Difference between revisions of "2025 AMC 8 Problems/Problem 10"
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==Problem== | ==Problem== | ||
In the figure below, <math>ABCD</math> is a rectangle with sides of length <math>AB = 5</math> inches and <math>AD</math> = 3 inches. Rectangle <math>ABCD</math> is rotated <math>90^\circ</math> clockwise around the midpoint of side <math>DC</math> to give a second rectangle. What is the total area, in square inches, covered by the two overlapping rectangles? | In the figure below, <math>ABCD</math> is a rectangle with sides of length <math>AB = 5</math> inches and <math>AD</math> = 3 inches. Rectangle <math>ABCD</math> is rotated <math>90^\circ</math> clockwise around the midpoint of side <math>DC</math> to give a second rectangle. What is the total area, in square inches, covered by the two overlapping rectangles? | ||
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| + | [[File:Amc8_2025_prob10.PNG]] | ||
<math>\textbf{(A)}\ 21 \qquad \textbf{(B)}\ 22.25 \qquad \textbf{(C)}\ 23 \qquad \textbf{(D)}\ 23.75 \qquad \textbf{(E)}\ 25</math> | <math>\textbf{(A)}\ 21 \qquad \textbf{(B)}\ 22.25 \qquad \textbf{(C)}\ 23 \qquad \textbf{(D)}\ 23.75 \qquad \textbf{(E)}\ 25</math> | ||
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==Solution 1== | ==Solution 1== | ||
The area of each rectangle is <math>5 \cdot 3 = 15</math>. Then the sum of the areas of the two regions is the sum of the areas of the two rectangles, minus the area of their overlap. To find the area of the overlap, we note that the region of overlap is a square, each of whose sides have length <math>2.5</math> (as they are formed by the midpoint of one of the long sides and a vertex). Then the answer is <math>15+15-2.5^2=\boxed{\textbf{(D)}~23.75}</math>. ~cxsmi | The area of each rectangle is <math>5 \cdot 3 = 15</math>. Then the sum of the areas of the two regions is the sum of the areas of the two rectangles, minus the area of their overlap. To find the area of the overlap, we note that the region of overlap is a square, each of whose sides have length <math>2.5</math> (as they are formed by the midpoint of one of the long sides and a vertex). Then the answer is <math>15+15-2.5^2=\boxed{\textbf{(D)}~23.75}</math>. ~cxsmi | ||
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| + | ==Video Solution 1 by SpreadTheMathLove== | ||
| + | https://www.youtube.com/watch?v=jTTcscvcQmI | ||
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| + | ==Video Solution by Thinking Feet== | ||
| + | https://youtu.be/PKMpTS6b988 | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2025|num-b=9|num-a=11}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 21:44, 30 January 2025
Contents
Problem
In the figure below, 
is a rectangle with sides of length 
inches and 
= 3 inches. Rectangle is rotated 
clockwise around the midpoint of side 
to give a second rectangle. What is the total area, in square inches, covered by the two overlapping rectangles?
Solution 1
The area of each rectangle is 
. Then the sum of the areas of the two regions is the sum of the areas of the two rectangles, minus the area of their overlap. To find the area of the overlap, we note that the region of overlap is a square, each of whose sides have length 
(as they are formed by the midpoint of one of the long sides and a vertex). Then the answer is 
. ~cxsmi
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution by Thinking Feet
See Also
| 2025 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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. 
