7. Let $A B C D$ be a tetrahedron such that edges $A B, A C$, and $A D$ are mutually perpendicular. Let the areas of triangles $A B C, A C D$, and $A D B$ be denoted by $x, y$, and $z$, respectively. In terms of $x, y$, and $z$, find the area of triangle $B C D$.
Solution: $\sqrt{x^{2}+y^{2}+z^{2}}$
Place $A, B, C$, and $D$ at $(0,0,0),(b, 0,0),(0, c, 0)$, and $(0,0, d)$ in Cartesian coordinate space, with $b, c$, and $d$ positive. Then the plane through $B, C$, and $D$ is given by the equation $\frac{x}{b}+\frac{y}{c}+\frac{z}{d}=1$. The distance from the origin to this plane is then
$$
\frac{1}{\sqrt{\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}}}=\frac{b c d}{\sqrt{b^{2} c^{2}+c^{2} d^{2}+d^{2} b^{2}}}=\frac{b c d}{2 \sqrt{x^{2}+y^{2}+z^{2}}} \text {. }
$$

Then if the area of $B C D$ is $K$, the volume of the tetrahedron is
$$
\frac{b c d}{6}=\frac{b c d K}{6 \sqrt{x^{2}+y^{2}+z^{2}}}
$$
implying $K=\sqrt{x^{2}+y^{2}+z^{2}}$.
Alternative Solution: The area of $B C D$ is also half the length of the cross product of the vectors $\overrightarrow{B C}=(0,-c, d)$ and $\overrightarrow{B D}=(-b, 0, d)$. This cross product is $(-c d,-d b,-b c)=-2(y, z, x)$, which has length $2 \sqrt{x^{2}+y^{2}+z^{2}}$. Thus the area of $B C D$ is $\sqrt{x^{2}+y^{2}+z^{2}}$.