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

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== Solution 2 ==
 
== Solution 2 ==
  
<a href="https://lh3.googleusercontent.com/drive-viewer/AK7aPaC4b8-zSQplLWs8EgZIMHADt5u3opYEWoo7U_YureTD7QeP3fvkN02D4mE-dpPdDZCySDANeWvi4NQtIMfsolf5JKW1qw=s2560?source=screenshot.guru"> <img src="https://lh3.googleusercontent.com/drive-viewer/AK7aPaC4b8-zSQplLWs8EgZIMHADt5u3opYEWoo7U_YureTD7QeP3fvkN02D4mE-dpPdDZCySDANeWvi4NQtIMfsolf5JKW1qw=s2560" /> </a>
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<a href="https://lh3.googleusercontent.com/drive-viewer/AK7aPaC4b8-zSQplLWs8EgZIMHADt5u3opYEWoo7U_YureTD7QeP3fvkN02D4mE-dpPdDZCySDANeWvi4NQtIMfsolf5JKW1qw=s2560?source=screenshot.guru"></a>
  
 
== See also ==
 
== See also ==

Revision as of 06:44, 28 September 2023

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?


fillText(A, 0,0) 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);

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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

<a href="https://lh3.googleusercontent.com/drive-viewer/AK7aPaC4b8-zSQplLWs8EgZIMHADt5u3opYEWoo7U_YureTD7QeP3fvkN02D4mE-dpPdDZCySDANeWvi4NQtIMfsolf5JKW1qw=s2560?source=screenshot.guru"></a>

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