Difference between revisions of "2002 AMC 12A Problems/Problem 23"
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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> | ||
+ | |||
+ | '''Solution 3''' | ||
+ | |||
+ | Note that because the perpendicular bisector and angle bisector meet at side <math>AC</math> and <math>CD = BD</math> as triangle <math>BDC</math> is isosceles, so <math>BD = 7</math>. By the angle bisector theorem, we can express <math>AB</math> and <math>BC</math> as <math>9x</math> and <math>7x</math> respectively. We try to find <math>x</math> through Stewart's Theorem. So | ||
+ | |||
+ | <math>16(7^2+9\cdot7) = (7x)^2 \cdot 9 + (9x)^2 \cdot 7</math> | ||
+ | |||
+ | <math>16(49+63) = (49 \cdot 9 + 81 \cdot 7)x^2</math> | ||
+ | |||
+ | <math>16(49+63) = 9(49+63) \cdot x^2</math> | ||
+ | |||
+ | <math>x^2 = \frac{16}{9}</math> | ||
+ | |||
+ | <math>x=\frac{4}{3}</math> | ||
+ | |||
+ | We plug this to find that the sides of <math>\triangle ABD</math> are <math>12,7,9</math>. By Heron's formula, the area is <math>\sqrt{(14)(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}</math>. ~skyscraper | ||
==See Also== | ==See Also== |
Revision as of 18:40, 6 October 2019
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.