Problem:
Let ABCD be a parallelogram and let AA′,BB′,CC, and DD′ be parallel rays in space on the same side of the plane determined by ABCD. If AA′=10, BB′=8,CC′=18,DD′=22, and M and N are the midpoints of A′C′ and B′D′, respectively, then MN=
Answer Choices:
A. 0
B. 1
C. 2
D. 3
E. 4
Solution:
Let O be the intersection of AC and BD. Then O is the midpoint of AC and BD, so OM and ON are the midlines in trapezoids ACC′A′ and BDD′B′, respectively. Hence OM=(10+18)/2=14 and ON=(8+22)/2=15. Since OM∥AA′,ON∥BB′, and AA′∥BB′, it follows that O,M, and N are collinear. Therefore,
MN=∣OM−ON∣=∣14−15∣=1
Note. In general, if AA′=a,BB′=b,CC′=c, and DD′=d, then MN=∣a−b+c−d∣/2.
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