2014 AMC 10B Problems/Problem 21
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 | |
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All AMC 10 Problems and Solutions |
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