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2022 AMC 8 Problems/Problem 18: Difference between revisions

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==Solution 1==
==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.
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 <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> It follows that the area of rhombus <math>ABCD</math> is <math>\frac{4\sqrt{5}\cdot2\sqrt{5}}{2}=20,</math> so the area of the rectangle is <math>20\cdot2=\boxed{\textbf{(C) } 40}.</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> It follows that the dimensions of the rectangle are <math>4\sqrt{5}</math> and <math>2\sqrt{5},</math> so the area of the rectangle is <math>4\sqrt{5}\cdot2\sqrt{5}=\boxed{\textbf{(C) } 40}.</math>


~MRENTHUSIASM
~MRENTHUSIASM
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~Fruitz
~Fruitz
==Video Solution (A Clever Explanation You’ll Get Instantly)==
https://youtu.be/tYWp6fcUAik?si=V8hv_zOn_zYOi9E5&t=2153
~hsnacademy
==Video Solution by Pi Academy (🚀Fast and EZ!🚀)==
https://youtu.be/Bxp0coLo5r4?si=49fvyWZBrsutgpBt
~ AMC8 Learning
==Video Solution==
https://youtu.be/5Vti6QS7TfU
~Education, the Study of Everything
==Video Solution==
https://youtu.be/Ij9pAy6tQSg?t=1564
~Interstigation
==Video Solution==
https://youtu.be/9-TlEV5SGqM
~savannahsolver


==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}}
[[Category:Introductory Geometry Problems]]

Latest revision as of 23:16, 4 August 2025

Problem

The midpoints of the four sides of a rectangle are $(-3,0), (2,0), (5,4),$ and $(0,4).$ What is the area of the rectangle?

$\textbf{(A) } 20 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 40 \qquad \textbf{(D) } 50 \qquad \textbf{(E) } 80$

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 $A=(-3,0), B=(2,0), C=(5,4),$ and $D=(0,4).$ Note that $A,B,C,$ and $D$ are the vertices of a rhombus whose diagonals have lengths $AC=4\sqrt{5}$ and $BD=2\sqrt{5}.$ It follows that the dimensions of the rectangle are $4\sqrt{5}$ and $2\sqrt{5},$ so the area of the rectangle is $4\sqrt{5}\cdot2\sqrt{5}=\boxed{\textbf{(C) } 40}.$

~MRENTHUSIASM

Solution 2

If a rectangle has area $K,$ then the area of the quadrilateral formed by its midpoints is $\frac{K}{2}.$

Define points $A,B,C,$ and $D$ as Solution 1 does. Since $A,B,C,$ and $D$ are the midpoints of the rectangle, the rectangle's area is $2[ABCD].$ Now, note that $ABCD$ is a parallelogram since $AB=CD$ and $\overline{AB}\parallel\overline{CD}.$ As the parallelogram's height from $D$ to $\overline{AB}$ is $4$ and $AB=5,$ its area is $4\cdot5=20.$ Therefore, the area of the rectangle is $20\cdot2=\boxed{\textbf{(C) } 40}.$

~Fruitz

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/tYWp6fcUAik?si=V8hv_zOn_zYOi9E5&t=2153 ~hsnacademy

Video Solution by Pi Academy (🚀Fast and EZ!🚀)

https://youtu.be/Bxp0coLo5r4?si=49fvyWZBrsutgpBt

~ AMC8 Learning

Video Solution

https://youtu.be/5Vti6QS7TfU

~Education, the Study of Everything

Video Solution

https://youtu.be/Ij9pAy6tQSg?t=1564

~Interstigation

Video Solution

https://youtu.be/9-TlEV5SGqM

~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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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