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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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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=\sqrt{80}</math> and <math>BD=\sqrt{20}.</math> It follows that the area of rhombus <math>ABCD</math> is <math>\frac{\sqrt{80}\cdot\sqrt{20}}{2}=20,</math> so the area of the rectangle is <math>20\cdot2=\boxed{\textbf{(C) } 40}.</math>
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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{\sqrt{80}\cdot\sqrt{20}}{2}=20,</math> so the area of the rectangle is <math>20\cdot2=\boxed{\textbf{(C) } 40}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 11:53, 29 January 2022

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 $A=(-3,0), B=(2,0), C=(5,4),$ and $D=(0,4)$ are the vertices of a rhombus whose diagonals have lengths $AC=\sqrt{80}$ and $BD=\sqrt{20}.$ It follows that the area of rhombus $ABCD$ is $\frac{\sqrt{80}\cdot\sqrt{20}}{2}=20,$ so the area of the rectangle is $20\cdot2=\boxed{\textbf{(C) } 40}.$

~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 $A, B, C, D$ are the midpoints of the rectangle, its area is $2\cdot[ABCD]$. Now, note that $ABCD$ is a parallelogram since $AB=CD$ and $AB\parallel CD$. As the parallelogram's altitude from $D$ to $AB$ is $4$ and $AB=5$, its area is $4\times 5=20$. Therefore, the area of the rectangle is $20\cdot2=\boxed{\textbf{(C) } 40}$.

~Fruitz

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

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