1957 AHSME Problems/Problem 49

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Problem

The parallel sides of a trapezoid are $3$ and $9$. The non-parallel sides are $4$ and $6$. 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:

[asy] defaultpen(linewidth(.8pt)); unitsize(2cm); pair A = origin; pair B = (2.25,0); pair C = (2,1); pair D = (1,1); pair E = waypoint(A--D,0.25); pair F = waypoint(B--C,0.25); draw(A--B--C--D--cycle); draw(E--F); label("6",midpoint(A--D),NW); label("3",midpoint(C--D),N); label("4",midpoint(C--B),NE); label("9",midpoint(A--B),S);[/asy]

$\textbf{(A)}\ 4: 3\qquad\textbf{(B)}\ 3: 2\qquad\textbf{(C)}\ 4: 1\qquad\textbf{(D)}\ 3: 1\qquad\textbf{(E)}\ 6: 1$

Solution

$\boxed{\textbf{(C) }4:1}$.

See Also

1957 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 48
Followed by
Problem 50
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