Difference between revisions of "2014 AMC 10B Problems/Problem 21"
Football017 (talk | contribs) m (reverted it back... I don't know how to do it) |
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label("$C$",C,SE); | label("$C$",C,SE); | ||
label("$D$",D,SW); | label("$D$",D,SW); | ||
+ | label("$E$",DD,N); | ||
+ | label("$F$",CC,N); | ||
draw(C--CC); draw(D--DD); | draw(C--CC); draw(D--DD); | ||
</asy> | </asy> | ||
Line 55: | Line 57: | ||
label("$C$",C,SE); | label("$C$",C,SE); | ||
label("$D$",D,SW); | label("$D$",D,SW); | ||
+ | label("$D$",D,SW); | ||
+ | label("$E$",DD,SE); | ||
+ | label("$F$",CC,SW); | ||
draw(C--CC); draw(D--DD); | draw(C--CC); draw(D--DD); | ||
label("21",(CC+DD)/2,N); | label("21",(CC+DD)/2,N); |
Revision as of 18:32, 26 December 2017
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
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
.
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.