Difference between revisions of "2018 AMC 8 Problems/Problem 20"
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==Solution 2== | ==Solution 2== | ||
− | We can extend | + | Let <math>a = DE</math> and <math>b =</math> the height of <math>\triangle ABC</math>. We can extend <math>\triangle ABC</math> To form a parallelogram, which would equal <math>3a \cdot 3b</math>. The smaller parallelogram is <math>a</math> times <math>2b</math>. 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{(\textbf{A}) \frac{4}{9}}</math>. |
By babyzombievillager with credits to many others who helped with the solution :D | By babyzombievillager with credits to many others who helped with the solution :D | ||
Line 35: | Line 35: | ||
==Solution 4 (Non-math solution) == | ==Solution 4 (Non-math solution) == | ||
− | If you have little time to calculate, divide DEFC into triangles that are equal to | + | If you have little time to calculate, divide DEFC into triangles that are equal to DAE by drawing lines through points D and F that are parallel to AB and a line through the middle of EF parallel to CB. Also cut triangle EFB into triangles similar to DAE. We see that there are 9 total triangles, and 4 of those are occupied by DEFC. Thus, 4/9. (although it could be wrong) |
==See Also== | ==See Also== |
Latest revision as of 00:28, 28 December 2020
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 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
Let and the height of . We can extend To form a parallelogram, which would equal . The smaller parallelogram is times . 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 with credits to many others who helped with the solution :D
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
Solution 4 (Non-math solution)
If you have little time to calculate, divide DEFC into triangles that are equal to DAE by drawing lines through points D and F that are parallel to AB and a line through the middle of EF parallel to CB. Also cut triangle EFB into triangles similar to DAE. We see that there are 9 total triangles, and 4 of those are occupied by DEFC. Thus, 4/9. (although it could be wrong)
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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