Difference between revisions of "2004 AMC 12A Problems/Problem 8"
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== Solutions == | == Solutions == | ||
===Solution 0=== | ===Solution 0=== | ||
− | Looking, we see that the area of <math>[\triangle EBA]</math> is 16 and the area of <math>[\triangle ABC]</math> is 12. Set the area of <math>[\triangle ADB]</math> to be x. We want to find <math>[\triangle ADE]</math> - <math>[\triangle CDB]</math>. So, that would be <math>[\triangle EBA]-[\triangle ADB]=16-x</math> and <math>[\triangle ABC]-[\triangle ADB]=12-x</math>. Therefore, <math></math>[\triangle ADE]-[\triangle DBC]=(16-x)-(12-x)=16-x-12+x=\boxed{\mathrm{(B)}\ 4}<math> | + | Looking, we see that the area of <math>[\triangle EBA]</math> is 16 and the area of <math>[\triangle ABC]</math> is 12. Set the area of <math>[\triangle ADB]</math> to be x. We want to find <math>[\triangle ADE]</math> - <math>[\triangle CDB]</math>. So, that would be <math>[\triangle EBA]-[\triangle ADB]=16-x</math> and <math>[\triangle ABC]-[\triangle ADB]=12-x</math>. Therefore, <math></math>[\triangle ADE]-[\triangle DBC]=(16-x)-(12-x)=16-x-12+x= \boxed{\mathrm{(B)}\ 4}<math> |
~ MathKatana | ~ MathKatana |
Revision as of 19:37, 4 April 2024
- 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 and
sharing common side
,
and
are right angles,
,
,
, and
and
intersect at
. What is the difference between the areas of
and
?
Solutions
Solution 0
Looking, we see that the area of is 16 and the area of
is 12. Set the area of
to be x. We want to find
-
. So, that would be
and
. Therefore, $$ (Error compiling LaTeX. Unknown error_msg)[\triangle ADE]-[\triangle DBC]=(16-x)-(12-x)=16-x-12+x= \boxed{\mathrm{(B)}\ 4}$~ MathKatana
=== Solution 1 ===
Since$ (Error compiling LaTeX. Unknown error_msg)AE \perp ABBC \perp AB
AE \parallel BC
AA\sim
\triangle ADE \sim \triangle CDB
\frac{4}{3}
4
h_{ADE} = 4 \cdot \frac{4}{7} = \frac{16}{7}
h_{CDB} = 3 \cdot \frac 47 = \frac {12}7
\frac{1}{2} \cdot 8 \cdot \frac {16}7 - \frac 12 \cdot 6 \cdot \frac{12}7 = 4$$ (Error compiling LaTeX. Unknown error_msg)\Rightarrow$$ (Error compiling LaTeX. Unknown error_msg)\boxed{\mathrm{(B)}\ 4}$.
=== Solution 2 ===
Let$ (Error compiling LaTeX. Unknown error_msg)[X]X
[\triangle BEA]=[\triangle ABD]+[\triangle ADE]
[\triangle BCA]=[\triangle ABD]+[\triangle BDC]
[\triangle ADE]-[\triangle BDC]=[\triangle BEA]-[\triangle BCA]=\frac{1}{2}\times8\times4-\frac{1}{2}\times6\times4= 16-12=4\Rightarrow\boxed{\mathrm{(B)}\ 4}
ABCDE
\overline{AC}
\overline{BE}
\overline{AC}
y = 1.5x
\overline{BE}
y = -2x + 8
1.5x = -2x + 8
3.5x = 8 $$ (Error compiling LaTeX. Unknown error_msg) x = 8 \times \frac{2}{7}$$ (Error compiling LaTeX. Unknown error_msg) = \frac{16}{7}
\frac{16}{7}
\triangle ADE
\triangle ADE
\frac{16}{7} \times 8 \times \frac{1}{2}$$ (Error compiling LaTeX. Unknown error_msg) = \frac{64}{7}
\triangle BDC
4-\frac{16}{7} = \frac{12}{7}
\triangle BDC
6 \times \frac{12}{7} \times \frac{1}{2} = \frac{36}{7}
\frac{64}{7} - \frac{36}{7} = 4
4
Area(\triangle ADE) - Area(\triangle BDC)
\triangle ABC
\triangle BAE
\triangle ADB$.
This means that$ (Error compiling LaTeX. Unknown error_msg)Area(\triangle BAE) - Area(\triangle ABC) = Area(\triangle ADE) - Area(\triangle BDC)Area(\triangle ADB)
Area(\triangle BAE) = 0.5 * 4 * 8 = 16$$ (Error compiling LaTeX. Unknown error_msg)Area(\triangle ABC) = 0.5 * 4 * 16 = 12$$ (Error compiling LaTeX. Unknown error_msg)Area(\triangle BDC) - Area(\triangle ADE) = 16 - 12 =\boxed{\mathrm{(B)}\ 4}$
Video Solution
Education, the Study of Everything
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.