1990 AHSME Problems/Problem 20

Problem

$[asy] draw((0,0)--(7,4.2)--(10,0)--(3,-5)--cycle,dot); draw((0,0)--(3,0)--(7,0)--(10,0),dot); draw((3,-5)--(3,0)--(7,0)--(7,4.2),dot); draw((3/sqrt(34),-5/sqrt(34))--(3/sqrt(34)+1/sqrt(1.36),-5/sqrt(34)+.6/sqrt(1.36))--(1/sqrt(1.36),.6/sqrt(1.36)),black+linewidth(.5)); draw((10-7/sqrt(74),0-5/sqrt(74))--(10-7/sqrt(74)-5/sqrt(74),0-5/sqrt(74)+7/sqrt(74))--(10-5/sqrt(74),7/sqrt(74)),black+linewidth(.5)); draw((3,-1)--(4,-1)--(4,0),black+linewidth(.5)); draw((6,0)--(6,1)--(7,1),black+linewidth(.5)); MP("A",(0,0),W);MP("B",(7,4.2),N);MP("C",(10,0),E);MP("D",(3,-5),S);MP("E",(3,0),N);MP("F",(7,0),S); [/asy]$

In the figure $ABCD$ is a quadrilateral with right angles at $A$ and $C$ . Points $E$ and $F$ are on $\overline{AC}$ , and $\overline{DE}$ and $\overline{BF}$ are perpendicual to $\overline{AC}$ . If $AE=3, DE=5,$ and $CE=7$ , then $BF=$

$\text{(A) } 3.6\quad \text{(B) } 4\quad \text{(C) } 4.2\quad \text{(D) } 4.5\quad \text{(E) } 5$

Solution

$\fbox{C}$

1990 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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1990 AHSME Problems/Problem 20

Problem

Solution

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