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

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Looking at the triangle <math>BCD</math>, we see that its perpendicular bisector reaches the vertex, therefore hinting it is isoceles. Let <math>x = \angle C</math> be <math>x</math>, so that <math>B=2x</math> from given and the previous deducted. Then <math>\angle ABD=x, \angle ADB=2x</math> 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 <math> \triangle ABD</math> and <math>\triangle ACB</math> are [[Similar (geometry)|similar]], so <math>\frac {16}{AB}=\frac {AB}{9} \Longrightarrow AB=12</math>.
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Looking at the triangle <math>BCD</math>, we see that its perpendicular bisector reaches the vertex, therefore hinting it is isoceles. Let <math>x = \angle C</math>, so that <math>B=2x</math> from given and the previous deducted. Then <math>\angle ABD=x, \angle ADB=2x</math> 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 <math> \triangle ABD</math> and <math>\triangle ACB</math> are [[Similar (geometry)|similar]], so <math>\frac {16}{AB}=\frac {AB}{9} \Longrightarrow AB=12</math>.
  
 
Then by using Heron's Formula on <math>ABD</math> (with sides <math>12,7,9</math>), we have <math>[\triangle ABD]= \sqrt{14(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}</math>.
 
Then by using Heron's Formula on <math>ABD</math> (with sides <math>12,7,9</math>), we have <math>[\triangle ABD]= \sqrt{14(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}</math>.

Revision as of 18:34, 28 February 2012

Problem

In triangle $ABC$ , side $AC$ and the perpendicular bisector of $BC$ meet in point $D$, and $BD$ bisects $\angle ABC$. 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

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

Looking at the triangle $BCD$, we see that its perpendicular bisector reaches the vertex, therefore hinting it is isoceles. Let $x = \angle C$, so that $B=2x$ from given and the previous deducted. Then $\angle ABD=x, \angle ADB=2x$ 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 $\triangle ABD$ and $\triangle ACB$ are similar, so $\frac {16}{AB}=\frac {AB}{9} \Longrightarrow AB=12$.

Then by using Heron's Formula on $ABD$ (with sides $12,7,9$), we have $[\triangle ABD]= \sqrt{14(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}$.

See Also

2002 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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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