Difference between revisions of "2019 AMC 10B Problems/Problem 8"
(added a solution) |
m (forgot to add credits to the solution (me)) |
||
Line 41: | Line 41: | ||
From the previous solutions, we know that the side length of the square is <math>2\sqrt{3}</math>, which means that the area of the square is <math>(2\sqrt{3})^2 = 12</math>. Since the side lengths of the equilateral triangles are integers and not expressions containing an integer and a radical, we know that the combined areas of the triangles will be in the form <math>x\sqrt{3}</math>. Therefore, the only solution that would work is <math>\boxed{\textbf{(B) }12 - 4\sqrt{3}}</math> because it is the only solution with a <math>12</math> subtracted by a multiple of the square root of <math>3</math>. | From the previous solutions, we know that the side length of the square is <math>2\sqrt{3}</math>, which means that the area of the square is <math>(2\sqrt{3})^2 = 12</math>. Since the side lengths of the equilateral triangles are integers and not expressions containing an integer and a radical, we know that the combined areas of the triangles will be in the form <math>x\sqrt{3}</math>. Therefore, the only solution that would work is <math>\boxed{\textbf{(B) }12 - 4\sqrt{3}}</math> because it is the only solution with a <math>12</math> subtracted by a multiple of the square root of <math>3</math>. | ||
+ | |||
+ | -RuiyangWu | ||
==Video Solution== | ==Video Solution== |
Revision as of 14:06, 3 January 2021
Contents
Problem
The figure below shows a square and four equilateral triangles, with each triangle having a side lying on a side of the square, such that each triangle has side length and the third vertices of the triangles meet at the center of the square. The region inside the square but outside the triangles is shaded. What is the area of the shaded region?
Solution 1
We notice that the square can be split into congruent smaller squares, with the altitude of the equilateral triangle being the side of this smaller square. Therefore, the area of each shaded part that resides within a square is the total area of the square subtracted from each triangle (which has already been split in half). #storm
When we split an equilateral triangle in half, we get two triangles. Therefore, the altitude, which is also the side length of one of the smaller squares, is . We can then compute the area of the two triangles as .
The area of the each small squares is the square of the side length, i.e. . Therefore, the area of the shaded region in each of the four squares is .
Since there are of these squares, we multiply this by to get as our answer.
Solution 2
We can see that the side length of the square is by considering the altitude of the equilateral triangle as in Solution 1. Using the Pythagorean Theorem, the diagonal of the square is thus . Because of this, the height of one of the four shaded kites is . Now, we just need to find the length of that kite. By the Pythagorean Theorem again, this length is . Now using , the area of one of the four kites is .
Solution 3
Based on the previous solutions, we know that the side length of the square is . That means that the area of the square is
We also know that the area of one of the triangles is . That means that all four triangles have a total area of
So the answer is .
Solution 4 (usage of answer choices)
From the previous solutions, we know that the side length of the square is , which means that the area of the square is . Since the side lengths of the equilateral triangles are integers and not expressions containing an integer and a radical, we know that the combined areas of the triangles will be in the form . Therefore, the only solution that would work is because it is the only solution with a subtracted by a multiple of the square root of .
-RuiyangWu
Video Solution
~IceMatrix
~savannahsolver
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.