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Difference between revisions of "1989 AHSME Problems/Problem 4"

(Created page with "==Problem== In the figure, <math>ABCD</math> is an isosceles trapezoid with side lengths <math>AD=BC=5</math>, <math>AB=4</math>, and <math>DC=10</math>. The point <math>C</math>...")
 
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</asy>
 
</asy>
 
<math>XY=AB=4</math> and <math>DX=YC=3</math>, hence <math>DY=7\implies DF=14\implies CF=\boxed{4.0}</math>.
 
<math>XY=AB=4</math> and <math>DX=YC=3</math>, hence <math>DY=7\implies DF=14\implies CF=\boxed{4.0}</math>.
 +
{{MAA Notice}}

Revision as of 12:47, 5 July 2013

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 $DY=7\implies DF=14\implies CF=\boxed{4.0}$. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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