An equilateral triangle is constructed on each side of a square with side length 2 as shown below. The four outer vertices are then joined to form a large square. Find the side length of the large square.
Originally I got the answer of 2√2, but that was incorrect (just a word of warning).
If you draw the altitude of that top triangle, you create a 1, sqrt(3), 2 right triangle.
That altitude is sqrt(3), then coming on down, the line through the square is 2,
then you've got another sqrt(3) at the bottom. So the diagonal is 2 + (2)*sqrt(3).
Let's simplify that to 2 * (1 + sqrt(3))
The diagonal of a square is the square root of 2 times a side.
So, a side is the diagonal divided by the square root of 2.
The side of the large square:
2 * (1 + sqrt(3))
————————
sqrt(2)
You could divide that sqrt(2) on the bottom into the 2 on the top, to
get (sqrt(2) * (1 + sqrt(3)) and multiply that out, do what you want.
.
In th top triangle, the angle of the large square is 90 degrees.
The angle of the equilateral triangle ==60 and splits the right angle into 3 angles:
[90 - 60] / 2 ==15 degrees - the size of the 2 smaller angles
So, on any of the 4 sides of the larger square, you have 4 highly obtuse isosceles triangles:
With sides: 2, 2 and the large obtuse angle between them==150 degrees
Use the law of cosines to get the side of the square==3.864
