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.

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

2022 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AJHSME/AMC 8 Problems and Solutions

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2022 AMC 8 Problems/Problem 18

Contents

Problem

Solution 1

Solution 2

See Also