Difference between revisions of "2009 AMC 12B Problems/Problem 9"
(→Solution 2) |
(→See also) |
||
Line 42: | Line 42: | ||
== See also == | == See also == | ||
{{AMC12 box|year=2009|ab=B|num-b=8|num-a=10}} | {{AMC12 box|year=2009|ab=B|num-b=8|num-a=10}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | [[Category:Area Problems]] |
Revision as of 23:16, 10 April 2013
Problem
Triangle has vertices
,
, and
, where
is on the line
. What is the area of
?
Solution
Solution 1
Because the line is parallel to
, the area of
is independent of the location of
on that line. Therefore it may be assumed that
is
. In that case the triangle has base
and altitude
, so its area is
.
Solution 2
The base of the triangle is . Its altitude is the distance between the point
and the parallel line
, which is
. Therefore its area is
. The answer is
.
See also
2009 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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 12 Problems and Solutions |