2011 UNCO Math Contest II Problems/Problem 5

Problem

Determine the area of the square $ABCD$ , with the given lengths along a zigzag line connecting $B$ and $D$ . $[asy] pair A,B,C,D,E,F,P,R; A=(0,4*sqrt(10)); B=(4*sqrt(10),4*sqrt(10)); C=(4*sqrt(10),0); D=(0,0); E=((9/5)*sqrt(10),(3/5)*sqrt(10)); F=(sqrt(10),3*sqrt(10)); draw(A--B--C--D--cycle,black); draw(D--E--F--B,black); P=unit(D-E)+E; R=unit(F-E)+E; draw(P--(P+R-E)--R,black); P=unit(B-F)+F; R=unit(E-F)+F; draw(P--(P+R-F)--R,black); MP("A",A,NW);MP("B",B,NE);MP("C",C,SE);MP("D",D,SW); MP("6",(D/2+E/2),NW);MP("8",(E/2+F/2),NE);MP("10",(F/2+B/2),S); [/asy]$

Solution

We see that, due to the Pythagorean Theorem, $BD=\sqrt{8^2+(10+6)^2}$ $BD=\sqrt{320}$ $BD=\sqrt{160}(\sqrt{2})$ Therefore, the side length of the square is equal to $\frac{BD}{\sqrt{2}}=\sqrt{160}$ , and the area of the square is $(\sqrt{160})^2=\boxed{160}$ .

Note

If the Pythagorean Theorem is hard to visualize initially, try "sliding" the side lengths to form a right triangle with hypotenuse $BD$ .

~megaboy6679

2011 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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2011 UNCO Math Contest II Problems/Problem 5

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Solution

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