Difference between revisions of "2018 AMC 8 Problems/Problem 20"
(→Solution 3) |
(→Solution) |
||
Line 17: | Line 17: | ||
<math>\textbf{(A) } \frac{4}{9} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{5}{9} \qquad \textbf{(D) } \frac{3}{5} \qquad \textbf{(E) } \frac{2}{3}</math> | <math>\textbf{(A) } \frac{4}{9} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{5}{9} \qquad \textbf{(D) } \frac{3}{5} \qquad \textbf{(E) } \frac{2}{3}</math> | ||
− | ==Solution== | + | ==Solution 1== |
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 {A}.</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 {A}.</math> |
Revision as of 19:16, 30 September 2019
Contents
[hide]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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.