# 2011 AIME II Problems/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.

## Solution

Drawing the square and examining the given lengths, $[asy] size(2inch, 2inch); currentpen = fontsize(8pt); pair A = (0, 0); dot(A); label("A", A, plain.SW); pair B = (3, 0); dot(B); label("B", B, plain.SE); pair C = (3, 3); dot(C); label("C", C, plain.NE); pair D = (0, 3); dot(D); label("D", D, plain.NW); pair E = (0, 1); dot(E); label("E", E, plain.W); pair F = (3, 2); dot(F); label("F", F, plain.E); label("\frac x3", E--A); label("\frac x3", F--C); label("x", A--B); label("x", C--D); label("\frac {2x}3", B--F); label("\frac {2x}3", D--E); label("30", B--E); label("30", F--E); label("30", F--D); draw(B--C--D--F--E--B--A--D); [/asy]$ you find that the three segments cut the square into three equal horizontal sections. Therefore, ($x$ being the side length), $\sqrt{x^2+(x/3)^2}=30$, or $x^2+(x/3)^2=900$. Solving for $x$, we get $x=9\sqrt{10}$, and $x^2=810.$

Area of the square is $\fbox{810.}$