Difference between revisions of "2024 AMC 10B Problems/Problem 10"
(→Solution 3) |
|||
Line 18: | Line 18: | ||
Let <math>[AFE]=1</math>. Since <math>\triangle AFE\sim\triangle CFB</math> with a scale factor of <math>2</math>, <math>[CFB]=4</math>. The scale factor of <math>2</math> also means that <math>\dfrac{AF}{FC}=\dfrac{1}{2}</math>, therefore since <math>\triangle BCF</math> and <math>\triangle BFA</math> have the same height, <math>[BFA]=2</math>. Since <math>ABCD</math> is a parallelogram, <cmath>[BCA]=[DAC]\implies4+2=1+[CDEF]\implies [CDEF]=5\implies\boxed{\text{(A) }5:4}</cmath> ~Tacos_are_yummy_1 | Let <math>[AFE]=1</math>. Since <math>\triangle AFE\sim\triangle CFB</math> with a scale factor of <math>2</math>, <math>[CFB]=4</math>. The scale factor of <math>2</math> also means that <math>\dfrac{AF}{FC}=\dfrac{1}{2}</math>, therefore since <math>\triangle BCF</math> and <math>\triangle BFA</math> have the same height, <math>[BFA]=2</math>. Since <math>ABCD</math> is a parallelogram, <cmath>[BCA]=[DAC]\implies4+2=1+[CDEF]\implies [CDEF]=5\implies\boxed{\text{(A) }5:4}</cmath> ~Tacos_are_yummy_1 | ||
− | ==Solution 3== | + | ==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 2. 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 2. 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> | ||
Revision as of 11:45, 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 2. 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
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.