Difference between revisions of "2011 AIME II Problems/Problem 2"
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Solution: (Needs better solution, I cannot remember exactly how I got the side length) | Solution: (Needs better solution, I cannot remember exactly how I got the side length) | ||
− | Drawing the square and examining the given lengths, you find that the three segments cut the square into three equal horizontal sections ( | + | Drawing the square and examining the given lengths, you find that the three segments cut the square into three equal horizontal sections (see diagram) |
Therefore, (x being the side length), <math>sqrt(x^2+(x/3)^2)=30</math>, or <math>x^2+(x/3)^2=900</math>. Solving for x, we get that x=9sqrt(10), and <math>x^2</math>=810 | Therefore, (x being the side length), <math>sqrt(x^2+(x/3)^2)=30</math>, or <math>x^2+(x/3)^2=900</math>. Solving for x, we get that x=9sqrt(10), and <math>x^2</math>=810 | ||
Area of the square is 810. | Area of the square is 810. | ||
+ | |||
+ | <geogebra>bb56f64482bd0dabba65a4dcad9d52a05d440f19</geogebra> |
Revision as of 22:48, 30 March 2011
Problem:
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.
Solution: (Needs better solution, I cannot remember exactly how I got the side length)
Drawing the square and examining the given lengths, you find that the three segments cut the square into three equal horizontal sections (see diagram)
Therefore, (x being the side length), , or
. Solving for x, we get that x=9sqrt(10), and
=810
Area of the square is 810.
<geogebra>bb56f64482bd0dabba65a4dcad9d52a05d440f19</geogebra>