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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?  
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<math>A) 14</math>  <math>B) 21</math>  <math>C)28</math>  <math>D)14\sqrt5</math>  <math>E)28\sqrt5</math>
 
<math>A) 14</math>  <math>B) 21</math>  <math>C)28</math>  <math>D)14\sqrt5</math>  <math>E)28\sqrt5</math>
  
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{{Solution}}
 
{{Solution}}
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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.
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<math>\frac {16}{AB}=\frac {AB}{9}</math>
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<math>AB=12</math>
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Then by using Heron's Formula on ABD (12,7,9 as sides), we have
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<math>\sqrt{14(2)(7)(5)}</math>
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<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 18:02, 17 January 2010

Problem

In triangle $ABC$ , side $AC$ and the perpendicular bisector of $BC$ meet in point $D$, and bisects $<ABC$. If $AD=9$ and $DC=7$, what is the area of triangle ABD?

$A) 14$ $B) 21$ $C)28$ $D)14\sqrt5$ $E)28\sqrt5$

Solution

This problem needs a solution. If you have a solution for it, please help us out 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.

$\frac {16}{AB}=\frac {AB}{9}$ $AB=12$

Then by using Heron's Formula on ABD (12,7,9 as sides), we have $\sqrt{14(2)(7)(5)}$ $14\sqrt5=E$


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
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