Difference between revisions of "2018 AMC 8 Problems/Problem 20"
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− | By similar triangles, we have <math> | + | By similar triangles, we have <math>ADE = \frac{1}{9}ABC</math>. Similarly, we see that <math>BEF = \frac{4}{9}ABC.</math> Using this information, we get <cmath>ACFE = \frac{5}{9}ABC.</cmath> Then, since <math>ADE = \frac{1}{9}ABC</math>, it follows that the <math>CDEF = \frac{4}{9}ABC</math>. Thus, the answer would be <math>\boxed{\textbf{(A) } \frac{4}{9}}</math>. |
Sidenote: <math>[ABC]</math> denotes the area of triangle <math>ABC</math>. Similarly, <math>[ABCD]</math> denotes the area of figure <math>ABCD</math>. | Sidenote: <math>[ABC]</math> denotes the area of triangle <math>ABC</math>. Similarly, <math>[ABCD]</math> denotes the area of figure <math>ABCD</math>. |
Revision as of 20:03, 8 November 2019
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 1
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
Solution 3
. We can substitute as and as , where is . Side having, distance , has parts also. And and are and respectfully. You can consider the height of and as and respectfully. The area of is because the area formula for a triangle is or . The area of will be . So the area of will be . The area of parallelogram will be . Parallelogram to . The answer is .
By: sap2018
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 |
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