Difference between revisions of "1979 AHSME Problems/Problem 24"
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− | Let <math>AB</math> and <math>CD</math> intersect at <math>E</math>. | + | Let <math>AB</math> and <math>CD</math> intersect at <math>E</math>. We know that <math>\angle AED=90^{\circ}</math> because <math>\angle E = 180 - (A+D)=180-90 = 90^{\circ}</math>. |
− | <math>\boxed{\textbf{E}}</math> | + | Since <math>\sin BCD = \frac{3}{5}</math>, we get <math>\sin ECB=\sin(180-BCD)=\sin BCD = \frac{3}{5}</math>. Thus, <math>EB=3</math> and <math>EC=4</math> from simple sin application. |
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+ | <math>AD</math> is the hypotenuse of right <math>\triangle AED</math>, with leg lengths <math>AB+BE=7</math> and <math>EC+CD=24</math>. Thus, <math>AD=\boxed{\textbf{E} 25}</math> | ||
== See also == | == See also == |
Revision as of 23:22, 21 June 2018
Problem 24
Sides , and
of (simple*) quadrilateral
have lengths
, and
, respectively.
If vertex angles
and
are obtuse and
, then side
has length
- A polygon is called “simple” if it is not self intersecting.
Solution
We know that . Since
and
are obtuse, we have
. It is known that
, so
. We simplify this as follows:
Since , we know that
. Now extend
and
as follows:
Let and
intersect at
. We know that
because
.
Since , we get
. Thus,
and
from simple sin application.
is the hypotenuse of right
, with leg lengths
and
. Thus,
See also
1979 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |
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