Difference between revisions of "2009 UNCO Math Contest II Problems/Problem 9"
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== Solution == | == Solution == | ||
+ | {{solution}} | ||
+ | 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> | ||
+ | If x is the base of one of the triangles, then the height will be <math>\frac{2x}{3}</math>. | ||
+ | By the pythaogrean theorem, longer side of the parallelogram has length <math>\sqrt(x^2+(\frac{2x}{3})^2)</math> | ||
+ | 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 == | == See also == | ||
− | {{ | + | {{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 what is the area of the square?
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 and each portion has area If x is the base of one of the triangles, then the height will be . By the pythaogrean theorem, longer side of the parallelogram has length 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 by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
All UNCO Math Contest Problems and Solutions |