Vectors & 3D Geometry
Segment connecting midpoints of opposite edges
MJAT_TS6_P2
Grade 12
Question:
In tetrahedron $PQRS$: $PQ=6$, $PR=4$, $PS=3$, $QR=4$, $QS=5$, $RS=2$. $A$=midpoint of $PQ$, $B$=midpoint of $RS$, $|AB|=K$. Then $10K^2=$
Step-by-Step Solution
Key Concept: Bimedian formula: $4|AB|^2=|PR|^2+|PS|^2+|QR|^2+|QS|^2-|PQ|^2-|RS|^2$.
$10K^2=\mathbf{65}$.
Correct Answer: 65