Difference between revisions of "2018 AMC 10A Problems/Problem 9"
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==Solution 5== | ==Solution 5== | ||
You can see that we can create a "stack" of 5 triangles congruent to the 7 small triangles shown here, arranged in a row above those 7, whose total area would be 5. Similarly, we can create another row of 3, and finally 1 more at the top, as follows. We know this cumulative area will be <math>7+5+3+1=16</math>, so to find the area of such trapezoid <math>BCED</math>, we just take <math>40-16=\boxed{24}</math>, like so. ∎ --anna0kear | You can see that we can create a "stack" of 5 triangles congruent to the 7 small triangles shown here, arranged in a row above those 7, whose total area would be 5. Similarly, we can create another row of 3, and finally 1 more at the top, as follows. We know this cumulative area will be <math>7+5+3+1=16</math>, so to find the area of such trapezoid <math>BCED</math>, we just take <math>40-16=\boxed{24}</math>, like so. ∎ --anna0kear | ||
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+ | ==Solution 6== | ||
+ | The combined area of the small triangles is <math>7</math>, and from the fact that each small triangle has an area of <math>1</math>, we can deduce that the larger triangle above has an area of <math>9</math> (as the sides of the triangles are in a proportion of <math>\frac{1}{3}</math>, so will their areas have a proportion that is the square of the proportion of their sides, or <math>\frac {1}{9}</math>). Thus, the combined area of the top triangle and the trapezoid immediately below is <math>7 + 9 = 16</math>. The area of trapezoid <math>BCED</math> is thus the area of triangle <math>ABC-16 =\boxed{24}</math>. --lepetitmoulin | ||
==See Also== | ==See Also== |
Revision as of 07:48, 24 March 2018
Contents
Problem
All of the triangles in the diagram below are similar to isosceles triangle , in which
. Each of the 7 smallest triangles has area 1, and
has area 40. What is the area of trapezoid
?
Solution 1
Let be the area of
. Note that
is comprised of the
small isosceles triangles and a triangle similar to
with side length ratio
(so an area ratio of
). Thus, we have
This gives
, so the area of
.
Solution 2
Let the base length of the small triangle be . Then, there is a triangle
encompassing the 7 small triangles and sharing the top angle with a base length of
. Because the area is proportional to the square of the side, let the base
be
. Then triangle
has an area of 16. So the area is
.
Solution 3
Notice .
Let the base of the small triangles of area 1 be
, then the base length of
. Notice,
, then
Thus,
Solution by ktong
Solution 4
The area of is 16 times the area of the small triangle, as they are similar and their side ratio is
. Therefore the area of the trapezoid is
.
Solution 5
You can see that we can create a "stack" of 5 triangles congruent to the 7 small triangles shown here, arranged in a row above those 7, whose total area would be 5. Similarly, we can create another row of 3, and finally 1 more at the top, as follows. We know this cumulative area will be , so to find the area of such trapezoid
, we just take
, like so. ∎ --anna0kear
Solution 6
The combined area of the small triangles is , and from the fact that each small triangle has an area of
, we can deduce that the larger triangle above has an area of
(as the sides of the triangles are in a proportion of
, so will their areas have a proportion that is the square of the proportion of their sides, or
). Thus, the combined area of the top triangle and the trapezoid immediately below is
. The area of trapezoid
is thus the area of triangle
. --lepetitmoulin
See Also
2018 AMC 10A (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 10 Problems and Solutions |
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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.