2012 AMC 10B Problems/Problem 19: Difference between revisions
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==Solution== | ==Solution== | ||
<asy> | <asy> | ||
unitsize(4); | |||
pair B=(0,0); | pair B=(0,0); | ||
pair A=(0,6); | pair A=(0,6); | ||
Revision as of 13:47, 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
unitsize(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 | ||
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