Difference between revisions of "2011 AIME II Problems/Problem 2"
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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> | + | 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 x^2=810 |
| + | |||
| + | Area of the square is 810. | ||
Revision as of 21:29, 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:
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 x^2=810
Area of the square is 810.
