Difference between revisions of "2002 AMC 12A Problems/Problem 23"
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This problem needs a picture. You can help by adding it. | 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 ( | + | 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 (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 ABD and ACB are similar. |
<math>\frac {16}{AB}=\frac {AB}{9}</math> | <math>\frac {16}{AB}=\frac {AB}{9}</math> | ||
Revision as of 01:17, 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 (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 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 | |
