Difference between revisions of "2022 AMC 8 Problems/Problem 18"
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+ | ==Video Solution by Math-X (First understand the problem!!!)== | ||
+ | https://youtu.be/oUEa7AjMF2A?si=8oNmauhAnW5T5vEX&t=3182 | ||
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+ | ~Math-X | ||
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+ | ==Video Solution (🚀Just 2 min!🚀)== | ||
+ | https://youtu.be/5Vti6QS7TfU | ||
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+ | ~Education, the Study of Everything | ||
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==Video Solution== | ==Video Solution== | ||
https://youtu.be/Ij9pAy6tQSg?t=1564 | https://youtu.be/Ij9pAy6tQSg?t=1564 |
Latest revision as of 01:46, 4 November 2024
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: Note that the diagonals of the rhombus have the same lengths as the sides of the rectangle.
Let and Note that and are the vertices of a rhombus whose diagonals have lengths and It follows that the dimensions of the rectangle are and so the area of the rectangle is
~MRENTHUSIASM
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 by Math-X (First understand the problem!!!)
https://youtu.be/oUEa7AjMF2A?si=8oNmauhAnW5T5vEX&t=3182
~Math-X
Video Solution (🚀Just 2 min!🚀)
~Education, the Study of Everything
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.