Difference between revisions of "2010 UNCO Math Contest II Problems/Problem 2"
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== Solution == | == Solution == | ||
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| + | Notice the rectangle is divided into 10 triangles of equal triangles. Therefore, the shaded area is <math>\frac 25</math> of the area of the rectangle, which is <math>\dfrac 25\cdot 75 \cdot 67 = \boxed {2010}.</math> | ||
== See also == | == See also == | ||
| − | {{ | + | {{UNCO Math Contest box|year=2010|n=II|num-b=1|num-a=3}} |
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
Latest revision as of 19:51, 13 October 2016
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
Notice the rectangle is divided into 10 triangles of equal triangles. Therefore, the shaded area is 
of the area of the rectangle, which 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 | ||
