Difference between revisions of "2004 AMC 12A Problems/Problem 8"
m (Replaced diagram with Asymptote) |
Hashtagmath (talk | contribs) |
||
(6 intermediate revisions by 4 users not shown) | |||
Line 1: | Line 1: | ||
{{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>? | ||
Line 32: | Line 33: | ||
<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 === | ||
+ | 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 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}\times8\times4-\frac{1}{2}\times6\times4= 16-12=4\Rightarrow\boxed{\mathrm{(B)}\ 4}</math> | ||
+ | |||
+ | === Solution 3 (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 | + | ==Video Solution== |
+ | https://youtu.be/DlA71MBSviU | ||
− | + | Education, the Study of Everything | |
− | |||
== See also == | == See also == |
Latest revision as of 17:38, 16 January 2021
- 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.