2014 AMC 10B Problems/Problem 21
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. Unknown error_msg).
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 |
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