Problem:
Quadrilateral ABCD is a trapezoid, AD=15,AB=50,BC=20, and the altitude is 12. What is the area of the trapezoid?
Answer Choices:
A. 600
B. 650
C. 700
D. 750
E. 800
Solution:
Let E and F be the feet of the perpendiculars from A and B to DC. In right △AED, DE2=152−122=225−144=81, so DE=9. In right △BFC,FC2=202−122=400−144=256, so FC=16.
Right △AED has area 21⋅9⋅12=54, right △BFC has area 21⋅16⋅12=96, and rectangle ABFE has area 50⋅12=600. The trapezoid ABCD has area 54+96+600=750.
OR
Begin as in the first solution and note that DC=DE+EF+FC=9+50+16= 75. Then the area of trapezoid is 21(AB+DC)⋅AE=21(50+75)⋅12=125⋅6=750.
Answer: D.
The problems on this page are the property of the MAA's American Mathematics Competitions
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