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) = 56</math>. | + | Since opposite sides of parallelograms are equal, the perimeter is <math>2 * (AD + DE) = |
+ | \boxed{\textbf{(C) }{56}}</math>. | ||
+ | ==Solution (Cheese)== | ||
+ | We can set point <math>F</math> to be on point <math>C</math>, and point <math>D</math> to be on point <math>A</math>. | ||
+ | |||
+ | This makes a degenerate parallelogram with sides of length <math>28</math> and <math>0</math>, so it has a perimeter of <math>28 + 28 = | ||
+ | \boxed{\textbf{(C) }{56}}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/CCjcMVtkVaQ | ||
+ | |||
+ | ~sugar_rush | ||
== See also == | == See also == | ||
Line 41: | Line 52: | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | |||
+ | {{AMC10 box|year=2013|ab=A|num-b=11|num-a=13}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 16:25, 12 August 2023
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 (Cheese)
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.