Difference between revisions of "2012 AMC 10B Problems/Problem 19"
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label("15",(G--D),N); | label("15",(G--D),N); | ||
label("6",(A--B),W); | label("6",(A--B),W); | ||
| − | label("2",(B--E), | + | label("2",(B--E),W); |
label("6",(D--C),W); | label("6",(D--C),W); | ||
draw(A--B); | draw(A--B); | ||
Revision as of 14:00, 31 December 2019
Problem
In rectangle 
, 
, 
, and 
is the midpoint of 
. Segment 
is extended 2 units beyond 
to point 
, and 
is the intersection of 
and 
. What is the area of 
?
Solution
![[asy] unitsize(8); pair B=(0,0); pair A=(0,6); pair C=(30,0); pair D=(30,6); pair G=(15,6); pair E=(0,-2); pair F=(15/2,0); dot(A); dot(B); dot(C); dot(D); dot(G); dot(E); dot(F); label("A",(0,6),NW); label("B",(0,0),W); label("C",(30,0),E); label("D",(30,6),NE); label("G",(15,6),N); label("E",(0,-2),SW); label("F",(15/2,0),N); label("15",(A--G),N); label("15",(G--D),N); label("6",(A--B),W); label("2",(B--E),W); label("6",(D--C),W); draw(A--B); draw(B--C); draw(C--D); draw(D--A); draw(E--D); draw(B--E); [/asy]](http://latex.artofproblemsolving.com/0/6/9/069b8c8e3400ee35d656ac4e18eb38f9da1907ab.png)
Note that the area of equals the area of 
.
Since 
. Now, 
, so 
and 
so 
Therefore, 
hence our answer is 
See Also
| 2012 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
