Difference between revisions of "2005 AMC 10B Problems/Problem 23"

Latest revision as of 18:22, 17 September 2023

Problem

In trapezoid $ABCD$ we have $\overline{AB}$ parallel to $\overline{DC}$ , $E$ as the midpoint of $\overline{BC}$ , and $F$ as the midpoint of $\overline{DA}$ . The area of $ABEF$ is twice the area of $FECD$ . What is $AB/DC$ ?

$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 8$

Solution 1

Since the heights of both trapezoids are equal, and the area of $ABEF$ is twice the area of $FECD$ ,

$\frac{AB+EF}{2}=2\left(\frac{DC+EF}{2}\right)$ .

$\frac{AB+EF}{2}=DC+EF$ , so

$AB+EF=2DC+2EF$ .

$EF$ is exactly halfway between $AB$ and $DC$ , so $EF=\frac{AB+DC}{2}$ .

$AB+\frac{AB+DC}{2}=2DC+AB+DC$ , so

$\frac{3}{2}AB+\frac{1}{2}DC=3DC+AB$ , and

$\frac{1}{2}AB=\frac{5}{2}DC$ .

$\frac{AB}{DC} = \boxed{\textbf{(C) }5}$ .

Solution 2

Mark $DC=z$ , $AB=x$ , and $FE=y.$ Note that the heights of trapezoids $ABEF$ & $FECD$ are the same. Mark the height to be $h$ .

Then, we have that $\tfrac{x+y}{2}\cdot h=2(\tfrac{y+z}{2} \cdot h)$ .

From this, we get that $x=2z+y$ .

We also get that $\tfrac{x+z}{2} \cdot 2h= 3(\tfrac{y+z}{2} \cdot h)$ .

Simplifying, we get that $2x=z+3y$

Notice that we want $\tfrac{AB}{DC}=\tfrac{x}{z}$ .

Dividing the first equation by $z$ , we get that $\tfrac{x}{z}=2+\tfrac{y}{z}\implies 3(\tfrac{x}{z})=6+3(\tfrac{y}{z})$ .

Dividing the second equation by $z$ , we get that $2(\tfrac{x}{z})=1+3(\tfrac{y}{z})$ .

Now, when we subtract the top equation from the bottom, we get that $\tfrac{x}{z}=\boxed{\textbf{(C) }5}$

Solution 3

Since the bases of the trapezoids along with the height are the same, the only thing that matters is the second base. Denote the length of the bigger trapezoid $x$ . The area of the smaller trapezoid is $A$ = $h \cdot \frac {b_1 + b_2}{2}$ . The area of the larger trapezoid is $A$ = $h \cdot \frac {b_1 + x}{2}$ . Since this problem asks for proportions, assume that $b_1 = 1$ and $b_2 = 2$ .

The smaller trapezoid has area $h$ while the larger trapezoid must have area $5h$ . We have the equation $\frac {x}{2} = 5$ . $x$ = 10, and our answer is $\boxed{\textbf{(C) }5}$

~Arcticturn

Video Solution

https://www.youtube.com/watch?v=fsNJbC3hGtk ~David

@@ Line 47: / Line 47: @@
 Since the bases of the trapezoids along with the height are the same, the only thing that matters is the second base. Denote the length of the bigger trapezoid <math>x</math>. The area of the smaller trapezoid is <math>A</math> = <math>h \cdot \frac {b_1 + b_2}{2}</math>. The area of the larger trapezoid is <math>A</math> = <math>h \cdot \frac {b_1 + x}{2}</math>. Since this problem asks for proportions, assume that <math>b_1 = 1</math> and <math>b_2 = 2</math>.
-The smaller trapezoid has area <math>h</math> while the larger trapezoid must have area <math>5h</math>. We have the equation <math>\frac {x}{2} = 5</math>. <math>x</math> = 10, and our answer is <math>\boxed {(C)5}</math>
+The smaller trapezoid has area <math>h</math> while the larger trapezoid must have area <math>5h</math>. We have the equation <math>\frac {x}{2} = 5</math>. <math>x</math> = 10, and our answer is <math>\boxed{\textbf{(C) }5}</math>
 ~Arcticturn
+== Video Solution ==
+https://www.youtube.com/watch?v=fsNJbC3hGtk   ~David
 == See Also ==
 {{AMC10 box|year=2005|ab=B|num-b=22|num-a=24}}
 {{MAA Notice}}

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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

Page

Toolbox

Search