1989 AHSME Problems/Problem 4

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 $DY=7\implies DF=14\implies CF=\boxed{4.0}$ .

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1989 AHSME Problems/Problem 4

Problem

Solution