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# I'll be eternally grateful if you can answer all these geometry questions!

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Feb 10, 2019

#1
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For the challenge...

The side length of the cube is $$66^{1/3}=\sqrt[3]{66}$$

The volume of a square pyramid is $$a^2\frac{h}{3}$$, where $$a$$ is the base edge and $$h$$ is the height.

Of course, they are all $$\sqrt[3]{66}$$.

So...

$$\sqrt[3]{66}^2\times \frac{\sqrt[3]{66}}{3}=66^{\frac{2}{3}}\times \frac{66^{\frac{1}{3}}}{3}=\frac{66}{3}=22$$ in^3 is the volume.

You are very welcome!

:P

(that was a lot of LaTeX for me: \sqrt[3]{66}^2\times \frac{\sqrt[3]{66}}{3}=66^{\frac{2}{3}}\times \frac{66^{\frac{1}{3}}}{3}=\frac{66}{3}=22 was what i entered)

Feb 10, 2019
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Do you have a diagram for the previous problems...

Feb 10, 2019
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No I don't have diagram, I think it's asking if we know what shape has those requirements.

Guest Feb 11, 2019