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

(Solution 1)
Line 11: Line 11:
 
=== Solution 1===
 
=== Solution 1===
 
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>.
 +
 +
=== Solution 2 ===
 +
Let <math>[X]</math> represent the area of figure <math>X</math>. Note that <math>[\triangle BEA]=[\triangle ABD]+[\triangle ADE]</math> and <math>[\triangle BCA]=[\triangle ABD]+[\triangle BDC]</math>.
 +
 +
<math>[\triangle ADE]-[\triangle BDC]=[\triangle BEA]-[\triangle BCA]=\frac{1}{2} 8*4-\frac{1}{2} 6*4= 16-12=4</math> <math>\Rightarrow</math> <math>\boxed{(B)}</math>.
 +
  
 
== See also ==
 
== See also ==

Revision as of 17:24, 24 December 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 1

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, $\frac{1}{2} \cdot 8 \cdot \frac {16}7 - \frac 12 \cdot 6 \cdot \frac{12}7 = 4$ $\Rightarrow$ $\boxed{(B)}$.

Solution 2

Let $[X]$ represent the area of figure $X$. Note that $[\triangle BEA]=[\triangle ABD]+[\triangle ADE]$ and $[\triangle BCA]=[\triangle ABD]+[\triangle BDC]$.

$[\triangle ADE]-[\triangle BDC]=[\triangle BEA]-[\triangle BCA]=\frac{1}{2} 8*4-\frac{1}{2} 6*4= 16-12=4$ $\Rightarrow$ $\boxed{(B)}$.


See also

2004 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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
Invalid username
Login to AoPS