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# A square with side length is drawn, along with four equilateral triangles, one on each side of the square. Four vertices of the ...

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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

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

Small square side     a = 1

Large square area   A = ? Feb 8, 2020
edited by Dragan  Feb 8, 2020
edited by Dragan  Feb 8, 2020
edited by Dragan  Feb 8, 2020
#4
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Melody, please delete the answers to this topic. This comes from an active AoPS homework problem. Thank you!

#5
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probably not the whole answers, maybe just delete part of the answers

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