Difference between revisions of "2019 AMC 8 Problems/Problem 2"
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== Video Solution == | == Video Solution == | ||
− | Solution detailing how to solve the problem:https://www.youtube.com/watch?v=eEqtoI8BQKE&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=3 | + | Solution detailing how to solve the problem: https://www.youtube.com/watch?v=eEqtoI8BQKE&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=3 |
==See also== | ==See also== |
Latest revision as of 21:00, 12 May 2021
Contents
Problem
Three identical rectangles are put together to form rectangle , as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 5 feet, what is the area in square feet of rectangle ?
Solutions
Solution 1
We can see that there are rectangles lying on top of the other and that is the same as the length of one rectangle. Now we know that the shorter side is , if we take the information of the problem, and the bigger side is , if we do . Now we get the sides of the big rectangles being and , so the area is . ~avamarora
Solution 2
Using the diagram we find that the larger side of the small rectangle is times the length of the smaller side. Therefore, the longer side is . So the area of the identical rectangles is . We have 3 identical rectangles that form the large rectangle. Therefore the area of the large rectangle is . ~~fath2012
Solution 3
We see that if the short sides are 5, the long side has to be because the long side is equal to the 2 short sides and because the rectangles are congruent. If that is to be, then the long side of the BIG rectangle(rectangle ) is because long side + short side of the small rectangle is . The short side of rectangle is because it is the long side of the short rectangle. Multiplying and together gets us which is . ~~mathboy282
Video Solution
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=eEqtoI8BQKE&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=3
See also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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.