Difference between revisions of "2002 AMC 12A Problems/Problem 23"
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==Problem== | ==Problem== | ||
− | In triangle <math>ABC</math> , side <math>AC</math> and the [[perpendicular bisector]] of <math>BC</math> meet in point <math>D</math>, and <math>BD</math> bisects <math>\angle ABC</math>. If <math>AD=9</math> and <math>DC=7</math>, what is the area of triangle ABD? | + | In triangle <math>ABC</math>, side <math>AC</math> and the [[perpendicular bisector]] of <math>BC</math> meet in point <math>D</math>, and <math>BD</math> bisects <math>\angle ABC</math>. If <math>AD=9</math> and <math>DC=7</math>, what is the area of triangle ABD? |
<math>\text{(A)}\ 14 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 14\sqrt5 \qquad \text{(E)}\ 28\sqrt5</math> | <math>\text{(A)}\ 14 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 14\sqrt5 \qquad \text{(E)}\ 28\sqrt5</math> |
Revision as of 08:44, 3 August 2020
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
,
Solution 3
Note that because the perpendicular bisector and angle bisector meet at side and
as triangle
is isosceles, so
. By the angle bisector theorem, we can express
and
as
and
respectively. We try to find
through Stewart's Theorem. So
We plug this to find that the sides of are
. By Heron's formula, the area is
. ~skyscraper
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.