Difference between revisions of "2002 AMC 12A Problems/Problem 23"
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In triangle <math>ABC</math> , side <math>AC</math> and the perpendicular bisector of <math>BC</math> meet in point <math>D</math>, and bisects <math><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 bisects <math><ABC</math>. If <math>AD=9</math> and <math>DC=7</math>, what is the area of triangle ABD? | ||
| − | <math> | + | <math>\text{(A)}\ 14 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 14\sqrt5 \qquad \text{(E)}\ 28\sqrt5</math> |
| − | \text{(A) } | ||
| − | \qquad | ||
| − | \text{(B) } | ||
| − | \qquad | ||
| − | \text{(C) } | ||
| − | \qquad | ||
| − | \text{(D) } | ||
| − | \qquad | ||
| − | \text{(E) } | ||
| − | </math> | ||
==Solution== | ==Solution== | ||
Revision as of 01:16, 27 June 2010
Problem
In triangle 
, side 
and the perpendicular bisector of 
meet in point 
, and bisects 
. If 
and 
, what is the area of triangle ABD?
Solution
This problem needs a picture. You can help by adding it.
Looking at the triangle BCD, we see that its perpendicular bisector reaches the vertex, therefore hinting it is isoceles. Let angle C be x. B=2x from given and the previous deducted. <ABD=x, <ADB=2x (outer angle). That means ABD and ACB are similar.
Then by using Heron's Formula on ABD (12,7,9 as sides), we have
See Also
| 2002 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 24 |
Followed by Last Problem |
| 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 | |
