Question

Let $ABCD$ be a tetrahedron with $AB = 41, BC = 36, CA = 7, DA = 18, DB = 27$ and $DC = 13$. Let the mid points of $AB$ and $CD$ are $E$ and $F$ respectively, then the value of $(EF)^2$ is

Answer 1

137

Answer 2

The square of the distance between the midpoints of two opposite edges of a tetrahedron ($E$ on $AB$, $F$ on $CD$) can be calculated using the formula: $(EF)^2 = \frac{1}{4} (AC^2 + AD^2 + BC^2 + BD^2 - AB^2 - CD^2)$ Given the edge lengths: $AB = 41$ $BC = 36$ $CA = 7$ $DA = 18$ $DB = 27$ $DC = 13$

Squaring the lengths: $AC^2 = 7^2 = 49$ $AD^2 = 18^2 = 324$ $BC^2 = 36^2 = 1296$ $BD^2 = 27^2 = 729$ $AB^2 = 41^2 = 1681$ $CD^2 = 13^2 = 169$

Substitute these values into the formula: $(EF)^2 = \frac{1}{4} (49 + 324 + 1296 + 729 - 1681 - 169)$ $(EF)^2 = \frac{1}{4} (373 + 2025 - 1850)$ $(EF)^2 = \frac{1}{4} (2398 - 1850)$ $(EF)^2 = \frac{1}{4} (548)$ $(EF)^2 = 137$