Elementary school pyramid competition problem with isosceles and equilateral triangles

Let $$SABC$$ be a pyramid with triangle as a base.
Height of triangle is $$9\sqrt{3}/2$$. And triangle $$SAB$$ is an equilateral triangle with side $$SA=6\sqrt{3}$$($$SA=SB=AB$$) and other sides are equal($$SC=AC=BC$$), what's the volume of the pyramid?
My approach: Sketched it, and have general idea how it looks like, but can't establish any meaningful relationships. I know three triangles with be isosceles and one equilateral.

• What is the base of the triangle when you say "height is $9\sqrt{3}/2$"? Surely with different bases, you'd have different heights.
– Jam
May 11, 2019 at 10:50
• If the equilateral triangle $SAB$ is the base of the triangle, the answer shouldn't be difficult. Use Pythagoras' theorem to find an altitude of $SAB$, find the area of the base with $\frac12bh$, where $b$ and $h$ are respectively a side length and altitude of $SAB$. Finally, find the volume of the pyramid with $\frac12BH$, where $B$ is the area of $SAB$ and $H$ is the height of the pyramid.
– Jam
May 11, 2019 at 11:02

Hint:

The area of a similar square-based pyramid with the same side length $$6\sqrt{3}$$ is equal to $$\frac{1}{3} \cdot 6\sqrt{3} \cdot \frac{9\sqrt{3}}{2}$$.

Now draw an equilateral triangle, and also draw a square with the same base length. How does the area of the two shapes compare? Then you can use this information to find the volume.

• Yes. But I can't seem to find base area, as it is isosceles triangle, only know one side.
– user354021
May 11, 2019 at 8:22
• My answer has changed to include an easier method. For the original method, you can always split the equilateral triangle into two right-angled triangles. By Pythagoras's theorem, $(3\sqrt{3})^2 + h^2 = (6\sqrt{3})^2$. Now solve for $h$. May 11, 2019 at 8:27

Hint:

You can also use the formula for the volume of a pyramid: $$V = \frac{1}{3} \cdot \text{base area} \cdot \text{height}$$. This can be proved by integration, or by using the similarity of the cross-sectional areas.

To find the base, you can split the equilateral triangles into two right-angled triangles. By Pythagoras's theorem, $$(3\sqrt{3})^2 + h^2 = (6\sqrt{3})^2$$. When you find $$h$$, you can find the area of the base, and hence the volume of the pyramid.

Let $$CK$$ be an altitude of $$\Delta ABC$$ and $$CH$$ be and altitude of the pyramid.

Thus, $$CS=BC=\sqrt{\left(\frac{9\sqrt3}{2}\right)^2+\left(3\sqrt3\right)^2}=\frac{3\sqrt{39}}{2}$$ and $$CH=\sqrt{SK^2-KH^2}=\sqrt{\left(\frac{9\sqrt{3}}{2}\right)^2-\left(\frac{1}{3}\cdot\frac{6\sqrt3\cdot\sqrt3}{2}\right)^2}=\frac{\sqrt{207}}{2}$$ and $$V_{SABC}=\frac{1}{3}\cdot\frac{\left(\frac{9\sqrt3}{2}\right)^2\sqrt3}{4}\cdot\frac{\sqrt{207}}{2}=\frac{243\sqrt{69}}{32}.$$ The case, when $$AN$$ is an altitude of the $$\Delta ABC$$ for you.