What is a simple formula for finding the volume of a triangular pyramid with a base side length of 18 and a height of 4?
please hurry!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
+36 V=Bh
So it would be the area of the base times the heighth. I think the answer is 72units^2
+1006 @cdromdj: You second answer (the more recent one) seems to be correct, since we are finding volume.
+128707 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
+118613 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. 
$$Volume \approx 187unit^3$$
