Difference between revisions of "2019 AMC 8 Problems/Problem 2"
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Revision as of 23:44, 19 November 2019
Solution 1
We know that the length of the shorter side of the 3 identical rectangles are all 5 so we can use that by seeing that the longer side of the right rectangle is the same as 2 of the shorter sides of the other 2 left rectangles. This means that 
which is the longer side of the right rectangle, and because all the rectangles are congruent, we see that each of the rectangles have a longer side of 10 and a shorter side of 5. Now the bigger rectangle has a shorter length of 10(because the shorter side of the bigger rectangle is the bigger side of the shorter rectangle, which is 10) and so the bigger side of the bigger rectangle is the bigger side of the smaller rectangle + the smaller side of the smaller rectangle, which is 
. Thus, the area is 
for choice 
~~Saksham27
See Also
| 2019 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 2 | |
| 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. 
