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
[hide]Problem
Determine the area of the square , with the given lengths along a zigzag line connecting and .
Solution
We see that, due to the Pythagorean Theorem, 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 |