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

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(→‎Solution)
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==Solution==
 
==Solution==
 
<asy>
 
<asy>
−
draw((0,0)--(10,0),black);
+
unitsize(0.25 cm);
−
draw((10,0)--(10.5,5)--(0,0),black);
+
pair A, B, C, D, M;
−
draw((10,0)--(5,2.4),black);
+
A = (0,0);
−
draw((5,2.4)--(5,0),black);
+
B = (88/9, 28*sqrt(5)/9);
−
draw((4.6,0)--(4.6,0.4)--(5.4,0.4)--(5.4,0),red);
+
C = (16,0);
−
label("A",(10.5,5),N);
+
D = 9/16*C;
−
label("D",(5,2.4),N);
+
M = (B + C)/2;
−
label("C",(0,0),W);
+
draw(A--B--C--cycle);
−
label("B",(10,0),E);
+
draw(B--D--M);
−
label("9",(7.5,3.7),N);
+
label("$A$", A, SW);
−
label("7",(2.5,1.6),N);
+
label("$B$", B, N);
 +
label("$C$", C, SE);
 +
label("$D$", D, S);
 
</asy>
 
</asy>
−
 
 
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>.
 
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>.
  

Revision as of 12:34, 21 July 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

[asy] unitsize(0.25 cm); pair A, B, C, D, M; A = (0,0); B = (88/9, 28*sqrt(5)/9); C = (16,0); D = 9/16*C; M = (B + C)/2; draw(A--B--C--cycle); draw(B--D--M); label("$A$", A, SW); label("$B$", B, N); label("$C$", C, SE); label("$D$", D, S); [/asy] 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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