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

Revision as of 08:44, 3 August 2020

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

Solution 1 $[asy] unitsize(0.25 cm); pair A, B, C, D, M; A = (0,0); B = (88/9, 28*sqrt(5)/9); C = (16,0); D = 9/16*C; M = (B + C)/2; draw(A--B--C--cycle); draw(B--D--M); label("$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 implying it is isosceles. 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 $[\triangle ABD]= \sqrt{14(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}$ .

Solution 2

Let M be the point of the perpendicular bisector on BC. By the perpendicular bisector theorem, $BD = DC = 7$ and $BM = MC$ . Also, by the angle bisector theorem, $\frac {AB}{BC} = \frac{9}{7}$ . Thus, let $AB = 9x$ and $BC = 7x$ . In addition, $BM = 3.5x$ .

Thus, $\cos\angle CBD = \frac {3.5x}{7} = \frac {x}{2}$ . Additionally, using the Law of Cosines and the fact that $\angle CBD = \angle ABD$ , $81 = 49 + 81x^2 - 2(9x)(7)\cos\angle CBD$

Substituting and simplifying, we get $x = 4/3$

Thus, $AB = 12$ . We now know all sides of $\triangle ABD$ . Using Heron's Formula on $\triangle ABD$ , $\sqrt{(14)(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}$

Solution 3

Note that because the perpendicular bisector and angle bisector meet at side $AC$ and $CD = BD$ as triangle $BDC$ is isosceles, so $BD = 7$ . By the angle bisector theorem, we can express $AB$ and $BC$ as $9x$ and $7x$ respectively. We try to find $x$ through Stewart's Theorem. So

$16(7^2+9\cdot7) = (7x)^2 \cdot 9 + (9x)^2 \cdot 7$

$16(49+63) = (49 \cdot 9 + 81 \cdot 7)x^2$

$16(49+63) = 9(49+63) \cdot x^2$

$x^2 = \frac{16}{9}$

$x=\frac{4}{3}$

We plug this to find that the sides of $\triangle ABD$ are $12,7,9$ . By Heron's formula, the area is $\sqrt{(14)(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}$ . ~skyscraper

@@ Line 1: / Line 1: @@
 ==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>\angle 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>

2002 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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Difference between revisions of "2002 AMC 12A Problems/Problem 23"

Revision as of 08:44, 3 August 2020

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