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.
![[asy] pair A,B,C,D; A==(0,0);B=(75,0);C=(75,67);D=(0,67); draw(A--B--C--D--cycle,black); filldraw(A--(D+.2(B-D))--C--(D+.4(B-D))--cycle,blue); filldraw(A--(D+.6(B-D))--C--(D+.8(B-D))--cycle,blue); draw(B--D,black); dot(D+.2(B-D));dot(D+.4(B-D));dot(D+.6(B-D));dot(D+.8(B-D)); MP("A",D,NW);MP("B",B,SE); [/asy]](http://latex.artofproblemsolving.com/c/3/d/c3d2f4704d94eb6ca2088440d227f21bcc331bbf.png)
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 | ||
