Difference between revisions of "2014 AMC 10B Problems/Problem 21"
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− | The area of <math>\Delta AED</math> is by Heron's, <math>4\sqrt{9(4)(3)(2)}=24\sqrt{6}</math>. This makes the length of the altitude from <math>D</math> onto <math>\overline{AE}</math> equal to <math>4\sqrt{6}</math>. One may now proceed as in Solution 1 to obtain an answer of <math>\boxed{\textbf{(B) }25}</math>. | + | The area of <math>\Delta AED</math> is by Heron's, <math>4\sqrt{9(4)(3)(2)}=24\sqrt{6}</math>. This makes the length of the altitude from <math>D</math> onto <math>\overline{AE}</math> equal to <math>4\sqrt{6}</math>. One may now proceed as in Solution <math>1</math> to obtain an answer of <math>\boxed{\textbf{(B) }25}</math>. |
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=B|num-b=20|num-a=22}} | {{AMC10 box|year=2014|ab=B|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:56, 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
.
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 |
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