Difference between revisions of "2022 AMC 8 Problems/Problem 18"
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If a rectangle has area <math>K,</math> then the area of the quadrilateral formed by its midpoints is <math>\frac{K}{2}.</math> | If a rectangle has area <math>K,</math> then the area of the quadrilateral formed by its midpoints is <math>\frac{K}{2}.</math> | ||
− | Define points <math>A,B,C,</math> and <math>D</math> as Solution 1 does. Since <math>A,B,C,</math> and <math>D</math> are the midpoints of the rectangle, the rectangle's area is <math>2[ABCD].</math> Now, note that <math>ABCD</math> is a parallelogram since <math>AB=CD</math> and <math>AB\parallel CD.</math> As the parallelogram's | + | Define points <math>A,B,C,</math> and <math>D</math> as Solution 1 does. Since <math>A,B,C,</math> and <math>D</math> are the midpoints of the rectangle, the rectangle's area is <math>2[ABCD].</math> Now, note that <math>ABCD</math> is a parallelogram since <math>AB=CD</math> and <math>AB\parallel CD.</math> As the parallelogram's height from <math>D</math> to <math>\overline{AB}</math> is <math>4</math> and <math>AB=5,</math> its area is <math>4\cdot5=20.</math> Therefore, the area of the rectangle is <math>20\cdot2=\boxed{\textbf{(C) } 40}.</math> |
~Fruitz | ~Fruitz |
Revision as of 13:19, 29 January 2022
Contents
Problem
The midpoints of the four sides of a rectangle are and
What is the
area of the rectangle?
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 and
Note that
and
are the vertices of a rhombus whose diagonals have lengths
and
It follows that the area of rhombus
is
so the area of the rectangle is
~MRENTHUSIASM
Solution 2
If a rectangle has area then the area of the quadrilateral formed by its midpoints is
Define points and
as Solution 1 does. Since
and
are the midpoints of the rectangle, the rectangle's area is
Now, note that
is a parallelogram since
and
As the parallelogram's height from
to
is
and
its area is
Therefore, the area of the rectangle is
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.