Difference between revisions of "2013 AMC 10A Problems/Problem 12"
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==Solution 2== | ==Solution 2== | ||
− | We can set <math>AD=0</math>, | + | We can set <math>AD=0</math>, by fakesolving, we get <math>56\implies \boxed{\textbf{(C)}}</math>. |
==Solution 3== | ==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 <math>\boxed{( | + | 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 <math>\boxed{(C)}</math>. |
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | Similar to Solution 1, we find that the internal triangles <math>\triangle BDE</math> and <math>\triangle EFC</math> are similar to <math>\triangle ABC</math>, and are therefore also isosceles triangles and similar to each other. We know that the perimeter of <math>ADEF</math> can be found by <math>2(l + w)</math>, and <math>DE</math> will be the length and the width will be <math>EF</math>. These can be found by looking at the long sides of <math>\triangle DBE</math> and <math>\triangle FEC</math> respectively. We then find the ratio of long sides to the short side of <math>\triangle ABC</math> to be <math>7/5</math> by <math>28/20</math>, which applies to the other triangles since they are similar. We set up an expression, calling <math>BE</math> as <math>x</math> and <math>EC</math> as <math>20 - x</math>, and substitute it into the perimeter equation knowing our long sides from the ratio: | ||
+ | |||
+ | <math> 2(7x/5 + 7/5(20-x)) </math> | ||
+ | |||
+ | <math> 2(7x/5 + 28 - 7x/5) </math> | ||
+ | |||
+ | <math> 2(28) </math> | ||
+ | |||
+ | <math> 56\implies \boxed{\textbf{(C)}}</math> | ||
+ | |||
+ | ~neeyakkid23 | ||
+ | |||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/8ki_yMyE6no | https://youtu.be/8ki_yMyE6no |
Latest revision as of 20:59, 10 September 2024
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 .
Solution 4
Similar to Solution 1, we find that the internal triangles and
are similar to
, and are therefore also isosceles triangles and similar to each other. We know that the perimeter of
can be found by
, and
will be the length and the width will be
. These can be found by looking at the long sides of
and
respectively. We then find the ratio of long sides to the short side of
to be
by
, which applies to the other triangles since they are similar. We set up an expression, calling
as
and
as
, and substitute it into the perimeter equation knowing our long sides from the ratio:
~neeyakkid23
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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