Difference between revisions of "2002 AMC 12A Problems/Problem 23"

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==Problem==
 
==Problem==
  
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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In triangle <math>ABC</math> , side <math>AC</math> and the perpendicular bisector of <math>BC</math> meet in point <math>D</math>, and <math>BD</math> bisects <math><ABC</math>. If <math>AD=9</math> and <math>DC=7</math>, what is the area of triangle ABD?  
  
 
<math>\text{(A)}\ 14 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 14\sqrt5 \qquad \text{(E)}\ 28\sqrt5</math>
 
<math>\text{(A)}\ 14 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 14\sqrt5 \qquad \text{(E)}\ 28\sqrt5</math>

Revision as of 19:31, 28 November 2010

Problem

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

$\text{(A)}\ 14 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 14\sqrt5 \qquad \text{(E)}\ 28\sqrt5$

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 (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.

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

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 (ProblemsAnswer KeyResources)
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