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]]}} | {{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 [[edge | 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 [[edge | 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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<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> | ||
− | + | == Solutions == | |
+ | ===Solution 1=== | ||
+ | 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>[\triangle ADE]-[\triangle DBC]=(16-x)-(12-x)=16-x-12+x= \boxed{\mathrm{(B)}\ 4}</math> | ||
− | + | ~ MathKatana | |
+ | === Solution 2 === | ||
Since <math>AE \perp AB</math> and <math>BC \perp AB</math>, <math>AE \parallel BC</math>. By alternate interior angles and <math>AA\sim</math>, 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{\mathrm{(B)}\ 4}</math>. | Since <math>AE \perp AB</math> and <math>BC \perp AB</math>, <math>AE \parallel BC</math>. By alternate interior angles and <math>AA\sim</math>, 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{\mathrm{(B)}\ 4}</math>. | ||
− | == Solution | + | === Solution 3 === |
− | |||
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>. | 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}\times8\times4-\frac{1}{2}\times6\times4= 16-12=4\Rightarrow\boxed{\mathrm{(B)}\ 4}</math> | <math>[\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}</math> | ||
+ | |||
+ | === Solution 4 (coordbash)=== | ||
+ | Put figure <math>ABCDE</math> on a graph. <math>\overline{AC}</math> goes from (0, 0) to (4, 6) and <math>\overline{BE}</math> goes from (4, 0) to (0, 8). <math>\overline{AC}</math> is on line <math>y = 1.5x</math>. <math>\overline{BE}</math> is on line <math>y = -2x + 8</math>. Finding intersection between these points, | ||
+ | |||
+ | <math>1.5x = -2x + 8</math>. | ||
+ | |||
+ | <math>3.5x = 8 </math> | ||
+ | |||
+ | <math> x = 8 \times \frac{2}{7}</math> | ||
+ | |||
+ | <math> = \frac{16}{7}</math> | ||
+ | |||
+ | This gives us the x-coordinate of D. | ||
+ | So, <math>\frac{16}{7}</math> is the height of <math>\triangle ADE</math>, then area of <math>\triangle ADE</math> is | ||
+ | <math>\frac{16}{7} \times 8 \times \frac{1}{2}</math> | ||
+ | <math> = \frac{64}{7}</math> | ||
+ | |||
+ | Now, the height of <math>\triangle BDC</math> is <math>4-\frac{16}{7} = \frac{12}{7}</math> | ||
+ | And the area of <math>\triangle BDC</math> is <math>6 \times \frac{12}{7} \times \frac{1}{2} = \frac{36}{7}</math> | ||
+ | |||
+ | This gives us <math>\frac{64}{7} - \frac{36}{7} = 4</math> | ||
+ | |||
+ | Therefore, the difference is <math>4</math> | ||
+ | |||
+ | === Solution 5 === | ||
+ | We want to figure out <math>[\triangle ADE] - [\triangle BDC]</math>. | ||
+ | Notice that <math>\triangle ABC</math> and <math>\triangle BAE</math> "intersect" and form <math>\triangle ADB</math>. | ||
+ | |||
+ | This means that <math>[\triangle BAE] - [\triangle ABC)] = [\triangle ADE] - [\triangle BDC]</math> because <math>[\triangle ADB]</math> cancels out, which can be seen easily in the diagram. | ||
+ | |||
+ | <math>[\triangle BAE] = 0.5 \cdot 4 \cdot 8 = 16</math> | ||
+ | |||
+ | <math>[\triangle ABC] = 0.5 \cdot 4 \cdot 16 = 12</math> | ||
+ | |||
+ | <math>[\triangle BDC] - [\triangle ADE] = 16 - 12 =\boxed{\mathrm{(B)}\ 4}</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/DlA71MBSviU | ||
+ | |||
+ | Education, the Study of Everything | ||
+ | |||
== See also == | == See also == |
Latest revision as of 22:24, 28 September 2024
- The following problem is from both the 2004 AMC 12A #8 and 2004 AMC 10A #9, so both problems redirect to this page.
Contents
[hide]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
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,
~ MathKatana
Solution 2
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 3
Let represent the area of figure . Note that and .
Solution 4 (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
Solution 5
We want to figure out . Notice that and "intersect" and form .
This means that because cancels out, which can be seen easily in the diagram.
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.