Difference between revisions of "2022 AMC 8 Problems/Problem 18"
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+ | ==Solution 2(Parallelograms, More Detailed)== | ||
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+ | Note that if a rectangle has area <math>A</math>, the area of the quadrilateral formed by its midpoints is <math>\frac{A}{2}</math>. Since <math>A, B, C, D</math> are the midpoints of the rectangle, its area would be <math>2\cdot[ABCD]</math>. Now, note that <math>ABCD</math> is a parallelogram since <math>AB=CD</math> and <math>AB||CD</math>. Note that the parallelogram's altitude from <math>D</math> to <math>AB</math> is <math>4</math> and <math>AB=5</math>. Thus, its area is <math>4\times 5=20</math>. The area of the rectangle follows as <math>20\cdot2=\boxed{\textbf{(C) } 40}.</math> | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2022|num-b=17|num-a=19}} | {{AMC8 box|year=2022|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:20, 29 January 2022
Problem
The midpoints of the four sides of a rectangle are and
What is the
area of the rectangle?
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 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(Parallelograms, More Detailed)
Note that if a rectangle has area , the area of the quadrilateral formed by its midpoints is
. Since
are the midpoints of the rectangle, its area would be
. Now, note that
is a parallelogram since
and
. Note that the parallelogram's altitude from
to
is
and
. Thus, its area is
. The area of the rectangle follows as
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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