Difference between revisions of "1990 AHSME Problems/Problem 20"

(Created page with "== Problem == In the figure <math>ABCD</math> is a quadrilateral with right angles at <math>A</math> and <math>C</math>. Points <math>E</math> and <math>F</math> are on <math>\o...")
 
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== Problem ==
 
== Problem ==
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<asy>
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draw((0,0)--(7,4.2)--(10,0)--(3,-5)--cycle,dot);
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draw((0,0)--(3,0)--(7,0)--(10,0),dot);
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draw((3,-5)--(3,0)--(7,0)--(7,4.2),dot);
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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));
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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));
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draw((3,-1)--(4,-1)--(4,0),black+linewidth(.5));
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draw((6,0)--(6,1)--(7,1),black+linewidth(.5));
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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);
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</asy>
  
 
In the figure <math>ABCD</math> is a quadrilateral with right angles at <math>A</math> and <math>C</math>. Points <math>E</math> and <math>F</math> are on <math>\overline{AC}</math>, and <math>\overline{DE}</math> and <math>\overline{BF}</math> are perpendicual to <math>\overline{AC}</math>. If <math>AE=3, DE=5, </math> and <math>CE=7</math>, then <math>BF=</math>
 
In the figure <math>ABCD</math> is a quadrilateral with right angles at <math>A</math> and <math>C</math>. Points <math>E</math> and <math>F</math> are on <math>\overline{AC}</math>, and <math>\overline{DE}</math> and <math>\overline{BF}</math> are perpendicual to <math>\overline{AC}</math>. If <math>AE=3, DE=5, </math> and <math>CE=7</math>, then <math>BF=</math>

Revision as of 01:13, 30 September 2014

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

$\fbox{C}$

See also

1990 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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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