Difference between revisions of "2019 AMC 10B Problems/Problem 8"
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Since there are four of these squares, we multiply this by 4 to get 4(3 - sqrt 3) = 12 - 4sqrt3 as our answer. This is choice B. | Since there are four of these squares, we multiply this by 4 to get 4(3 - sqrt 3) = 12 - 4sqrt3 as our answer. This is choice B. | ||
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==See Also== | ==See Also== | ||
Revision as of 16:41, 14 February 2019
Problem
Solution
We notice that the square can be split into 4 congruent squares with the altitude of the equilateral triangle being the side of the square. Therefore, the area of each shaded part that resides within a square is the total area of the square subtracted from each triangle (Note that it HAS been split in 1/2).
When we split a equilateral triangle in 1/2, we get 2 triangles with a 30-60-90 relationship. Therefore, we get that the altitude and a side length of a square is sqrt 3 (Please help with LaTex!!).
We can then compute the area of the two triangles using the base-height-area relationship and get 2 * (sqrt 3) / 2 = sqrt 3.
The area of the small squares is the altitude square = sqrt 3 ^ 2 = 3. Therefore, the area of the shaded region in each of the four squares is (3 - sqrt 3).
Since there are four of these squares, we multiply this by 4 to get 4(3 - sqrt 3) = 12 - 4sqrt3 as our answer. This is choice B.
~ Awesome2.1
See Also
| 2019 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
