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Difference between revisions of "2024 AMC 10B Problems/Problem 10"
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==Solution 1==
 
==Solution 1==
 
Assume <math>ABCD</math> is a square, sidelength 1 on Cartesian coordinate system...
 
Assume <math>ABCD</math> is a square, sidelength 1 on Cartesian coordinate system...
 +
 +
==Solution 2==
 +
Let <math>AB = CD</math> have length <math>b</math> and let the altitude of the parallelogram perpendicular to <math>\overline{AD}</math> have length <math>h</math>.
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The area of the parallelogram is <math>bh</math> and the area of <math>\triangle ABE</math> equals <math>\frac{(b/2)(h)}{2} = \frac{bh}{4}</math>. Thus, the area of quadrilateral <math>BCDE</math> is <math>bh - \frac{bh}{4} = \frac{3bh}{4}</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>.
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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>\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 03:05, 14 November 2024

Problem

Quadrilateral $ABCD$ is a parallelogram, and $E$ is the midpoint of the side $AD$. 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

Assume $ABCD$ is a square, sidelength 1 on Cartesian coordinate system...

Solution 2

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 $\textbf{(A) } 5:4$.

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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