The length of the altitude $h$ of a regular hexagon pyramid is three times the length of the side $s$ of the base. What is the volume of the pyramid in terms of $h$?
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What have you tried? Please show some effort by showing your steps and being clear on what you're stuck on. – Toby Mak Oct 25 '17 at 10:25
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@TobyMak h=3s then the regular hexagon pyramid affected if the side of the base is 1/3 of the altitude of the pyramid..so we will derive this formula ... – Johnpaul Cerdeña Oct 25 '17 at 10:32
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You can put that in your question. Don't worry about the downvotes - since many questions are just posted here with no effort, people tend to downvote them because it looks like you haven't shown any. – Toby Mak Oct 25 '17 at 10:35
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@TobyMak ok ..in my solution it's take something wrong in my final answer..wait I'll take a picture.. – Johnpaul Cerdeña Oct 25 '17 at 10:48
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Hint:
Remember the formula for a pyramid is $\frac{1}{3} * a * h$. Since we know that $h=3s$, then we have $a * \frac{1}{3} * 3s$, or $as$, where $a$ is the base area, $h$ is the height of the pyramid, and $s$ is the side length of the base.
Now to find the base area, split the hexagon up into $6$ smaller equilateral triangles. You can either use the Pythagorean theorem or trigonometry to calculate the area of one of them. Put this in terms of $h$ and simplify the expression.
Toby Mak
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