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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>
 +
\text{(A) }1992
 +
\qquad
 +
\text{(B) }1999
 +
\qquad
 +
\text{(C) }2001
 +
\qquad
 +
\text{(D) }2002
 +
\qquad
 +
\text{(E) }2004
 +
</math>
  
 
==Solution==
 
==Solution==
  
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This problems 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 (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.

Revision as of 01:19, 18 February 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?

$\text{(A) }1992 \qquad \text{(B) }1999 \qquad \text{(C) }2001 \qquad \text{(D) }2002 \qquad \text{(E) }2004$

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.

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

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)
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Problem 24 		Followed by
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All AMC 12 Problems and Solutions
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