# 2009 UNCO Math Contest II Problems/Problem 9

## Problem

A square is divided into three pieces of equal area by two parallel lines as shown. If the distance between the two parallel lines is $8$ what is the area of the square? $[asy] draw((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); draw((1,0)--(0,2/3),black); draw((1,1/3)--(0,1),black); [/asy]$

## Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it. Let x be the length of a side. Then the square has area $x^2$ and each portion has area $x^2 \times\frac{1}{3}$ If x is the base of one of the triangles, then the height will be $\frac{2x}{3}$. By the pythaogrean theorem, longer side of the parallelogram has length $\sqrt(x^2+(\frac{2x}{3})^2)$ Thus sqrt(13)*x/3*8 = x^2/3. Solving this gives x = 8*sqrt(13). Thus, the area of the square is 64*13 = 832.

## See also

 2009 UNCO Math Contest II (Problems • Answer Key • Resources) Preceded byProblem 8 Followed byProblem 10 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 All UNCO Math Contest Problems and Solutions
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