Difference between revisions of "2013 AMC 10A Problems/Problem 12"
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==Solution 2== | ==Solution 2== | ||
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==Solution 3== | ==Solution 3== |
Revision as of 10:09, 30 July 2022
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 1
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 .
Solution 2
We can set , by fakesolving, we get .
Solution 3
Drawing the diagram with a ruler and compass (and scaling back by x4), we can draw approximate parallel lines. This yields about 14, but we need to multiply by 4 to get 56, or .
Video Solution
~savannahsolver
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 |
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