Difference between revisions of "2010 UNCO Math Contest II Problems/Problem 2"
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== Solution == | == Solution == | ||
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+ | First, we notice that the shaded region can be split into <math>4</math> congruent triangles, all of which have bases of <math>\frac{1}{5}</math> of a diagonal, and a height of <math>\frac{1}{2}</math> of a diagonal. Therefore, the total area that we are looking for, if the diagonal is represented by <math>d</math>, is <math>4(\frac{1}{2}(\frac{d}{5})(\frac{d}{2})) \Rightarrow \frac{d^2}{5}</math>. Since we can find <math>d^2</math> by the Pythagorean Theorem to be <math>67^2 + 75^2</math>, our final answer is <math>\frac{67^2}{5} + 5^3\times3^2 = \boxed{\textbf{2022.8}}</math> | ||
== See also == | == See also == |
Revision as of 21:13, 9 February 2015
Problem
The rectangle has dimensions .
The diagonal
is divided into five
segments of equal length. Find the
total area of the shaded regions.
Solution
First, we notice that the shaded region can be split into congruent triangles, all of which have bases of
of a diagonal, and a height of
of a diagonal. Therefore, the total area that we are looking for, if the diagonal is represented by
, is
. Since we can find
by the Pythagorean Theorem to be
, our final answer is
See also
2010 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
All UNCO Math Contest Problems and Solutions |