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Difference between revisions of "2024 AMC 10B Problems/Problem 10"

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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>.
 
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>.
  
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[[File:2024 AMC 10B 10.png|300px|right]]
 
==Solution 2==
 
==Solution 2==
 
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
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[[File:2024 AMC 10B 10.png|400px|left]]
 
  
 
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==
 
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==

Revision as of 09:00, 15 November 2024

Problem

Quadrilateral $ABCD$ is a parallelogram, and $E$ is the midpoint of the side $\overline{AD}$. Let $F$ be the intersection of lines $EB$ and $AC$. What is the ratio of the area of quadrilateral $CDEF$ to the area of $\triangle CFB$?

$\textbf{(A) } 5:4 \qquad\textbf{(B) } 4:3 \qquad\textbf{(C) } 3:2 \qquad\textbf{(D) } 5:3 \qquad\textbf{(E) } 2:1$

Solution 1

Let $AB = CD$ have length $b$ and let the altitude of the parallelogram perpendicular to $\overline{AD}$ have length $h$.

The area of the parallelogram is $bh$ and the area of $\triangle ABE$ equals $\frac{(b/2)(h)}{2} = \frac{bh}{4}$. Thus, the area of quadrilateral $BCDE$ is $bh - \frac{bh}{4} = \frac{3bh}{4}$.

We have from $AA$ that $\triangle CBF \sim \triangle AEF$. Also, $CB/AE = 2$, so the length of the altitude of $\triangle CBF$ from $F$ is twice that of $\triangle AEF$. This means that the altitude of $\triangle CBF$ is $2h/3$, so the area of $\triangle CBF$ is $\frac{(b)(2h/3)}{2} = \frac{bh}{3}$.

Then, the area of quadrilateral $CDEF$ equals the area of $BCDE$ minus that of $\triangle CBF$, which is $\frac{3bh}{4} - \frac{bh}{3} = \frac{5bh}{12}$. Finally, the ratio of the area of $CDEF$ to the area of triangle $CFB$ is $\frac{\frac{5bh}{12}}{\frac{bh}{3}} = \frac{\frac{5}{12}}{\frac{1}{3}} = \frac{5}{4}$, so the answer is $\boxed{\textbf{(A) } 5:4}$.

2024 AMC 10B 10.png

Solution 2

Let $[AFE]=1$. Since $\triangle AFE\sim\triangle CFB$ with a scale factor of $2$, $[CFB]=4$. The scale factor of $2$ also means that $\dfrac{AF}{FC}=\dfrac{1}{2}$, therefore since $\triangle BCF$ and $\triangle BFA$ have the same height, $[BFA]=2$. Since $ABCD$ is a parallelogram, \[[BCA]=[DAC]\implies4+2=1+[CDEF]\implies [CDEF]=5\implies\boxed{\text{(A) }5:4}\] ~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

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