the square base) and $l$ is the slant height. Substitute $s = 18$ meters and $l = 41$ meters, we get $A_{triangle}=\frac{1}{2}\times18\times41=369$ square meters.

Step5: Find the surface area of the pyramid

The surface area of a square - based pyramid $SA=B + 4A_{triangle}$. Substitute $B = 324$ square meters and $A_{triangle}=369$ square meters. Then $SA=324+4\times369=324 + 1476=1800$ square meters.

Answer:

1800