Difference between revisions of "2012 AMC 10B Problems/Problem 19"
(→Solution) |
(→Solution) |
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| Line 16: | Line 16: | ||
dot(C); | dot(C); | ||
dot(D); | dot(D); | ||
| + | label("A",(0,6),S); | ||
| + | label("B",(0,0),W): | ||
| + | label("C",(30,0),E); | ||
| + | label("D",(30,6),N); | ||
| + | draw(A--B); | ||
| + | draw(B--C); | ||
| + | draw(C--D); | ||
| + | draw(D--A); | ||
</asy> | </asy> | ||
Revision as of 13:50, 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);
label("A",(0,6),S);
label("B",(0,0),W):
label("C",(30,0),E);
label("D",(30,6),N);
draw(A--B);
draw(B--C);
draw(C--D);
draw(D--A);
(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. 
