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2002 AMC 12A Problems/Problem 23: Difference between revisions

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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 <math>BD</math> bisects <math>\angle 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 <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>


==Solution==
==Solution==
{{image}}


This problem needs a picture. You can help by adding it.
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>.


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.
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>.
 
<math>\frac {16}{AB}=\frac {AB}{9}</math>
<math>AB=12</math>
 
Then by using Heron's Formula on ABD (12,7,9 as sides), we have
<math>\sqrt{14(2)(7)(5)}</math>
<math>14\sqrt5=D</math>


==See Also==
==See Also==
{{AMC12 box|year=2002|ab=A|num-b=22|num-a=24}}


{{AMC12 box|year=2002|ab=A|num-b=22|num-a=24}}
[[Category:Introductory Geometry Problems]]
[[Category:Similar Triangles Problems]]
[[Category:Area Problems]]

Revision as of 00:11, 20 August 2011

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 $x = \angle C$ be $x$, 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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