Track / Stage 3 / 27 of 260 #27 of 1964

Problem 27

AMC 10/12, early questions

Geometry Difficulty 3.1 Find the answer

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 $ABCD$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

Drawing the square and examining the given lengths,

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 $\fbox{810}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.