Difference between revisions of "2005 AMC 10B Problems/Problem 23"
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In trapezoid <math>ABCD</math> we have <math>\overline{AB}</math> parallel to <math>\overline{DC}</math>, <math>E</math> as the midpoint of <math>\overline{BC}</math>, and <math>F</math> as the midpoint of <math>\overline{DA}</math>. The area of <math>ABEF</math> is twice the area of <math>FECD</math>. What is <math>AB/DC</math>? | In trapezoid <math>ABCD</math> we have <math>\overline{AB}</math> parallel to <math>\overline{DC}</math>, <math>E</math> as the midpoint of <math>\overline{BC}</math>, and <math>F</math> as the midpoint of <math>\overline{DA}</math>. The area of <math>ABEF</math> is twice the area of <math>FECD</math>. What is <math>AB/DC</math>? | ||
− | <math>\ | + | <math>\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 8 </math> |
− | == Solution == | + | == Solution 1== |
− | Since the | + | Since the heights of both trapezoids are equal, and the area of <math>ABEF</math> is twice the area of <math>FECD</math>, |
− | <math>\frac{AB+EF}{2}=2\left(\frac{ | + | <math>\frac{AB+EF}{2}=2\left(\frac{DC+EF}{2}\right)</math>. |
− | <math>\frac{AB+EF}{2}= | + | <math>\frac{AB+EF}{2}=DC+EF</math>, so |
− | <math>AB+EF= | + | <math>AB+EF=2DC+2EF</math>. |
− | <math>EF</math> is exactly halfway between <math>AB</math> and <math> | + | <math>EF</math> is exactly halfway between <math>AB</math> and <math>DC</math>, so <math>EF=\frac{AB+DC}{2}</math>. |
− | <math>AB+\frac{AB+ | + | <math>AB+\frac{AB+DC}{2}=2DC+AB+DC</math>, so |
− | <math>\frac{3}{2}AB+\frac{1}{2} | + | <math>\frac{3}{2}AB+\frac{1}{2}DC=3DC+AB</math>, and |
− | <math>\frac{1}{2}AB=\frac{5}{2} | + | <math>\frac{1}{2}AB=\frac{5}{2}DC</math>. |
− | + | <math>\frac{AB}{DC} = \boxed{\textbf{(C) }5}</math>. | |
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Mark <math>DC=z</math>, <math>AB=x</math>, and <math>FE=y.</math> | ||
+ | Note that the heights of trapezoids <math>ABEF</math> & <math>FECD</math> are the same. Mark the height to be <math>h</math>. | ||
+ | |||
+ | Then, we have that <math>\tfrac{x+y}{2}\cdot h=2(\tfrac{y+z}{2} \cdot h)</math>. | ||
+ | |||
+ | From this, we get that <math>x=2z+y</math>. | ||
+ | |||
+ | We also get that <math>\tfrac{x+z}{2} \cdot 2h= 3(\tfrac{y+z}{2} \cdot h)</math>. | ||
+ | |||
+ | Simplifying, we get that <math>2x=z+3y</math> | ||
+ | |||
+ | Notice that we want <math>\tfrac{AB}{DC}=\tfrac{x}{z}</math>. | ||
+ | |||
+ | Dividing the first equation by <math>z</math>, we get that <math>\tfrac{x}{z}=2+\tfrac{y}{z}\implies 3(\tfrac{x}{z})=6+3(\tfrac{y}{z})</math>. | ||
+ | |||
+ | Dividing the second equation by <math>z</math>, we get that <math>2(\tfrac{x}{z})=1+3(\tfrac{y}{z})</math>. | ||
+ | |||
+ | Now, when we subtract the top equation from the bottom, we get that <math>\tfrac{x}{z}=\boxed{\textbf{(C) }5}</math> | ||
+ | |||
+ | ==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 <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{\textbf{(C) }5}</math> | ||
+ | |||
+ | ~Arcticturn | ||
+ | |||
+ | == Video Solution == | ||
+ | https://www.youtube.com/watch?v=fsNJbC3hGtk ~David | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2005|ab=B|num-b=22|num-a=24}} | {{AMC10 box|year=2005|ab=B|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:22, 17 September 2023
Problem
In trapezoid we have
parallel to
,
as the midpoint of
, and
as the midpoint of
. The area of
is twice the area of
. What is
?
Solution 1
Since the heights of both trapezoids are equal, and the area of is twice the area of
,
.
, so
.
is exactly halfway between
and
, so
.
, so
, and
.
.
Solution 2
Mark ,
, and
Note that the heights of trapezoids
&
are the same. Mark the height to be
.
Then, we have that .
From this, we get that .
We also get that .
Simplifying, we get that
Notice that we want .
Dividing the first equation by , we get that
.
Dividing the second equation by , we get that
.
Now, when we subtract the top equation from the bottom, we get that
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 . The area of the smaller trapezoid is
=
. The area of the larger trapezoid is
=
. Since this problem asks for proportions, assume that
and
.
The smaller trapezoid has area while the larger trapezoid must have area
. We have the equation
.
= 10, and our answer is
~Arcticturn
Video Solution
https://www.youtube.com/watch?v=fsNJbC3hGtk ~David
See Also
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 |
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