Difference between revisions of "2018 AMC 10A Problems/Problem 9"
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==Solution 7== | ==Solution 7== | ||
− | You can assume for the base of one of the smaller triangles to be <math>\frac{1}{a}</math> and the height to be <math>2a</math>, giving an area of 1. The larger triangle above the 7 smaller ones then has base <math>\frac{3}{a}</math> and height <math>6a</math>, giving it an area of <math>9</math>. Then the area of triangle <math>ADE</math> is <math>16</math> and <math>40-16=\boxed{24}</math>. | + | You can assume for the base of one of the smaller triangles to be <math>\frac{1}{a}</math> and the height to be <math>2a</math>, giving an area of 1. The larger triangle above the 7 smaller ones then has base <math>\frac{3}{a}</math> and height <math>6a</math>, giving it an area of <math>9</math>. Then the area of triangle <math>ADE</math> is <math>16</math> and <math>40-16=\boxed{24}</math>. |
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+ | ==Solution 8== | ||
+ | You can construct another trapezoid directly above the one shown, with it's bottom length as the top length of the original. Its area would then be 9/16 of the original. Repeating this process infinitely gives us the sequence <math>7\cdot\left(1+\left(\frac{9}{16}\right)+\left(\frac{9}{16}\right)^2+\left(\frac{9}{16}\right)^3\dots\right)</math> . Using the infinite geometric series sum formula gives us <math>7\cdot\left(\frac{1}{1-\frac{9}{16}}\right)=7\cdot\frac{16}{7}=16</math>. The triangle's area would thus be <math>40-16=\boxed{24}</math>. | ||
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+ | -ConcaveTriangle | ||
==Video Solution== | ==Video Solution== |
Revision as of 01:08, 22 January 2021
- The following problem is from both the 2018 AMC 12A #8 and 2018 AMC 10A #9, so both problems redirect to this page.
Contents
[hide]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 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.
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 .
Solution 7
You can assume for the base of one of the smaller triangles to be and the height to be , giving an area of 1. The larger triangle above the 7 smaller ones then has base and height , giving it an area of . Then the area of triangle is and .
Solution 8
You can construct another trapezoid directly above the one shown, with it's bottom length as the top length of the original. Its area would then be 9/16 of the original. Repeating this process infinitely gives us the sequence . Using the infinite geometric series sum formula gives us . The triangle's area would thus be .
-ConcaveTriangle
Video Solution
~savannahsolver
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.