Difference between revisions of "2013 AMC 10A Problems/Problem 12"
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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>. | ||
| − | + | The opposite sides of parallelograms are equal(you can prove this fact simply by drawing the diagonal of the parallelogram and proving that the two resulting triangles are congruent by SSS), so the perimeter is <math>2 * (AD + DE) = 56\implies \boxed{\textbf{(C) }}</math>. | |
==See Also== | ==See Also== | ||
Revision as of 14:36, 5 November 2019
Problem
In 
, 
and 
. Points 
and 
are on sides 
, 
, and 
, respectively, such that 
and 
are parallel to and , respectively. What is the perimeter of parallelogram 
?
![[asy] size(180); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); real r=5/7; pair A=(10,sqrt(28^2-100)),B=origin,C=(20,0),D=(A.x*r,A.y*r); pair bottom=(C.x+(D.x-A.x),C.y+(D.y-A.y)); pair E=extension(D,bottom,B,C); pair top=(E.x+D.x,E.y+D.y); pair F=extension(E,top,A,C); draw(A--B--C--cycle^^D--E--F); dot(A^^B^^C^^D^^E^^F); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,W); label("$E$",E,S); label("$F$",F,dir(0)); [/asy]](http://latex.artofproblemsolving.com/e/c/4/ec45f0635e288a5e5da73a19319f4eb7045b5eab.png)
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, 
.
The opposite sides of parallelograms are equal(you can prove this fact simply by drawing the diagonal of the parallelogram and proving that the two resulting triangles are congruent by SSS), so the perimeter is 
.
See Also
| 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 | ||
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
