Question

The volume of a pyramid that fits exactly inside a cube is 18 cubic feet. What is the volume of the cube?

6 cubic feet

18 cubic feet

54 cubic feet

72 cubic feet

Answer 1

@whpalmer4 can you help with this one

Answer 2

So this is a pyramid with a square base, and its height is exactly the same as the length of a side of the base, is that how you see it?

Answer 3

yes

Answer 4

where do we go from there

Answer 5

@whpalmer4

Answer 6

do you remember the formula for the volume of such a pyramid?

Answer 7

no i dont @whpalmer4

Answer 8

okay, I didn't either, so I looked it up :-)

\[V = \frac{1}{3}Ah\]where \(A\) is the area of the base.

But as we have a square base, with side length \(s\), and our height is equal to the side length of the base, what is our formula going to be?

Answer 9

V=1/3 (64)(8) ? @whpalmer4

Answer 10

oh, we don't know any numbers...

\[A = s^2\] if the side length is \(s\), right?

and \[h = s\] if the height is equal to the side length (it should be, since we are fitting this into a cube where it just fits)

so \[V = \frac{1}{3}Ah = \frac{1}{3}s^2*s = \frac{1}{3}s^3\]right?

Answer 11

im confused, do you mean 18? @whpalmer4

Answer 12

Yeah, that's the next step. I'm just finding the formula for the volume of a rectangular prism that is fit inside a cube, so we could do this problem with any size cube or pyramid we like.

So, do you agree with how I derived that formula?

Answer 13

yes @whpalmer4

Answer 14

Good. that's the important part here.

Now, we know that for this particular pyramid, V = 18 ft^3, so

\[V = 18 \text{ ft}^3 = \frac{1}{3}{s^3}\]If we rearrange that a bit, we have
\[54 \text{ ft}^3 = s^3\]

Now, what is the volume of the cube enclosing our pyramid? Remember, we've already decided it has side length \(s\)...

Answer 15

54? @whpalmer4

Answer 16

No, your answer should be "\(54\text{ ft}^3\)", and said with some certainty :-)

Answer 17

thanks so much @whpalmer4

Answer 18

Notice that we don't ever have to actually figure out the side length or height, which is good, because it's kind of a clumsy number, \(s = 3\sqrt[3]{2} \approx 3.77976\)

We can test our answer, now that I did so:

\[V_{cube} = (3\sqrt[3]{2})^3 = 3^3*2 = 27*2 = 54\checkmark\]\[V_{pyramid} = \frac{1}{3}(3\sqrt[3]{2})^2*3\sqrt[3]{2} = \frac{1}{3}*9*3*2 = 18\checkmark\]

Answer 19

Any questions?