Difference between revisions of "2014 AMC 10B Problems/Problem 21"
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− | The two diagonals are <math>\overline{AC}</math> and <math>\overline{BD}</math>. Using the Pythagorean theorem again on <math>\bigtriangleup AFC</math> and <math>\bigtriangleup BED</math>, we can find these lengths to be <math>\sqrt{96+529} = 25</math> and <math>\sqrt{96+961} = \sqrt{1057}</math>. Since <math>\sqrt{96+529}<\sqrt{96+961}</math>, <math>25</math> is the shorter length, so the answer is <math>\boxed{\textbf{(B) }25}</math>. | + | The two diagonals are <math>\overline{AC}</math> and <math>\overline{BD}</math>. Using the Pythagorean theorem again on <math>\bigtriangleup AFC</math> and <math>\bigtriangleup BED</math>, we can find these lengths to be <math>\sqrt{96+529} = 25</math> and <math>\sqrt{96+961} = \sqrt{1057}</math>. Since <math>\sqrt{96+529}<\sqrt{96+961}</math>, <math>25</math> is the shorter length*, so the answer is <math>\boxed{\textbf{(B) }25}</math>. |
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+ | *Or, alternatively, one can notice that the two triangles have the same height but <math>\bigtriangleleup AFC</math> has a shorter base than <math>\bigtriangleup BED</math>. | ||
==Solution 2== | ==Solution 2== |
Revision as of 18:59, 27 December 2019
Contents
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 1
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 and
.
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 . Since , is the shorter length*, so the answer is .
- Or, alternatively, one can notice that the two triangles have the same height but $\bigtriangleleup AFC$ (Error compiling LaTeX. ! Undefined control sequence.) has a shorter base than .
Solution 2
The area of is by Heron's, . This makes the length of the altitude from onto equal to . One may now proceed as in Solution to obtain an answer of .
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.