Difference between revisions of "2011 AIME II Problems/Problem 2"
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| − | Solution: | + | 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 (I dont know how to make a diagram so somebody please insert one) | Drawing the square and examining the given lengths, you find that the three segments cut the square into three equal horizontal sections (I dont know how to make a diagram so somebody please insert one) | ||
| − | Therefore, (x being the side length), <math>sqrt(x^2+(x/3)^2)=30</math>, or <math>x^2+(x/3)^2=900</math>. | + | 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. | ||
Revision as of 21:30, 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 (I dont know how to make a diagram so somebody please insert one)
Therefore, (x being the side length), 
, or 
. Solving for x, we get that x=9sqrt(10), and 
=810
Area of the square is 810.
