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>. |
+ | |||
+ | *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 19:59, 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. Unknown error_msg) 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.