Difference between revisions of "2022 AMC 8 Problems/Problem 18"
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| + | Alternatively, we can find the area of the rhombus by utilizing the [[Shoelace Theorem]]. We then get <math>ABCD=\frac{|-12-28|}{2}=20</math>. Subsequently, the area of the rectangle is <math>20\cdot2=\boxed{\textbf{(C) } 40}.</math> | ||
==Solution 2== | ==Solution 2== | ||
Revision as of 12:21, 7 February 2022
Contents
Problem
The midpoints of the four sides of a rectangle are 
and 
What is the
area of the rectangle?
Solution 1
The midpoints of the four sides of every rectangle are the vertices of a rhombus whose area is half the area of the rectangle.
Let 
and 
Note that 
and 
are the vertices of a rhombus whose diagonals have lengths 
and 
It follows that the area of rhombus 
is 
so the area of the rectangle is 
~MRENTHUSIASM
Alternatively, we can find the area of the rhombus by utilizing the Shoelace Theorem. We then get 
. Subsequently, the area of the rectangle is
Solution 2
If a rectangle has area 
then the area of the quadrilateral formed by its midpoints is 
Define points and as Solution 1 does. Since and are the midpoints of the rectangle, the rectangle's area is ![$2[ABCD].$](http://latex.artofproblemsolving.com/1/e/2/1e2ceebf98aecc6c1a157ef5aeeeacf3ad98408d.png)
Now, note that is a parallelogram since 
and 
As the parallelogram's height from to 
is 
and 
its area is 
Therefore, the area of the rectangle is
~Fruitz
See Also
| 2022 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
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