Difference between revisions of "2022 AMC 8 Problems/Problem 18"
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The midpoints of the four sides of every rectangle are the vertices of a rhombus whose area is half the area of the rectangle. | The midpoints of the four sides of every rectangle are the vertices of a rhombus whose area is half the area of the rectangle. | ||
| − | Note that <math>A=(-3,0), B=(2,0), C=(5,4),</math> and <math>D=(0,4)</math> are the vertices of a rhombus whose diagonals have lengths <math>AC=\sqrt{80}</math> and <math>BD=\sqrt{20}.</math> It follows that the area of rhombus <math>ABCD</math> is <math>\frac{ | + | Note that <math>A=(-3,0), B=(2,0), C=(5,4),</math> and <math>D=(0,4)</math> are the vertices of a rhombus whose diagonals have lengths <math>AC=\sqrt{80}</math> and <math>BD=\sqrt{20}.</math> It follows that the area of rhombus <math>ABCD</math> is <math>\frac{\sqrt{80}\cdot\sqrt{20}}{2}=20,</math> so the area of the rectangle is <math>20\cdot2=\boxed{\textbf{(C) } 40}.</math> |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
Revision as of 12:38, 28 January 2022
Problem
The midpoints of the four sides of a rectangle are 
and 
What is the
area of the rectangle?
Solution
The midpoints of the four sides of every rectangle are the vertices of a rhombus whose area is half the area of the rectangle.
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
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
