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.

Solution 2

unitsize(135);
defaultpen(linewidth(.8pt)+fontsize(10pt));
pair A, B, C, D, E, F;
A=(0, 0); B=(1, 0); C=(1, -1); D=(0,-1);
draw((A--B--C--D--cycle);
draw((1,0)--(0,2/3),black);
draw((1,1/3)--(0,1),black);
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2009 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
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2009 UNCO Math Contest II Problems/Problem 9

Contents

Problem

Solution

Solution 2

See also