Difference between revisions of "2002 AMC 12A Problems/Problem 23"
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Looking at the triangle <math>BCD</math>, we see that its perpendicular bisector reaches the vertex, therefore implying it is isosceles. Let <math>x = \angle C</math>, so that <math>B=2x</math> from given and the previous deducted. Then <math>\angle ABD=x, \angle ADB=2x</math> 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 <math> \triangle ABD</math> and <math>\triangle ACB</math> are [[Similarity (geometry)|similar]], so <math>\frac {16}{AB}=\frac {AB}{9} \Longrightarrow AB=12</math>. | Looking at the triangle <math>BCD</math>, we see that its perpendicular bisector reaches the vertex, therefore implying it is isosceles. Let <math>x = \angle C</math>, so that <math>B=2x</math> from given and the previous deducted. Then <math>\angle ABD=x, \angle ADB=2x</math> 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 <math> \triangle ABD</math> and <math>\triangle ACB</math> are [[Similarity (geometry)|similar]], so <math>\frac {16}{AB}=\frac {AB}{9} \Longrightarrow AB=12</math>. | ||
− | Then by using Heron's Formula on <math>ABD</math> (with sides <math>12,7,9</math>), we have <math>[\triangle ABD]= \sqrt{14(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}</math>. | + | Then by using [[Heron's Formula]] on <math>ABD</math> (with sides <math>12,7,9</math>), we have <math>[\triangle ABD]= \sqrt{14(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}</math>. |
'''Solution 2''' | '''Solution 2''' | ||
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Substituting and simplifying, we get <math>x = 4/3</math> | Substituting and simplifying, we get <math>x = 4/3</math> | ||
− | Thus, <math>AB = 12</math>. We now know all sides of <math> \triangle ABD</math>. Using Heron's Formula on <math>\triangle ABD</math>, <math>\sqrt{(14)(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}</math> | + | Thus, <math>AB = 12</math>. We now know all sides of <math> \triangle ABD</math>. Using [[Heron's Formula]] on <math>\triangle ABD</math>, <math>\sqrt{(14)(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}</math> |
==See Also== | ==See Also== |
Revision as of 21:29, 30 December 2016
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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