Difference between revisions of "2018 AMC 10A Problems/Problem 9"
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Solution by ktong | Solution by ktong | ||
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+ | ===Solution 4=== | ||
+ | The area of <math>ADE</math> is 16 times the area of the small triangle, as they are similar and their side ratio is <math>4:1</math>. Therefore the area of the trapezoid is <math>40-16=\boxed{24}</math>. | ||
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+ | ===Solution 5=== | ||
+ | Let <math>x</math> be the area of <math>ADE</math>. Note that <math>x</math> is comprised of the <math>7</math> small isosceles triangles and a triangle similar to <math>ADE</math> with side length ratio <math>3:4</math> (so an area ratio of <math>9:16</math>). Thus, we have <cmath>x=7+\dfrac{9}{16}x</cmath> This gives <math>x=16</math>, so the area of <math>DBCE=40-x=\boxed{24}</math>. | ||
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+ | ===Solution 6=== | ||
+ | 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 | ||
== See Also == | == See Also == |
Revision as of 23:19, 10 February 2018
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 ?
Contents
Solutions
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
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 6
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
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 |
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