2005 AMC 10B Problems/Problem 23
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 | |
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