Finding the volume of tetrahedron $ABCD$ if $AB=AC=AD=5$, $BC=3$, $CD = 4$, and $BD = 5$ I am trying to solve this question with out using calculus:

Calculate the volume of a tetrahedron $ABCD$ where $AB = AC = AD = 5$ and $BC = 3, CD = 4$, and $BD = 5$.

I managed to find the area of the base triangle which came up to be $3/4$ ($\sqrt{91}$), I am stuck on finding the height of the tetrahedron...
 A: The $3-4-5$ triangle $BCD$ is very well-known right triangle with hypotenuse $5$. As in any right triangle the center $O$ of the circumscribed circle is the mid-point of the hypotenuse. Since all three edges $AB,AC,AD$ are equal the point $O$ is the foot of the altitude drawn from $A$ to the $(BCD)$-plane, so that the height $OA$ is equal $\sqrt{5^2-\left(\frac 52\right)^2}$, and the volume is
$$
\frac 13\frac{3\cdot4}2\frac {5\sqrt3}2=5\sqrt3.
$$
A: I have another analytical geometry explanation for the height of the tetrahedron:
Consider that $C(0,0,0), B(3,0,0), D(0,4,0).$
Let $A(x,y,z)$. The constraints are:
$$\begin{cases}AB^2&=&25\\AC^2&=&25\\AD^2&=&25\end{cases} \ \ \iff \ \ \begin{cases}(x-3)^2+y^2+z^2&=&25& \ \ (1) \\
x^2+(y-4)^2+z^2&=&25& \ \ (2) \\
x^2+y^2+z^2&=&25& \ \ (3)
\end{cases}$$
Subtracting (3) from (1) gives $-6x+9=0$, i.e., $x=\frac32$;
Subtracting (3) from (2) gives $-8y+16=0$, i.e., $y=2$.
Plugging these two values of $x$ and $y$ in (3) gives :
$$\frac{9}{4}+4+z^2=25 \ \ \ \implies \ \ \ z=\frac52 \sqrt{3} \ \tag{4}$$
which is the height of the tetrahedron.
Remark: we could have taken the negative value for $z$  in (4): the result would have been a tetrahedron with apex below the base instead of being above it...
