Difference between revisions of "2024 AMC 10B Problems/Problem 10"
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We have from <math>AA</math> that <math>\triangle CBF \sim \triangle AEF</math>. Also, <math>CB/AE = 2</math>, so the length of the altitude of <math>\triangle CBF</math> from <math>F</math> is twice that of <math>\triangle AEF</math>. This means that the altitude of <math>\triangle CBF</math> is <math>2h/3</math>, so the area of <math>\triangle CBF</math> is <math>\frac{(b)(2h/3)}{2} = \frac{bh}{3}</math>. | We have from <math>AA</math> that <math>\triangle CBF \sim \triangle AEF</math>. Also, <math>CB/AE = 2</math>, so the length of the altitude of <math>\triangle CBF</math> from <math>F</math> is twice that of <math>\triangle AEF</math>. This means that the altitude of <math>\triangle CBF</math> is <math>2h/3</math>, so the area of <math>\triangle CBF</math> is <math>\frac{(b)(2h/3)}{2} = \frac{bh}{3}</math>. | ||
− | Then, the area of quadrilateral <math>CDEF</math> equals the area of <math>BCDE</math> minus that of <math>\triangle CBF</math>, which is <math>\frac{3bh}{4} - \frac{bh}{3} = \frac{5bh}{12}</math>. Finally, the ratio of the area of <math>CDEF</math> to the area of triangle <math>CFB</math> is <math>\frac{\frac{5bh}{12}}{\frac{bh}{3}} = \frac{\frac{5}{12}}{\frac{1}{3}} = \frac{5}{4}</math>, so the answer is <math>\textbf{(A) } 5:4</math>. | + | Then, the area of quadrilateral <math>CDEF</math> equals the area of <math>BCDE</math> minus that of <math>\triangle CBF</math>, which is <math>\frac{3bh}{4} - \frac{bh}{3} = \frac{5bh}{12}</math>. Finally, the ratio of the area of <math>CDEF</math> to the area of triangle <math>CFB</math> is <math>\frac{\frac{5bh}{12}}{\frac{bh}{3}} = \frac{\frac{5}{12}}{\frac{1}{3}} = \frac{5}{4}</math>, so the answer is <math>\boxed{\textbf{(A) } 5:4}</math>. |
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=B|num-b=9|num-a=11}} | {{AMC10 box|year=2024|ab=B|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 03:26, 14 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 triangle ?
Solution 1
Assume is a square, sidelength 1 on Cartesian coordinate system...
Solution 2
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 .
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 |
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