Difference between revisions of "2003 AMC 10B Problems/Problem 20"
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2003|ab=B|num-b=19|num-a=21}} | {{AMC10 box|year=2003|ab=B|num-b=19|num-a=21}} | ||
+ | {{MAA Notice}} |
Revision as of 12:10, 4 July 2013
Problem
In rectangle and
. Points
and
are on
so that
and
. Lines
and
intersect at
. Find the area of
.
Solution
![[asy] unitsize(8mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(5,0), C=(5,3), D=(0,3); pair F=(1,3), G=(3,3); pair E=(5/3,5); draw(A--B--C--D--cycle); draw(A--E); draw(B--E); pair[] ps={A,B,C,D,E,F,G}; dot(ps); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",E,N); label("$F$",F,SE); label("$G$",G,SW); label("$1$",midpoint(D--F),N); label("$2$",midpoint(G--C),N); label("$5$",midpoint(A--B),S); label("$3$",midpoint(A--D),W); [/asy]](http://latex.artofproblemsolving.com/1/a/c/1ac3fc6b18ce26a1ce9b9a1e0e9c2c73bed94a30.png)
because
The ratio of
to
is
since
and
from subtraction. If we let
be the height of
\begin{align*}\frac{2}{5} &= \frac{h-3}{h}\\ 2h &= 5h-15\\ 3h &= 15\\ h &= 5 (Error compiling LaTeX. Unknown error_msg)
The height is so the area of
is
.
See Also
2003 AMC 10B (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 | ||
All AMC 10 Problems and Solutions |
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