Difference between revisions of "2022 AMC 8 Problems/Problem 18"

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==Solution==
 
==Solution==
 
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 <math>(-3,0), (2,0), (5,4),</math> and <math>(0,4)</math> are the vertices of a square whose diagonal is <math>5.</math> The area of this square is <math>\frac{5\cdot5}{2}=\frac{25}{2},</math> so the area of the rectangle is <math>\frac{25}{2}\cdot2=\boxed{\textbf{(B) } 25}.</math>
 
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 13:17, 28 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

~MRENTHUSIASM

See Also

2022 AMC 8 (ProblemsAnswer KeyResources)
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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