2013 AMC 12A Problems/Problem 9
Problem
In ,
and
. Points
and
are on sides
,
, and
, respectively, such that
and
are parallel to
and
, respectively. What is the perimeter of parallelogram
?
Solution
Note that because and
are parallel to the sides of
, the internal triangles
and
are similar to
, and are therefore also isosceles triangles.
It follows that . Thus,
.
Since opposite sides of parallelograms are equal, the perimeter is .
Solution
We can set point to be on point
, and point
to be on point
. This makes a degenerate parallelogram with sides of length
and
, so it has a perimeter of
.
Video Solution
~sugar_rush
See also
2013 AMC 12A (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 |
2013 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.