A triangular pyramid has five sides of length $2$, and another side of length $\sqrt6$. What is the volume of the pyramid?
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Found what? – user2345215 Feb 27 '14 at 21:45
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volume of pyramid... – whyguy Feb 27 '14 at 21:46
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he is new , his writing is wrong, but this is quite interesting question – mahavir Feb 27 '14 at 21:47
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I am new, what is wrong ? – whyguy Feb 27 '14 at 21:48
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Tetrahedron: Heron-type formula for the volume of a tetrahedron – Kaster Feb 27 '14 at 22:01
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Let $M$ be the middle of $SC$. Then $|SM| = |MC| = 1$
Because $|AC| = |AS|$, we have that $AMS$ is a right triangle, so by the Pythagorean theorem we have $|AM| = \sqrt{|AS|^2 - |MS|^2} = \sqrt{2^2 - 1} = \sqrt3$, similarly $|BM| = \sqrt3$.
But then $|AM|^2 + |BM|^2 = 3 + 3 = 6 = |AB|^2$, so by the converse of the Pythagorean theorem the triangle $AMB$ is right.
So the volume can be calculated as (base $ACS$ and height $MB$) $$V = \frac13\cdot\frac12|CS||MA|\cdot|MB| = \frac16\cdot2\cdot\sqrt3\cdot\sqrt3 = \frac13\cdot3 = 1$$
user2345215
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Why V= 1/3(|CS|⋅1/2|MA||MB|) ...
I know, V= 1/3 Sb x h = 1/3 (b x l / 2) x SO
– whyguy Feb 28 '14 at 12:36 -
@user2685833 Well I calculate it as a sum of 2 pyramids, MABS and MABC, each with height $\frac12|CS|$, so the sum of heights is $|CS|$. But you could also calculate it as a pyramid with base ACS and height MB like this: $\frac13(\frac12|MA||CS|\cdot|MB|)$ which gives the same result. That's probably easier. – user2345215 Feb 28 '14 at 12:54