Difference between revisions of "2018 AMC 8 Problems/Problem 20"
m (→Solution 2) |
(→Solution 2) |
||
Line 24: | Line 24: | ||
==Solution 2== | ==Solution 2== | ||
− | We can extend it into a parallelogram, so it would equal <math>3a \cdot 3b</math>. The smaller parallelogram is 1 a times 2 b. The smaller parallelogram is <math>\frac{2}{9}</math> of the larger parallelogram, so the answer would be <math>\frac{2}{9} \cdot 2</math>, since the triangle is <math>\frac{1}{2}</math> of the parallelogram, so the answer is <math>\boxed{ | + | We can extend it into a parallelogram, so it would equal <math>3a \cdot 3b</math>. The smaller parallelogram is 1 a times 2 b. The smaller parallelogram is <math>\frac{2}{9}</math> of the larger parallelogram, so the answer would be <math>\frac{2}{9} \cdot 2</math>, since the triangle is <math>\frac{1}{2}</math> of the parallelogram, so the answer is <math>\boxed{(A) \frac{4}{9}}.</math> |
Revision as of 13:50, 15 July 2019
Contents
Problem 20
In a point is on with and Point is on so that and point is on so that What is the ratio of the area of to the area of
Solution
By similar triangles, we have . Similarly, we see that Using this information, we get Then, since , it follows that the . Thus, the answer would be
Sidenote: denotes the area of triangle . Similarly, denotes the area of figure .
Solution 2
We can extend it into a parallelogram, so it would equal . The smaller parallelogram is 1 a times 2 b. The smaller parallelogram is of the larger parallelogram, so the answer would be , since the triangle is of the parallelogram, so the answer is
By babyzombievillager
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.