Four equilateral triangles are constructed on the sides of a square with side length 1 as shown below. The four outer vertices are then joined to form a large square. Find the area of the large square.

Guest May 3, 2020

#1**+1 **

If the side of an equilateral triangle is "a", the altitude of that triangle is: a·sqrt(3)/2.

Since each of the equilateral triangles have a base of 1, the height of each of these triangles is: sqrt(3)/2.

So the distance from one vertex of the large square to its opposite vertex is sqrt(3)/2 + 1 + sqrt(3)/2 or 1 + sqrt(3).

This is the length of each diagonal of a square.

If "d" is the length of a diagonal of a square, its area is: A = ½·d^{2}.

So, the area of this square is: ½·( 1 + sqrt(3) )^{2} or **2 + sqrt(3)**.

geno3141 May 3, 2020

#1**+1 **

Best Answer

If the side of an equilateral triangle is "a", the altitude of that triangle is: a·sqrt(3)/2.

Since each of the equilateral triangles have a base of 1, the height of each of these triangles is: sqrt(3)/2.

So the distance from one vertex of the large square to its opposite vertex is sqrt(3)/2 + 1 + sqrt(3)/2 or 1 + sqrt(3).

This is the length of each diagonal of a square.

If "d" is the length of a diagonal of a square, its area is: A = ½·d^{2}.

So, the area of this square is: ½·( 1 + sqrt(3) )^{2} or **2 + sqrt(3)**.

geno3141 May 3, 2020