Difference between revisions of "2011 UNCO Math Contest II Problems/Problem 5"
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== Solution == | == Solution == | ||
| − | <cmath>\ | + | |
| + | We see that, due to the [[Pythagorean Theorem]], <cmath>BD=\sqrt{8^2+(10+6)^2}</cmath> | ||
| + | <cmath>BD=\sqrt{320}</cmath> | ||
| + | <cmath>BD=\sqrt{160}(\sqrt{2})</cmath> | ||
| + | Therefore, the side length of the square is equal to <math>\frac{BD}{\sqrt{2}}=\sqrt{160}</math>, and the area of the square is <math>(\sqrt{160})^2=\boxed{160}</math>. | ||
| + | |||
| + | === Note === | ||
| + | If the Pythagorean Theorem is hard to visualize initially, try "sliding" the side lengths to form a right triangle with hypotenuse <math>BD</math>. | ||
| + | |||
| + | ~megaboy6679 | ||
== See Also == | == See Also == | ||
Latest revision as of 01:16, 11 November 2024
Contents
Problem
Determine the area of the square 
,
with the given lengths along a zigzag
line connecting 
and 
.
![[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]](http://latex.artofproblemsolving.com/3/3/b/33b0c4051e323f1a71acaa592ff599061bda52fc.png)
Solution
We see that, due to the Pythagorean Theorem, ![\[BD=\sqrt{8^2+(10+6)^2}\]](http://latex.artofproblemsolving.com/f/8/f/f8fc7c36ba100be7132f4b06df458799d99af713.png)
![\[BD=\sqrt{320}\]](http://latex.artofproblemsolving.com/e/4/c/e4ca1537eb0742731c110dae28f265840a68ca4f.png)
![\[BD=\sqrt{160}(\sqrt{2})\]](http://latex.artofproblemsolving.com/b/c/a/bca84c1759528d0036f22bf232f17519f2ac5709.png)
Therefore, the side length of the square is equal to 
, and the area of the square is 
.
Note
If the Pythagorean Theorem is hard to visualize initially, try "sliding" the side lengths to form a right triangle with hypotenuse 
.
~megaboy6679
See Also
| 2011 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| 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 | ||
