Problem:
In trapezoid ABCD with bases AB and CD, we have AB=52,BC=12, CD=39, and DA=5. The area of ABCD is
Answer Choices:
A. 182
B. 195
C. 210
D. 234
E. 260
Solution:
First drop perpendiculars from D and C to AB. Let E and F be the feet of the perpendiculars to AB from D and C, respectively, and let
h=DE=CF,x=AE, and y=FB
Then
25=h2+x2,144=h2+y2, and 13=x+y
So
144=h2+y2=h2+(13−x)2=h2+x2+169−26x=25+169−26x
which gives x=50/26=25/13, and
h=52−(1325)2=51−16925=5169144=1360
Hence
Area (ABCD)=21(39+52)⋅1360=210.
OR
Extend AD and BC to intersect at P. Since △PDC and △PAB are similar, we have
PD+5PD=5239=PC+12PC
So PD=15 and PC=36. Note that 15,36, and 39 are three times 5,12, and 13, respectively, so ∠APB is a right angle. The area of the trapezoid is the difference of the areas of △PAB and △PDC, so
Area(ABCD)=21(20)(48)−21(15)(36)=210.
OR
Draw the line through D parallel to BC, intersecting AB at E. Then BCDE is a parallelogram, so DE=12,EB=39, and AE=52−39=13. Thus DE2+AD2=AE2, and △ADE is a right triangle. Let h be the altitude from D to AE, and note that
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