# 1989 AHSME Problems/Problem 4

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## Problem

In the figure, $ABCD$ is an isosceles trapezoid with side lengths $AD=BC=5$, $AB=4$, and $DC=10$. The point $C$ is on $\overline{DF}$ and $B$ is the midpoint of hypotenuse $\overline{DE}$ in right triangle $DEF$. Then $CF=$

$[asy] defaultpen(fontsize(10)); pair D=origin, A=(3,4), B=(7,4), C=(10,0), E=(14,8), F=(14,0); draw(B--C--F--E--B--A--D--B^^C--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F); pair point=(5,3); label("A", A, N); label("B", B, N); label("C", C, S); label("D", D, S); label("E", E, dir(point--E)); label("F", F, dir(point--F)); markscalefactor=0.05; draw(rightanglemark(E,F,D), linewidth(0.7));[/asy]$

$\textrm{(A)}\ 3.25\qquad\textrm{(B)}\ 3.5\qquad\textrm{(C)}\ 3.75\qquad\textrm{(D)}\ 4.0\qquad\textrm{(E)}\ 4.25$

## Solution

Drop perpendiculars from $A$ and $B$; then the triangle $DBY$ is similar to $DEF$ but with corresponding sides of half the length. $[asy] defaultpen(fontsize(10)); pair D=origin, A=(3,4), B=(7,4), C=(10,0), E=(14,8), F=(14,0); draw(B--C--F--E--B--A--D--B^^C--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F); pair point=(5,3); label("A", A, N); label("B", B, N); label("C", C, S); label("D", D, S); label("E", E, dir(point--E)); label("F", F, dir(point--F)); markscalefactor=0.05; draw(rightanglemark(E,F,D), linewidth(0.7)); pair X=(3,0), Y=(7,0); draw(A--X^^B--Y,dotted); dot(X^^Y); label("X", X, S); label("Y", Y, S); [/asy]$ $XY=AB=4$ and $DX=YC=3$, hence $DY=7\implies DF=14\implies CF=\boxed{4.0}$.

## See also

 1989 AHSME (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 5 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 All AHSME Problems and Solutions

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