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Problem 52 Hard Difficulty

Find the volume of the described solid $ S $.
A pyramid with height $ h $ and base an equilateral triangle with side $ a $ (a tetrahedron).


$V=\frac{\sqrt{3}}{12} a^{2} h$

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Welch'S G.

December 1, 2020

Find the volume V of the described solid S. The base of S is a circular disk with radius 2r. Parallel cross-sections perpendicular to the base are squares.

Video Transcript

Okay, The first thing you have to recognize before we start plucking into the point. So formula is that a is 1/2 ass times asked sign of 60 degrees. No, I don't wanna need your unit circle for this in order to get 1/2 a squared squirt of three over to which can finally be simplified into squirt of three over four. So again, if you know your unit circle If you know what sign of 60 degrees is, then you can drive this value, which we're gonna need later on, because now we're putting into point slope form. Remember, point Slope form is wise and which is our slope? Temps maximize X one plus one. This gives us a over to H times X November. This is half the length of what we need. Half the length of a cross section. We must double it. When we multiply this by two, we cancel off the two on the bottom, which gives us a over age times axe. Pretty straightforward. Now our volume, remember, is from zero h of the original value we found was sward of free over four. And then remember it was us squared, but Now we actually know what our justice. We can simply plug in a over h times acts and then square it. So you understand, we ended up with scored a few over for escort at the beginning and that we're plugging him, but our asses so this value overhears us. We just did that. And slide two. Okay, pull out our constant constant. Anything that isn't X to this whole thing over here is our constant. Now we have X squared. We know it can integrate this using the power, which means increased next one of my one divide by the new exponents. Plug in. Now again, the 12 came because I pulled out the constant once again. You've learned this in a previous chapter that Constance could be pulled out. We know the zero doesn't affect the problem, leaving a squirt of three over 12 a squared times. H is V