+0  
 
0
117
5
avatar

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

 Feb 8, 2020
 #1
avatar
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.

 Feb 8, 2020
 #2
avatar
+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

                                                                                                  s2 = 2 + sqrt 3                    s2  is also the area of the larger square

 Feb 8, 2020
 #3
avatar+531 
+2

Equilateral triangle height  h = sin( 60°) = 0.866

Small square side     a = 1

Large square area   A = ? laugh

 Feb 8, 2020
edited by Dragan  Feb 8, 2020
edited by Dragan  Feb 8, 2020
edited by Dragan  Feb 8, 2020
 #4
avatar+64 
0

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

 #5
avatar+2847 
0

probably not the whole answers, maybe just delete part of the answers

CalculatorUser  Feb 9, 2020
edited by CalculatorUser  Feb 9, 2020

30 Online Users

avatar
avatar
avatar