Difference between revisions of "2024 AMC 10B Problems/Problem 10"
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==Problem== | ==Problem== | ||
− | Quadrilateral <math>ABCD</math> is a parallelogram, and <math>E</math> is the midpoint of the side <math>AD</math>. Let <math>F</math> be the intersection of lines <math>EB</math> and <math>AC</math>. What is the ratio of the area of | + | Quadrilateral <math>ABCD</math> is a parallelogram, and <math>E</math> is the midpoint of the side <math>\overline{AD}</math>. Let <math>F</math> be the intersection of lines <math>EB</math> and <math>AC</math>. What is the ratio of the area of |
− | quadrilateral <math>CDEF</math> to the area of | + | quadrilateral <math>CDEF</math> to the area of <math>\triangle CFB</math>? |
<math>\textbf{(A) } 5:4 \qquad\textbf{(B) } 4:3 \qquad\textbf{(C) } 3:2 \qquad\textbf{(D) } 5:3 \qquad\textbf{(E) } 2:1</math> | <math>\textbf{(A) } 5:4 \qquad\textbf{(B) } 4:3 \qquad\textbf{(C) } 3:2 \qquad\textbf{(D) } 5:3 \qquad\textbf{(E) } 2:1</math> |
Revision as of 00:43, 15 November 2024
Contents
[hide]Problem
Quadrilateral is a parallelogram, and is the midpoint of the side . Let be the intersection of lines and . What is the ratio of the area of quadrilateral to the area of ?
Solution 1
Let have length and let the altitude of the parallelogram perpendicular to have length .
The area of the parallelogram is and the area of equals . Thus, the area of quadrilateral is .
We have from that . Also, , so the length of the altitude of from is twice that of . This means that the altitude of is , so the area of is .
Then, the area of quadrilateral equals the area of minus that of , which is . Finally, the ratio of the area of to the area of triangle is , so the answer is .
Solution 2
Let . Since with a scale factor of , . The scale factor of also means that , therefore since and have the same height, . Since is a parallelogram, ~Tacos_are_yummy_1
Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
https://youtu.be/QLziG_2e7CY?feature=shared
~ Pi Academy
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=24EZaeAThuE
See also
2024 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.