Difference between revisions of "2018 AMC 10A Problems/Problem 9"
m (→See Also) |
(→Solution 2) |
||
Line 26: | Line 26: | ||
===Solution 2=== | ===Solution 2=== | ||
Let the base length of the small triangle be <math>x</math>. Then, there is a triangle <math>ADE</math> encompassing the 7 small triangles and sharing the top angle with a base length of <math>4x</math>. Because the area is proportional to the square of the side, let the base <math>BC</math> be <math>\sqrt{40}x</math>. Then triangle <math>ADE</math> has an area of 16. So the area is <math>40 - 16 = \boxed{24}</math>. | Let the base length of the small triangle be <math>x</math>. Then, there is a triangle <math>ADE</math> encompassing the 7 small triangles and sharing the top angle with a base length of <math>4x</math>. Because the area is proportional to the square of the side, let the base <math>BC</math> be <math>\sqrt{40}x</math>. Then triangle <math>ADE</math> has an area of 16. So the area is <math>40 - 16 = \boxed{24}</math>. | ||
+ | |||
+ | ===Solution 3=== | ||
+ | Notice <math>\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]</math>. | ||
+ | Let the base of the small triangles of area 1 be <math>x</math>, then the base length of <math>\Delta ADE=4x</math>. Notice, <math>\big(\frac{DE}{BC}\big)^2=\frac{1}{40}\implies \frac{x}{BC}=\frac{1}{\sqrt{40}}</math>, then <math>4x=\frac{4BC}{\sqrt{40}}\implies \big[ADE\big]=\big(\frac{4}{\sqrt{40}}\big)^2\cdot \big[ABC\big]=\frac{2}{5}\big[ABC\big]</math> | ||
+ | Thus, <math>\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]=\big[ABC\big]\big(1-\frac{2}{5}\big)=\frac{3}{5}\cdot 40=\boxed{24}</math> | ||
+ | |||
+ | Solution by ktong | ||
+ | |||
== See Also == | == See Also == | ||
{{AMC10 box|year=2018|ab=A|num-b=8|num-a=10}} | {{AMC10 box|year=2018|ab=A|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:44, 9 February 2018
All of the triangles in the diagram below are similar to iscoceles triangle , in which . Each of the 7 smallest triangles has area 1, and has area 40. What is the area of trapezoid ?
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.