Difference between revisions of "2012 AMC 10B Problems/Problem 19"
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==Solution== | ==Solution== | ||
+ | <asy> | ||
+ | unitlength(4); | ||
+ | pair B=(0,0) | ||
+ | pair A=(0,6) | ||
+ | pair C=(30,0); | ||
+ | pair D=(30,6); | ||
+ | dot(A); | ||
+ | dot(B): | ||
+ | dot(C); | ||
+ | dot(D); | ||
+ | </asy> | ||
Note that the area of <math>BFDG</math> equals the area of <math>ABCD-\triangle AGB-\triangle DCF</math>. | Note that the area of <math>BFDG</math> equals the area of <math>ABCD-\triangle AGB-\triangle DCF</math>. |
Revision as of 14:46, 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
unitlength(4); pair B=(0,0) pair A=(0,6) pair C=(30,0); pair D=(30,6); dot(A); dot(B): dot(C); dot(D); (Error making remote request. Unknown error_msg)
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.