Video Transcript
𝑀𝐴𝐵𝐶𝐷 is a right pyramid whose base 𝐴𝐵𝐶𝐷 is a square of side length six centimeters, and 𝑀𝐴 equals 𝑀𝐵 equals 𝑀𝐶 equals 𝑀𝐷 equals 49 centimeters. Find the pyramid’s volume rounded to the nearest hundredth.
Let’s begin by sketching this pyramid. We’re told that its base 𝐴𝐵𝐶𝐷 is a square with a side length of six centimeters. We’re also told that this is a right pyramid, which means that its vertex will be vertically above the center of its base. And each of these line segments or edges 𝑀𝐴, 𝑀𝐵, 𝑀𝐶, and 𝑀𝐷 are of length 49 centimeters.
We’re asked to calculate the pyramid’s volume, so we need to recall the formula for doing so. It’s one-third multiplied by the base area multiplied by the height. Now the base area is no problem because the base is a square with a side length of six centimeters. But what about the pyramid’s height? Let’s add that height to our diagram. And remember because this is a right pyramid, the vertex is directly above the center of the base, which we’ll label as point 𝑋. We can create a right triangle in the interior of the pyramid using this height, the edge length 𝑀𝐴, and the length of the line segment 𝐴𝑋, which connects one corner of the pyramid’s base to the center of the base.
We can calculate the length of this line segment by first considering a right triangle in the base of the pyramid. 𝐴𝐶, the diagonal of the square base 𝐴𝐵𝐶𝐷, is the hypotenuse of this right triangle. So by applying the Pythagorean theorem, which states that in a right triangle the sum of the squares of the two shorter sides is equal to the square of the hypotenuse, we have 𝐴𝐶 squared is equal to six squared plus six squared. Six squared is 36, so 𝐴𝐶 squared is equal to 72. And 𝐴𝐶 is equal to the square root of 72, which in simplified form is six root two.
As point 𝑋 is in the center of the base, the length of 𝐴𝑋 is the same as the length of 𝑋𝐶. And so we can halve the length of 𝐴𝐶 to give 𝐴𝑋 is equal to six root two over two; it’s three root two centimeters.
Returning to our diagram of the pyramid, we can now apply the Pythagorean theorem a second time, this time in the right triangle composed of the lengths 𝑋𝑀, 𝑋𝐴, and 𝐴𝑀. Doing so gives the equation three root two squared plus ℎ squared is equal to 49 squared. Evaluating the squares and then subtracting 18 from each side, we find that ℎ squared is equal to 2383. ℎ is therefore the square root of 2383, which is 48.815 continuing.
Having calculated the height of this pyramid, we’re now able to work out its volume. We have one-third multiplied by the base area, that’s six squared, multiplied by the height, and we’ll use the exact value of the square root of 2383. Six squared is 36, and one-third of 36 is 12. So we have 12 root 2383. Evaluating this on a calculator gives 585.7917 continuing. And then rounding to the nearest hundredth as required gives 585.79.
The units for this volume are centimeters cubed. And so we found that the volume of the given right pyramid 𝑀𝐴𝐵𝐶𝐷 to the nearest hundredth is 585.79 cubic centimeters.
