Difference between revisions of "2022 AMC 8 Problems/Problem 18"
MRENTHUSIASM (talk | contribs) m (→Solution) |
MRENTHUSIASM (talk | contribs) |
||
| Line 7: | Line 7: | ||
==Solution== | ==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 <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{AC\cdot BD}{2}=20,</math> so the area of the rectangle is <math>20\cdot2=\boxed{\textbf{(C) } 40}.</math> | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
Revision as of 12:31, 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. 
