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Difference between revisions of "2004 AMC 12A Problems/Problem 8"

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== Solution ==
 
== Solution ==
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=== Solution 1 ===
 
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If we let <math>[\ldots]</math> denote area, <math>[ABE] - [ABC] = [ADE] + [ABD] - [ABD] - [BDC] = [ADE] - [BDC]</math>. Using the given, <math>[ABE] = \frac 12 \cdot 8 \cdot 4</math> and <math>[ABC] = \frac 12 \cdot 6 \cdot 4</math>, and their difference is <math>16 - 12 = 4\ \mathrm{(B)}</math>.
 
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=== Solution 2 ===
 
=== Solution 2 ===
 
Since <math>AE \perp AB</math> and <math>BC \perp AB</math>, <math>AE \parallel BC</math>. By [[alternate interior angles]] and AA~, we find that <math>\triangle ADE \sim \triangle CDB</math>, with side length ratio <math>\frac{4}{3}</math>. Their heights also have the same ratio, and since the two heights add up to <math>4</math>, we have that <math>h_{ADE} = 4 \cdot \frac{4}{7} = \frac{16}{7}</math> and <math>h_{CDB} = 3 \cdot \frac 47 = \frac {12}7</math>. Subtracting the areas, <math>\frac{1}{2} \cdot 8 \cdot \frac {16}7 - \frac 12 \cdot 6 \cdot \frac{12}7 = 4</math> <math>\Rightarrow</math> <math>\boxed{(B)}</math>.
 
Since <math>AE \perp AB</math> and <math>BC \perp AB</math>, <math>AE \parallel BC</math>. By [[alternate interior angles]] and AA~, we find that <math>\triangle ADE \sim \triangle CDB</math>, with side length ratio <math>\frac{4}{3}</math>. Their heights also have the same ratio, and since the two heights add up to <math>4</math>, we have that <math>h_{ADE} = 4 \cdot \frac{4}{7} = \frac{16}{7}</math> and <math>h_{CDB} = 3 \cdot \frac 47 = \frac {12}7</math>. Subtracting the areas, <math>\frac{1}{2} \cdot 8 \cdot \frac {16}7 - \frac 12 \cdot 6 \cdot \frac{12}7 = 4</math> <math>\Rightarrow</math> <math>\boxed{(B)}</math>.

Revision as of 19:56, 21 June 2012

The following problem is from both the 2004 AMC 12A #8 and 2004 AMC 10A #9, so both problems redirect to this page.

Problem

In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$, $\angle{EAB}$ and $\angle{ABC}$ are right angles, $AB=4$, $BC=6$, $AE=8$, and $\overline{AC}$ and $\overline{BE}$ intersect at $D$. What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$?

AMC10 2004A 9.gif

$\mathrm {(A)}\ 2 \qquad \mathrm {(B)}\ 4 \qquad \mathrm {(C)}\ 5 \qquad \mathrm {(D)}\ 8 \qquad \mathrm {(E)}\ 9 \qquad$

Solution

Solution 2

Since $AE \perp AB$ and $BC \perp AB$, $AE \parallel BC$. By alternate interior angles and AA~, we find that $\triangle ADE \sim \triangle CDB$, with side length ratio $\frac{4}{3}$. Their heights also have the same ratio, and since the two heights add up to $4$, we have that $h_{ADE} = 4 \cdot \frac{4}{7} = \frac{16}{7}$ and $h_{CDB} = 3 \cdot \frac 47 = \frac {12}7$. Subtracting the areas, $\frac{1}{2} \cdot 8 \cdot \frac {16}7 - \frac 12 \cdot 6 \cdot \frac{12}7 = 4$ $\Rightarrow$ $\boxed{(B)}$.

See also

2004 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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
2004 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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 10 Problems and Solutions
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