1957 AHSME Problems/Problem 49
Problem
The parallel sides of a trapezoid are and . The non-parallel sides are and . A line parallel to the bases divides the trapezoid into two trapezoids of equal perimeters. The ratio in which each of the non-parallel sides is divided is:
Solution
Let the points be labeled as in the new diagram above, with and (from the problem). Because and , and . Solving these equations for and , respectively yields and . Let . Thus, because , . Solving this equation for yields . Similarly, , and solving this equation for yields .
Now, we can set the perimeters of and equal to each other to solve for :
See Also
1957 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 48 |
Followed by Problem 50 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.