# Difference between revisions of "2011 AIME II Problems/Problem 2"

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− | Problem | + | == 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. | 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); | |

− | Therefore, (x being the side length), <math>sqrt | + | 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, (<math>x</math> being the side length), <math>\sqrt{x^2+(x/3)^2}=30</math>, or <math>x^2+(x/3)^2=900</math>. Solving for <math>x</math>, we get <math>x=9\sqrt{10}</math>, and <math>x^2=810.</math> | ||

− | < | + | Area of the square is <math>\fbox{810.}</math> |

## Revision as of 00:05, 3 April 2011

## 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, you find that the three segments cut the square into three equal horizontal sections. Therefore, ( being the side length), , or . Solving for , we get , and

Area of the square is