Difference between revisions of "2010 UNCO Math Contest II Problems/Problem 2"

Latest revision as of 19:51, 13 October 2016

Problem

The rectangle has dimensions $67 \times 75$ . The diagonal $AB$ is divided into five segments of equal length. Find the total area of the shaded regions.

$[asy] pair A,B,C,D; A==(0,0);B=(75,0);C=(75,67);D=(0,67); draw(A--B--C--D--cycle,black); filldraw(A--(D+.2(B-D))--C--(D+.4(B-D))--cycle,blue); filldraw(A--(D+.6(B-D))--C--(D+.8(B-D))--cycle,blue); draw(B--D,black); dot(D+.2(B-D));dot(D+.4(B-D));dot(D+.6(B-D));dot(D+.8(B-D)); MP("A",D,NW);MP("B",B,SE); [/asy]$

Solution

Notice the rectangle is divided into 10 triangles of equal triangles. Therefore, the shaded area is $\frac 25$ of the area of the rectangle, which is $\dfrac 25\cdot 75 \cdot 67 = \boxed {2010}.$

@@ Line 19: / Line 19: @@
 == Solution ==
-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>
+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 ==

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

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Difference between revisions of "2010 UNCO Math Contest II Problems/Problem 2"

Latest revision as of 19:51, 13 October 2016

Problem

Solution

See also