Difference between revisions of "2019 AMC 10B Problems/Problem 8"

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==Solution==
 
==Solution==
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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).
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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!!).
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We can then compute the area of the two triangles using the base-height-area relationship and get 2 * (sqrt 3) / 2 = sqrt 3.
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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).
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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.
  
 
==See Also==
 
==See Also==

Revision as of 16:40, 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.

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
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

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