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. | ||
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| + | Note that <math>(-3,0), (2,0), (5,4),</math> and <math>(0,4)</math> are the vertices of a square whose diagonal is <math>5.</math> The area of this square is <math>\frac{5\cdot5}{2}=\frac{25}{2},</math> so the area of the rectangle is <math>\frac{25}{2}\cdot2=\boxed{\textbf{(B) } 25}.</math> | ||
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| + | ~MRENTHUSIASM | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2022|num-b=17|num-a=19}} | {{AMC8 box|year=2022|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:16, 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 square whose diagonal is 
The area of this square 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. 
