A square with side length 1 is drawn, along with four equilateral triangles, one on each side of the square. Four vertices of the triangles are connected to form a larger square, as shown below. Find the area of the larger square.

https://latex.artofproblemsolving.com/a/c/f/acf7dc68fda04761f93b7bb7207a6455cab20e87.png

Guest Feb 8, 2020

#1**0 **

The area of the small square is 1, the area of each equilateral triangle is sqrt(3)/4, and the area of each thin triangle is (sqrt(3) - 1)/2. So the total area is 1 + 4*sqrt(3)/4 + 4*(sqrt(3) - 1)/2 = 3*sqrt(3) - 1.

Guest Feb 8, 2020

#2**+1 **

Height of equilateral triangle = s * sqrt 3 /2 s = 1 in this case

Diagonal of larger square = small square height + 2 eqilateral triangle heights

= 1 + 2 * sqrt3/2

= 1 + sqrt3

Pythagorean theorom for the larger squre side length s^2 + s^2 = (1+ sqrt3)^{2}

2 s^2 = 1 + 2 sqrt3 + 3

2 s^2 = 4 + 2 sqrt 3

s^{2} = 2 + sqrt 3 s^{2} is also the area of the larger square

Guest Feb 8, 2020

#4**0 **

Melody, please delete the answers to this topic. This comes from an active AoPS homework problem. Thank you!

MathProblemSolver101 Feb 8, 2020