Difference between revisions of "1998 AHSME Problems/Problem 26"
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Revision as of 21:18, 9 February 2008
Problem
In quadrilateral , it is given that
, angles
and
are right angles,
, and
. Then
Solution
Solution 1
Let the extensions of and
be at
. Since
,
and
is a 30-60-90 triangle. Also, $\triangle ABE \similar \triangle CDE$ (Error compiling LaTeX. Unknown error_msg), so
is also a 30-60-90 triangle.
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Thus , and
. By the Pythagorean Theorem on
,
.
Solution 2
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Opposite angles add up to , so
is a cyclic quadrilateral. Also,
, from which it follows that
is a diameter of the circumscribing circle. We can apply the extended version of the Law of Sines on
:
By the Law of Cosines on :
So .
See also
1998 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 27 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |