Difference between revisions of "2001 AMC 12 Problems/Problem 22"
(→Problem) |
(→Solution 2) |
||
Line 64: | Line 64: | ||
As <math>E</math> is the midpoint of <math>CD</math>, the height from <math>E</math> onto <math>AC</math> is <math>1/2</math> of the height from <math>D</math> onto <math>AC</math>. Therefore we have <math>[EHJ] = \frac{6}{35} \cdot \frac 12 \cdot [ACD] = \frac 3{35} \cdot 35 = \boxed{3}</math>. | As <math>E</math> is the midpoint of <math>CD</math>, the height from <math>E</math> onto <math>AC</math> is <math>1/2</math> of the height from <math>D</math> onto <math>AC</math>. Therefore we have <math>[EHJ] = \frac{6}{35} \cdot \frac 12 \cdot [ACD] = \frac 3{35} \cdot 35 = \boxed{3}</math>. | ||
+ | |||
+ | === Solution 3 === | ||
+ | Because we see that there are only lines and there is a rectangle, we can coordbash (place this figure on coordinates). We can find <math>H</math> and <math>J</math> by intersecting lines, and then we calculate the area of <math>EHJ</math> using shoelace formula. This yields <math>\boxed{3}</math>. | ||
== See Also == | == See Also == |
Revision as of 15:09, 12 August 2020
Problem
In rectangle , points and lie on so that and is the midpoint of . Also, intersects at and at . The area of the rectangle is . Find the area of triangle .
Solution
Solution 1
Note that the triangles and are similar, as they have the same angles. Hence .
Also, triangles and are similar, hence .
We can now compute as . We have:
- .
- is of , as these two triangles have the same base , and is of , therefore also the height from onto is of the height from . Hence .
- is of , as the base is of the base , and the height from is of the height from . Hence .
- is of for similar reasons, hence .
Therefore .
Solution 2
As in the previous solution, we note the similar triangles and prove that is in and in of .
We can then compute that .
As is the midpoint of , the height from onto is of the height from onto . Therefore we have .
Solution 3
Because we see that there are only lines and there is a rectangle, we can coordbash (place this figure on coordinates). We can find and by intersecting lines, and then we calculate the area of using shoelace formula. This yields .
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.