<s>[INST] Origin Triangle Tetrahedron Volume I have a problem that goes like this:
Triangle $ ABC$, with sides of length $ 5$, $ 6$, and $ 7$, has one vertex on the positive $ x$-axis, one on the positive $ y$-axis, and one on the positive $ z$-axis. Let $ O$ be the origin. What is the volume of tetrahedron $ OABC$?
I really can't wrap my head around this problem. I thought I could find the area of the triangle base by using Heron's Formula, and then multiply that by $1/3$ of the height, but I've got no way of finding the height.
The posted solution for this question was very confusing.
It says:
wlog use $A(a,0,0), B(0,b,0), C(0,0,c)$
(i) wlog set $ 5 = \sqrt{a^2+b^2} \qquad 6 = \sqrt{b^2+c^2} \qquad 7 = \sqrt{c^2+a^2}$
(ii) Square for $ 25 = a^2+b^2 \qquad 36 = b^2+c^2 \qquad 49 = c^2+a^2$
(iii) Adding and dividing by 2 gives $ a^2+b^2+c^2 = 55$
(iv) Subtracting each element of (ii) from (iii) gives $ a^2 = 19 \qquad b^2 = 6 \qquad c^2 = 30$
(v) Multiplying gives $ a^2b^2c^2 = 19 \cdot 6 \cdot 30 = 19 \cdot 5 \cdot 6^2$
(vi) Squareroot and divide by 6 to get $ V = \frac{1}{6} abc = \sqrt{95} \implies \boxed{\text{C}}$
What I don't understand about this solution is why they are using these $a, b$, and $c$ coordinates and taking the square roots of two coordinates added together and setting them equal to the triangle's side lengths.
Perhaps I just don't properly understand how tetrahedrons and $x$-$y$-$z$ coordinates work. I'd really appreciate some help wrapping my head around this problem.
Thanks :)
 [/INST] Maybe this will help you to visualize

$OA^2 + OB^2 = 5^2\\ 
OB^2 + OC^2 = 6^2\\
OA^2 + OC^2 = 7^2$
And even though it looks quadratic, it is really a system of linear equations.
If it feels strange to treat variables with squared terms as linear equations, rewrite it as:
$x + y = 25\\ 
y + z = 36\\
x + z = 49$
And once you have $x, y, z$ finding the volume is $\frac 16 \sqrt {xyz}$
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