Difference between revisions of "2024 AMC 10B Problems/Problem 10"
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==Solution 3 (Techniques)== | ==Solution 3 (Techniques)== | ||
We assert that <math>ABCD</math> is a square of side length <math>6</math>. Notice that <math>\triangle AFE\sim\triangle CFB</math> with a scale factor of <math>2</math>. Since the area of <math>\triangle ABC</math> is <math>18 \implies</math> the area of <math>\triangle CFB</math> is <math>12</math>, so the area of <math>\triangle AFE</math> is <math>3</math>. Thus the area of <math>CDEF</math> is <math>18-3=15</math>, and we conclude that the answer is <math>\frac{15}{12}\implies\boxed{\text{(A) }5:4}</math> | We assert that <math>ABCD</math> is a square of side length <math>6</math>. Notice that <math>\triangle AFE\sim\triangle CFB</math> with a scale factor of <math>2</math>. Since the area of <math>\triangle ABC</math> is <math>18 \implies</math> the area of <math>\triangle CFB</math> is <math>12</math>, so the area of <math>\triangle AFE</math> is <math>3</math>. Thus the area of <math>CDEF</math> is <math>18-3=15</math>, and we conclude that the answer is <math>\frac{15}{12}\implies\boxed{\text{(A) }5:4}</math> | ||
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+ | ==Solution 4== | ||
+ | Let <math>ABCE</math> be a square with side length <math>1</math>, to assist with calculations. We can put this on the coordinate plane with the points <math>D = (0,0)</math>, <math>C = (1, 0)</math>, <math>B = (1, 1)</math>, and <math>A = (0, 1)</math>. We have <math>E = (0, 0.5)</math>. Therefore, the line <math>EB</math> has slope <math>0.5</math> and y-intercept <math>0.5</math>. The equation of the line is then <math>y = 0.5x + 0.5</math>. The equation of line <math>AC</math> is <math>y = -x + 1</math>. The intersection is when the lines are equal to each other, so we solve the equation. <math>0.5x + 0.5 = -x + 1</math>, so <math>x = \frac{1}{3}</math>. Therefore, plugging it into the equation, we get <math>y= \frac{2}{3}</math>. Using the shoelace theorem, we get the area of <math>CDEF</math> to be <math>\frac{5}{12}</math> and the area of <math>CFB</math> to be <math>\frac{1}{3}</math>, so our ratio is <math>\frac{\frac{5}{12}}{\frac{1}{3}} = \boxed{(A) 5:4}</math> | ||
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | ==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== |
Revision as of 12:36, 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
Solution 3 (Techniques)
We assert that is a square of side length . Notice that with a scale factor of . Since the area of is the area of is , so the area of is . Thus the area of is , and we conclude that the answer is
Solution 4
Let be a square with side length , to assist with calculations. We can put this on the coordinate plane with the points , , , and . We have . Therefore, the line has slope and y-intercept . The equation of the line is then . The equation of line is . The intersection is when the lines are equal to each other, so we solve the equation. , so . Therefore, plugging it into the equation, we get . Using the shoelace theorem, we get the area of to be and the area of to be , so our ratio is
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.