Difference between revisions of "2009 UNCO Math Contest II Problems/Problem 9"

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== Solution ==
 
== Solution ==
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{{solution}}
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Let x be the length of a side. Then the square has area <math>x^2</math> and each portion has area <math>x^2 \times\frac{1}{3}</math>
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If x is the base of one of the triangles, then the height will be <math>\frac{2x}{3}</math>.
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By the pythaogrean theorem, longer side of the parallelogram has length <math>\sqrt(x^2+(\frac{2x}{3})^2)</math>
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Thus sqrt(13)*x/3*8 = x^2/3. Solving this gives x = 8*sqrt(13).
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Thus, the area of the square is 64*13 = 832.
  
 
== See also ==
 
== See also ==
{{UNC Math Contest box|year=2009|n=II|num-b=8|num-a=10}}
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{{UNCO Math Contest box|year=2009|n=II|num-b=8|num-a=10}}
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Latest revision as of 00:17, 30 December 2017

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions
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