Question

One of the vertices of an equilateral triangle is on the vertex of a square and two other vertices are on the not adjacent sides of the same square. Find the side of the triangle if the side of the square is 10 ft.

Answer 1

10√6 - 10√2

Explanation:

The shorter leg of the small triangle created by the square and the equilatoral triangle = x

Other leg = 10

One of the triangles is a 45-45-90 triangle, so the hypotenuse, which also happens to be a side of the equilatoral triangle can be denoted as √2(10-x).

Using pythagoearn theorem, we know that relative to the other 2 of the 15-75-90 triangles, their hypotenueses which also happen to be sides of the equilatoral triangle are √(100+x^2)

So using transitivity, we can create this equation:

√(100+x^2)=√2(10-x)

Solving this using quadratic equations, we get that x= 20+10√3, x=20-10√3

The first equation can't be true since x has to be <10, so that means

x=20-10√3

We has substitute this into another equation to get a side length of the equilatoral triangle.

The answer is 10√6 - 10√2

Answer 2

• 10.35 ft

Explanation:

Refer to attachment

Let the side of the triangle be a

The line segment x as per Pythagorean is: √(a² - 10²)

Therefore the vertices of the triangle are equidistant from the vertex of the square, so the distance x is same for both vertices

From the triangles we have side of the triangle:

• a = √(100 + x²) and
• a = √(10 - x)² + (10 - x)²

Comparing the two we have:

• 100 + x² = 2(10 - x)²
• 100 + x² = 200 - 40x + 2x²
• x² - 40x + 100 = 0
• x² - 2*20x + 20² = 300
• (x - 20)² = 300
• x - 20 = ±10√3
• x = 20 ± 10√3
• x = 20 ± 17.32
• x = 37.32 - this is excluded as it can't be greater than side of the square
• x = 2.68 ft

So the value of a is:

• a = √(100 + x²)
• a = √100 + 2.68² = √107.1824 = 10.35 ft