Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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...")
 
m (→‎Problem)
Line 1: Line 1:
 
== Problem ==
 
== 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 <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 00:13, 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 (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1990_AHSME_Problems/Problem_20&oldid=63906"