Area of largesz equilateral triangle inside square

Question

What is the area of the largest equilateral triangle which fits inside a square of side a? 

Answer 1

Hi Chek,

 

I have a different approach and a different answer than the other tutor.

 

Let's imagine a square of side a. By symmetry, let me draw imagine an equilateral triangle such that the upper vertex touches the midpoint of the higher horizontal leg of the square, imagining that I draw it such that the upper and lower legs are parallel to a 8 1/2 X 11 sheet of paper, and that the other sides are perpendicular to these upper and lower sides. So, the base of my equilateral triangle should be parallel to the lower horizontal leg.

 

I know that each vertex of an equilateral triangle is 60 deg. So, let me draw legs of this triangle from this upper vertex until I intersect the side legs of the square.

 

 

        -------------------------------------------

        |                       /\                            |

        |                     / | \                          |

        |                   /   |   \                        |

        |                 /     | θ  \                      |

        |               /       |       \                    |

        |             /         |         \                  |

        |           /           |           \                |

        |         /             | h          \              |

        |      /                |               \            |

             /                  |                 \

           /                    |                   \

           ________________________

                   a/2               a/2

 

If we draw out this equilateral triangle, we can draw a vertical down the center as I show, and the base leg on either side of the vertical is of size a/2, where the entire side of the square is a/2 + a/2 = a. Imagine the height of this vertical to be h. The angle of each vertex of the triangle is 60 deg, so θ = 30 deg. We can find h by

 

tan(30) = opposite/adjacent = (a/2)/h,

 

so h = a/(2tan(30)) = (a/2)√3

 

So, the area of the inscribed equilateral triangle is A = 1/2 base × height = (1/2)(a)(h) = (1/2)(a)((a/2)√3)

A = (a²/4)√3

Answer 2

Draw and label of diagram!

area is (0.5)a²