Difference between revisions of "2011 AIME II Problems/Problem 2"
| Line 6: | Line 6: | ||
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 (see diagram) | + | Drawing the square and examining the given lengths, you find that the three segments cut the square into three equal horizontal sections (see geoboard diagram [that thing is actually pretty cool]) |
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 | ||
Revision as of 22:49, 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 geoboard diagram [that thing is actually pretty cool])
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>
