Difference between revisions of "2002 AMC 12A Problems/Problem 23"
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− | Let M be the point of the perpendicular bisector on BC. By the perpendicular bisector theorem, <math>BD = DC = 7</math> and <math>BM = | + | Let M be the point of the perpendicular bisector on BC. By the perpendicular bisector theorem, <math>BD = DC = 7</math> and <math>BM = MC</math>. Also, by the angle bisector theorem, <math>\frac {AB}{BC} = \frac{9}{7}</math>. Thus, let <math>AB = 9x</math> and <math>BC = 7x</math>. In addition, <math>BM = 3.5x</math>. |
Thus, <math>\cos\angle CBD = \frac {3.5x}{7} = \frac {x}{2}</math>. Additionally, using the Law of Cosines and the fact that <math>\angle CBD = \angle ABD</math>, <math>81 = 49 + 81x^2 - 2(9x)(7)\cos\angle CBD</math> | Thus, <math>\cos\angle CBD = \frac {3.5x}{7} = \frac {x}{2}</math>. Additionally, using the Law of Cosines and the fact that <math>\angle CBD = \angle ABD</math>, <math>81 = 49 + 81x^2 - 2(9x)(7)\cos\angle CBD</math> |
Revision as of 23:19, 8 August 2015
Problem
In triangle , side
and the perpendicular bisector of
meet in point
, and
bisects
. If
and
, what is the area of triangle ABD?
Solution
Solution 1
Looking at the triangle
, we see that its perpendicular bisector reaches the vertex, therefore implying it is isosceles. Let
, so that
from given and the previous deducted. Then
because any exterior angle of a triangle has a measure that is the sum of the two interior angles that are not adjacent to the exterior angle. That means
and
are similar, so
.
Then by using Heron's Formula on (with sides
), we have
.
Solution 2
Let M be the point of the perpendicular bisector on BC. By the perpendicular bisector theorem, and
. Also, by the angle bisector theorem,
. Thus, let
and
. In addition,
.
Thus, . Additionally, using the Law of Cosines and the fact that
,
Substituting and simplifying, we get
Thus, . We now know all sides of
. Using Heron's Formula on
,
See Also
2002 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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 12 Problems and Solutions |
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