Difference between revisions of "2002 AMC 12A Problems/Problem 23"
(→See Also) |
m (wikify) |
||
Line 1: | Line 1: | ||
==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> | ||
<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> | ||
==Solution== | ==Solution== | ||
+ | {{image}} | ||
− | + | Looking at the triangle <math>BCD</math>, we see that its perpendicular bisector reaches the vertex, therefore hinting it is isoceles. Let <math>x = \angle C</math> be <math>x</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 [[Similar (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>. | |
− | |||
− | <math> | ||
− | <math> | ||
− | |||
− | |||
− | <math>\sqrt{14(2)(7)(5)} | ||
− | |||
==See Also== | ==See Also== | ||
+ | {{AMC12 box|year=2002|ab=A|num-b=22|num-a=24}} | ||
− | + | [[Category:Introductory Geometry Problems]] | |
+ | [[Category:Similar Triangles Problems]] | ||
+ | [[Category:Area Problems]] |
Revision as of 01:11, 20 August 2011
Problem
In triangle , side
and the perpendicular bisector of
meet in point
, and
bisects
. If
and
, what is the area of triangle ABD?
Solution
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
Looking at the triangle , we see that its perpendicular bisector reaches the vertex, therefore hinting it is isoceles. Let
be
, 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
.
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 |