The six edges of tetrahedron $ABCD$ measure $7,13,18,27,36$ and $41$ units. If the length of edge $AB$ is $41$,then find the length of edge $CD$.
I dont really know how to approach this problem.If you can give me only hints,that would be best. Thanks in advance
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$CD = 13$.
First, $AB$ is lying on two faces $ABC$ and $ABD$.
On triangle $ABC$, at least one of the edges $AC$ or $BC$ need to be at least $\frac{41}{2}$ or the triangular inequality will be violated.
Same thing happens to $ABD$, at least one of the edges $AD$ or $BD$ need to be at least $\frac{41}{2}$.
Aside from $AB$, there are only two edges with length $> \frac{41}{2}$. One with length $27$ and the other one $36$. These two edges have to go to different faces.
Relabeling $A \leftrightarrow B$ and $C \leftrightarrow D$ if necessary, we can assume $AC = 27$. By triangular inequality on triangle $ABC$, we have $$BC \ge | AB - AC | = 41 - 27 = 14.$$ So $BC$ can only be $18$.
For the remaining longest edge $36$, it can be $AD$ or $BD$.
If $BD = 36$, then by triangular inequality on triangle $BCD$, we have $$CD \ge | BD - BC | = |36 - 18| = 18$$ This is impossible as the remaining edges $7, 13$ are not long enough. This means $BD \ne 36$ and hence $AD = 36$. By triangular inequality on triangle $ACD$, we have $$CD \ge |AC - AD| = 36-27 = 9.$$ This implies $CD = 13$.
Hint: If $ABC$ is a triangle, then $|AB|+|BC|\geq|AC|$.