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 : \begin{align*} 3+(x-3)+x+(\frac{2x}3-2) &= x+[6-(x-3)]+9+[4-(\frac{2x}3-2)] \\ \frac{8x}3-2 &= 9+9+4-\frac{2x}3+2 \\ \frac{10x}3 &= 26 \\ \frac{5x}3 &= 13 \\ 5x &= 39 \\ x &= \frac{39}5 \end{align*} To find the ratio , we substitute into this expression to find our answer: \begin{align*} \frac{x-3}{6-(x-3)} &= \frac{x-3}{9-x} \\ &= \frac{\tfrac{39-15}5}{\tfrac{45-39}5} \\ &= \frac{24}6 \\ &= \frac 4 1 \end{align*} Thus, our answer is .
See Also
1957 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 48 |
Followed by Problem 50 | |
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