Difference between revisions of "2014 AMC 10B Problems/Problem 21"
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label("21",(CC+DD)/2,N); | label("21",(CC+DD)/2,N); | ||
label("$2$",(A+DD)/2,N); | label("$2$",(A+DD)/2,N); | ||
− | label("$ | + | label("$10$",(CC+B)/2,N); |
label("$\sqrt{96}$",(C+CC)/2,W); | label("$\sqrt{96}$",(C+CC)/2,W); | ||
label("$\sqrt{96}$",(D+DD)/2,E); | label("$\sqrt{96}$",(D+DD)/2,E); |
Revision as of 13:04, 21 February 2014
Problem
Trapezoid has parallel sides of length and of length . The other two sides are of lengths and . The angles and are acute. What is the length of the shorter diagonal of ?
Solution
In the diagram, . Denote and . In right triangle , we have from the Pythagorean theorem: . Note that since , we have . Using the Pythagorean theorem in right triangle , we have .
We isolate the term in both equations, getting $\begin{align*}h^2 &= 100-x^2
\\h^2 &= 196-(12-x)^2\end{align}$ (Error compiling LaTeX. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.).
Setting these equal, we have . Now, we can determine that .
The two diagonals are and . Using the Pythagorean theorem again on and , we can find these lengths to be and . Obviously, is the shorter length, and thus the answer is .
See Also
2014 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.