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

See Also

2011 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions