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 1== | == Solution 1== |
Revision as of 16:35, 16 December 2021
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
Hence, the answer is
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
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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