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The question is:

The surface area of a tetrahedron is $\sqrt{3}$. Find its volume. The tetrahedron is a regular tetrahedron with equilateral triangular faces.

I have tried looking up the centroid of a triangle which I believe is part of the solution to this question.

The volume formula must be derived and not just simply stated aswell.

Thank You.![Here is what I have so far] https://i.stack.imgur.com/Y5IP7.jpg)

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  • $\begingroup$ Do you mean a regular tetrahedron with equilateral triangular faces? $\endgroup$
    – user553944
    Jun 19 '18 at 11:50
  • $\begingroup$ Welcome to stackexchange. You are more likely to get help rather than downvotes and votes to close if you edit the question to show what you tried and where you are stuck. $\endgroup$ Jun 19 '18 at 11:52
  • $\begingroup$ Yes, a regular tetrahedron with equilateral triangular faces. $\endgroup$
    – KSAspProg
    Jun 19 '18 at 11:52
  • $\begingroup$ @KSAspProg Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$
    – user
    Aug 5 '18 at 20:47
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Total surface area: $$4S=\sqrt{3} \Rightarrow 4\cdot \frac{\sqrt{3}a^2}{4}=\sqrt{3} \Rightarrow a=1.$$ The volume: $$V=\frac13 Sh=\frac13\cdot \frac{\sqrt{3}a^2}{4}\cdot \sqrt{\frac23}=\frac{1}{6\sqrt{2}},$$ where: $$h=\sqrt{a^2-R^2}=\sqrt{1-\left(\frac{a^3}{4S}\right)^2}=\sqrt{1-\left(\frac{1}{\sqrt{3}}\right)^2}=\sqrt{\frac23},$$ where $R$ is the radius of the circumscribed circle.

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Surface area of (I am assuming) an equilateral tetrahedron is 4 equilateral triangles.

  1. Find $T(a)$, the area of 1 equilateral triangle with side $a$.
  2. Area of the tetrahedron is 4 of those, can you find $a$ based on the surface area being $\sqrt3$?
  3. Based on $a$, find the volume.
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HINT

Recall that

  • Surface area $S=\frac12Bh=\frac12 a\frac{\sqrt 3}{2}a= \frac{\sqrt 3}{4}a^2$

  • Volume $V=\frac13SH=\frac13\cdot\frac{\sqrt 3}{4}a^2\cdot\frac{\sqrt 6}{3}a=\frac{\sqrt 2}{12}a^3$

To find $H$ let apply Pythagorean theorem to the right triangle with

  • apex $\frac{\sqrt 3}{2}a$
  • base $\frac{\sqrt 3}{6}a$
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  • $\begingroup$ You are very quick to answer questions by newbies that show no work. $\endgroup$ Jun 19 '18 at 11:53
  • $\begingroup$ @EthanBolker indeed I only give hint in order to let some work by the asker, but in a minute a full answer will be given I guess $\endgroup$
    – user
    Jun 19 '18 at 11:54
  • $\begingroup$ Thanks for that but the s.a. and volume formulae must be derived and not just stated and used.... I am a little stuck with the derivation part. $\endgroup$
    – KSAspProg
    Jun 19 '18 at 11:55
  • $\begingroup$ @KSAspProg I've added some detail more, the only difficult part is find H $\endgroup$
    – user
    Jun 19 '18 at 11:57
  • $\begingroup$ @gimusi thanks you - much appreciated. $\endgroup$
    – KSAspProg
    Jun 19 '18 at 12:00
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There are several ingenious ways to find the volume of a regular tetrahedron without even using trigonometry! The simplest is to dissect a cube by joining every second diagonal to leave a large central regular tetrahedron and four corner pyramids created by the vertices of the cube. The edge of the regular tetrahedron is thus $\sqrt2$. The four right pyramids each have a volume of one third of the base times the height, ie, $\frac 1 6$. Thus, the volume of the four pyramids is $\frac 2 3$ leaving $\frac 1 3$ for the volume of the tetrahedron. Now since the volume of any tetrahedron is V=Ka³, (a = edge length), K is found to be $\frac 1 {6\sqrt2}$. Thus, you can now easily answer the question.

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If we consider $e_1,e_2,e_3,e_4$ in $\mathbb{R}^4$ we have that these points (lying on the hyperplane $x+y+z+w=1$) are the vertices of a regular tetrahedron with edge length $a=\sqrt{2}$ and volume $\frac{1}{3}$.
In particular, in terms of the edge length we have $V=\frac{a^3}{6\sqrt{2}}$ and $S=\frac{3\sqrt{3}\,a^2}{2}$. If $S=\sqrt{3}$, then $a=\sqrt{2/3}$ and $V=\frac{1}{9\sqrt{3}}$.

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