Difference between revisions of "2002 AMC 12A Problems/Problem 23"
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==Solution== | ==Solution== | ||
| − | + | This problems needs a picture. You can help by adding it. | |
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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. | 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. | ||
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<math>\sqrt{14(2)(7)(5)}</math> | <math>\sqrt{14(2)(7)(5)}</math> | ||
<math>14\sqrt5=E</math> | <math>14\sqrt5=E</math> | ||
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==See Also== | ==See Also== | ||
{{AMC12 box|year=2002|ab=A|num-b=24|after=Last<br>Problem}} | {{AMC12 box|year=2002|ab=A|num-b=24|after=Last<br>Problem}} | ||
Revision as of 18:03, 17 January 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 problems 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 | |
