2004 AMC 12A Problems/Problem 8
- The following problem is from both the 2004 AMC 12A #8 and 2004 AMC 10A #9, so both problems redirect to this page.
Contents
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 1
Since and
,
. By alternate interior angles and
, we find that
, with side length ratio
. Their heights also have the same ratio, and since the two heights add up to
, we have that
and
. Subtracting the areas,
.
Solution 2
Let represent the area of figure
. Note that
and
.
Solution 3 (coordbash)
Put figure on a graph.
goes from (0, 0) to (4, 6) and
goes from (4, 0) to (0, 8).
is on line
.
is on line
. Finding intersection between these points,
.
This gives us the x-coordinate of D.
So, is the height of
, then area of
is
Now, the height of is
And the area of
is
This gives us
Therefore, the difference is
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 |
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