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

Revision as of 11:45, 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 $\boxed{\textbf{(A) } 5:4}$ .

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

Solution 3

We assert that $ABCD$ is a square of side length $6$ . Notice that $\triangle AFE\sim\triangle CFB$ with a scale factor of 2. Since the area of $\triangle ABC$ is $18 \implies$ the area of $\triangle CFB$ is $12$ , so the area of $\triangle AFE$ is $3$ . Thus the area of $CDEF$ is $18-3=15$ , and we conclude that the answer is $\frac{15}{12}\implies\boxed{\text{(A) }5:4}$

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

@@ Line 17: / Line 17: @@
 ==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
+==Solution 3==
+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>
 ==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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