2022 AMC 8 Problems/Problem 18

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.

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 area of rhombus $ABCD$ is $\frac{4\sqrt{5}\cdot2\sqrt{5}}{2}=20,$ 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 $A,B,C,$ 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

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

~Interstigation

See Also

2022 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AJHSME/AMC 8 Problems and Solutions

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