Difference between revisions of "2011 AIME II Problems/Problem 2"
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== Problem 2 == | == Problem 2 == | ||
− | On square ABCD, point E lies on side AD and point F lies on side BC, so that BE=EF=FD=30. Find the area of the square ABCD. | + | On [[square]] <math>ABCD</math>, point <math>E</math> lies on side <math>AD</math> and point <math>F</math> lies on side <math>BC</math>, so that <math>BE=EF=FD=30</math>. Find the area of the square <math>ABCD</math>. |
== Solution == | == Solution == | ||
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you find that the three segments cut the square into three equal horizontal sections. Therefore, (<math>x</math> being the side length), <math>\sqrt{x^2+(x/3)^2}=30</math>, or <math>x^2+(x/3)^2=900</math>. Solving for <math>x</math>, we get <math>x=9\sqrt{10}</math>, and <math>x^2=810.</math> | you find that the three segments cut the square into three equal horizontal sections. Therefore, (<math>x</math> being the side length), <math>\sqrt{x^2+(x/3)^2}=30</math>, or <math>x^2+(x/3)^2=900</math>. Solving for <math>x</math>, we get <math>x=9\sqrt{10}</math>, and <math>x^2=810.</math> | ||
− | Area of the square is <math>\fbox{810 | + | Area of the square is <math>\fbox{810}</math>. |
+ | |||
+ | ==See also== | ||
+ | {{AIME box|year=2011|n=II|num-b=1|num-a=3}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Revision as of 09:32, 23 August 2011
Problem 2
On square , point lies on side and point lies on side , so that . Find the area of the square .
Solution
Drawing the square and examining the given lengths, you find that the three segments cut the square into three equal horizontal sections. Therefore, ( being the side length), , or . Solving for , we get , and
Area of the square is .
See also
2011 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |