2010 UNCO Math Contest II Problems/Problem 2

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

First, we notice that the shaded region can be split into $4$ congruent triangles, all of which have bases of $\frac{1}{5}$ of a diagonal, and a height of $\frac{1}{2}$ of a diagonal. Therefore, the total area that we are looking for, if the diagonal is represented by $d$ , is $4(\frac{1}{2}(\frac{d}{5})(\frac{d}{2})) \Rightarrow \frac{d^2}{5}$ . Since we can find $d^2$ by the Pythagorean Theorem to be $67^2 + 75^2$ , our final answer is $\frac{67^2}{5} + 5^3\times3^2 = \boxed{\textbf{2022.8}}$

2010 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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2010 UNCO Math Contest II Problems/Problem 2

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