2018 AMC 8 Problems/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
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 .
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
. 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)
Video Solution (Meta-Solving Technique)
|2018 AMC 8 (Problems • Answer Key • Resources)|
|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|