Problem:
Trapezoid ABCD has parallel sides AB of length 33 and CD of length 21. The other two sides are of lengths 10 and 14. The angles at A and B are acute. What is the length of the shorter diagonal of ABCD?
Answer Choices:
A. 106
B. 25
C. 810
D. 182
E. 26
Solution:
Assume without loss of generality that DA=10 and BC=14. Let M and N be the feet of the perpendicular segments to AB from D and C, respectively. The four points A,M,N,B appear on AB in that order. Let x=DM=CN,y=AM, and z=NB. Then x2+y2=102=100,x2+z2=142=196, and y+21+z=33. Therefore z=12−y, and it follows that 196−x2=12−100−x2. Squaring and simplifying gives 24100−x2=48, so x2=96 and y=100−96=2. The square of the length of the shorter diagonal, AC, is (y+21)2+x2=232+96=625, so AC=(B)25.
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