Question

a rectangular pyramid has a square base with 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

The area of the square base $A = s^{2}$, where $A = 324$ square meters. So, $s=\sqrt{324}=18$ 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. Then $4320=\frac{1}{3}\times324\times h$. Solving for $h$:

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Step3: Find the slant height of the triangular face

We can use the Pythagorean theorem in the right - triangle formed by half of the base side and the height of the pyramid to find the slant height $l$. Half of the base side is $\frac{s}{2}=9$ meters and $h = 40$ meters. Then $l=\sqrt{9^{2}+40^{2}}=\sqrt{81 + 1600}=\sqrt{1681}=41$ meters.

Step4: Calculate the surface area

The surface area of a rectangular pyramid with a square base $S=B + 4\times(\frac{1}{2}sl)$, where $B$ is the base area, $s$ is the side - length of the base, and $l$ is the slant height.
$B = 324$ square meters, $s = 18$ meters, and $l = 41$ meters.

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Answer:

1800