Difference between revisions of "2005 AMC 10B Problems/Problem 23"
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Note that the heights of trapezoids <math>ABEF</math> & <math>FECD</math> are the same. Mark the height to be <math>h</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>\ | + | Then, we have that <math>\dfrac{x+y}{2}\cdot h=2(\dfrac{y+z}{2} \cdot h)</math>. |
From this, we get that <math>x=2z+y</math>. | From this, we get that <math>x=2z+y</math>. | ||
− | We also get that <math>\ | + | We also get that <math>\dfrac{x+z}{2} \cdot h= 3(\dfrac{y+z}{2} \cdot h)</math>. |
Simplifying, we get that <math>2x=z+3y</math> | Simplifying, we get that <math>2x=z+3y</math> | ||
− | Notice that we want <math>\ | + | Notice that we want <math>\dfrac{AB}{DC}=\frac{x}{z}</math>. |
− | Dividing the first equation by <math>z</math>, we get that <math>\ | + | Dividing the first equation by <math>z</math>, we get that <math>\dfrac{x}{z}=2+\dfrac{y}{z}\implies 3(\dfrac{x}{z})=6+3(\dfrac{y}{z})</math>. |
− | Dividing the second equation by <math>z</math>, we get that <math>2(\ | + | Dividing the second equation by <math>z</math>, we get that <math>2(\dfrac{x}{z})=1+3(\dfrac{y}{z})</math>. |
− | Now, when we subtract the top equation from the bottom, we get that <math>\ | + | Now, when we subtract the top equation from the bottom, we get that <math>\dfrac{x}{z}=5</math> |
Hence, the answer is <math>\boxed{5}</math> | Hence, the answer is <math>\boxed{5}</math> |
Revision as of 21:21, 26 December 2020
Contents
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 height 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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.