Difference between revisions of "2004 AMC 12A Problems/Problem 8"
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+ | {{duplicate|[[2004 AMC 12A Problems|2004 AMC 12A #8]] and [[2004 AMC 10A Problems/Problem 9|2004 AMC 10A #9]]}} | ||
== Problem == | == Problem == | ||
In the overlapping [[triangle]]s <math>\triangle{ABC}</math> and <math>\triangle{ABE}</math> sharing common [[side]] <math>AB</math>, <math>\angle{EAB}</math> and <math>\angle{ABC}</math> are [[right angle]]s, <math>AB=4</math>, <math>BC=6</math>, <math>AE=8</math>, and <math>\overline{AC}</math> and <math>\overline{BE}</math> intersect at <math>D</math>. What is the difference between the areas of <math>\triangle{ADE}</math> and <math>\triangle{BDC}</math>? | In the overlapping [[triangle]]s <math>\triangle{ABC}</math> and <math>\triangle{ABE}</math> sharing common [[side]] <math>AB</math>, <math>\angle{EAB}</math> and <math>\angle{ABC}</math> are [[right angle]]s, <math>AB=4</math>, <math>BC=6</math>, <math>AE=8</math>, and <math>\overline{AC}</math> and <math>\overline{BE}</math> intersect at <math>D</math>. What is the difference between the areas of <math>\triangle{ADE}</math> and <math>\triangle{BDC}</math>? | ||
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+ | [[Image:AMC10_2004A_9.gif|center]] | ||
<math>\mathrm {(A)}\ 2 \qquad \mathrm {(B)}\ 4 \qquad \mathrm {(C)}\ 5 \qquad \mathrm {(D)}\ 8 \qquad \mathrm {(E)}\ 9 \qquad</math> | <math>\mathrm {(A)}\ 2 \qquad \mathrm {(B)}\ 4 \qquad \mathrm {(C)}\ 5 \qquad \mathrm {(D)}\ 8 \qquad \mathrm {(E)}\ 9 \qquad</math> | ||
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== Solution == | == Solution == | ||
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=== Solution 1 === | === Solution 1 === | ||
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>. | 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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== See also == | == See also == | ||
+ | * [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131320 AoPS topic] | ||
{{AMC12 box|year=2004|ab=A|num-b=7|num-a=9}} | {{AMC12 box|year=2004|ab=A|num-b=7|num-a=9}} | ||
+ | {{AMC10 box|year=2004|ab=A|num-b=8|num-a=10}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] |
Revision as of 21:57, 23 April 2008
- 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 ?
Solution
Solution 1
If we let denote area, . Using the given, and , and their difference is .
Solution 2
Since and , . By alternate interior angles and AA~, 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, .
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 |