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. | ||
− | Let <math>A=(-3,0), B=(2,0), C=(5,4),</math> and <math>D=(0,4).</math> Note that <math>A,B,C,</math> and <math>D</math> are the vertices of a rhombus whose diagonals have lengths <math>AC=4\sqrt{5}</math> and <math>BD=2\sqrt{5}.</math> | + | Let <math>A=(-3,0), B=(2,0), C=(5,4),</math> and <math>D=(0,4).</math> Note that <math>A,B,C,</math> and <math>D</math> are the vertices of a rhombus whose diagonals have lengths <math>AC=4\sqrt{5}</math> and <math>BD=2\sqrt{5}.</math> Since the diagonals of rhombus <math>ABCD</math> are equal to the sides of the rectangle, the rectangle's area is <math>4\sqrt{5}\cdot2\sqrt{5}=40</math>. |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
+ | |||
+ | ~Edited by jayrenfu | ||
==Solution 2== | ==Solution 2== |
Revision as of 18:32, 8 January 2023
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 Since the diagonals of rhombus are equal to the sides of the rectangle, the rectangle's area is .
~MRENTHUSIASM
~Edited by jayrenfu
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 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
Video Solution
https://youtu.be/Ij9pAy6tQSg?t=1564
~Interstigation
Video Solution by Ismail.maths
https://www.youtube.com/watch?v=JHBcnevL5_U
~Ismail.maths93
Video Solution
https://youtu.be/hs6y4PWnoWg?t=188
~STEMbreezy
Video Solution
~savannahsolver
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.