Difference between revisions of "2013 AMC 12A Problems/Problem 9"
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It follows that <math>BD = DE</math>. Thus, <math>AD + DE = AD + DB = AB = 28</math>. | It follows that <math>BD = DE</math>. Thus, <math>AD + DE = AD + DB = AB = 28</math>. | ||
− | Since opposite sides of parallelograms are equal, the perimeter is <math>2 * (AD + DE) = </math>\boxed{\textbf{(C) }{56}}$. | + | Since opposite sides of parallelograms are equal, the perimeter is <math>2 * (AD + DE) = |
+ | </math>\boxed{\textbf{(C) }{56}}$. | ||
== See also == | == See also == |
Revision as of 10:56, 1 January 2020
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 \boxed{\textbf{(C) }{56}}$.
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 |
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