2005 AMC 10B Problems/Problem 23
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
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 
= 3(\frac{y+z}{2} \cdot h)$.
Simplifying, we get that$ (Error compiling LaTeX. Unknown error_msg)2x=z+3y
\frac{AB}{DC}=\frac{x}{z}$.
Dividing the first equation by$ (Error compiling LaTeX. Unknown error_msg)z
\frac{x}{z}=2+\frac{y}{z}\implies 3(\frac{x}{z})=6+3(\frac{y}{z})$.
Dividing the second equation by$ (Error compiling LaTeX. Unknown error_msg)z2(\frac{x}{z}=1+3(\frac{y}{z}$.
Now, when we subtract the top equation from the bottom, we get that$ (Error compiling LaTeX. Unknown error_msg)\frac{x}{z}=5
\boxed{5}$
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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