2018 AMC 10A Problems/Problem 9
- The following problem is from both the 2018 AMC 10A #9 and 2018 AMC 12A #8, so both problems redirect to this page.
Contents
Problem
All of the triangles in the diagram below are similar to isosceles triangle , in which . Each of the smallest triangles has area and has area . 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 . The ratio of the area of triangle to triangle is . The problem says the area of triangle is , so the area of triangle is . So the area of trapezoid 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
Solution 9 (weird)
Note that the area of an isosceles triangle is equivalent to the square of its height. Using this information, the height of the smallest isosceles triangle is , and thus its base is
Let be the height of the top triangle. We can set up a height-to-base similarity ratio, using the top triangle and . The top triangle has a base of , and The height of is , therefore our ratio is , which yields as our answer.
To find the area of the trapezoid, we can take the area of and subtract the area of whose base is and height . It follows that the area of , and subtracting this from gives us .
-Benedict T (countmath1)
Video Solution (HOW TO THINK CREATIVELY!)
~Education, the Study of Everything
Video Solutions
~Whiz
~IceMatrix
~savannahsolver
https://youtu.be/4_x1sgcQCp4?t=2959
~pi_is_3.14
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.