PLEASE. PLEASE. PLEASE HELP ME.

What are ALL the dimentions?

wet der herck. im gonna fail my math homework!!!!

V=Bh

So it would be the area of the base times the heighth. I think the answer is 72units^2

NO 72units^3 sorry

@cdromdj: You second answer (the more recent one) seems to be correct, since we are finding volume.

wat. i got 240 and the online calculator got 120.

The "formula" is .......  (1/3) * (Area of the base)  * (Height)

The height is easy.

The area of the base is a little more complicated.

It's given by (√(3)/4)(s²), where s is the side length of the pyramid.

So we have  (1/3) * (√(3)/4)(18²) * (4) = (1/√(3))(324) = (324 / √(3)) ≈ 187.06 cubic units

What  CPhill said

I think CPhill has assumed that the base is an equilateral triangle of side length 18 units.

This is not actually stated in the question but i guess it is a reasonable assumption.

All the angels are 60 degrees

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in triangle ABC 

  $${\mathtt{Area}} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\mathtt{a}}{\mathtt{\,\times\,}}{\mathtt{b}}{\mathtt{\,\times\,}}{\mathtt{sinC}}$$

$${\mathtt{Area}} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\mathtt{18}}{\mathtt{\,\times\,}}{\mathtt{18}}{\mathtt{\,\times\,}}{\mathtt{sin60}}$$

$${\mathtt{Area}} = {\frac{{\mathtt{162}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}}{{\mathtt{2}}}}$$

$${\mathtt{Area}} = {\mathtt{81}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}$$

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$${\mathtt{Volume}} = {\frac{{\mathtt{1}}}{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{Area}}{\mathtt{\,\times\,}}{\mathtt{height}}$$

$${\mathtt{Volume}} = {\frac{{\mathtt{1}}}{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{81}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{4}}$$

$${\mathtt{Volume}} = {\mathtt{108}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}$$     units cubed.   

$$Volume \approx 187unit^3$$