AIME 2011 II · 第 2 题

AIME 2011 II — Problem 2

专题: Contest Math
难度: L4
来源: AIME

题目详情

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 $ABCD$ .

解析

Solution

Drawing the square and examining the given lengths,

$AIME diagram$

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 $x^2+(x/3)^2=900$ . Multiplying both sides by $9$ and simplifying, we find that $10x^2=8100$ . Dividing by ten gives $x^2=810.$

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