An equilateral triangle has sides that are 10 feet long. What is the height of this triangle?

Refer to the diagram below.
ABC is an equilateral triangle with each side 10 feet long.
AD is the height of ΔABC.

BD = DC = (10 ft) / 2 = 5 ft

In ΔABD :
AD² + BD² = AB² (Pythagoras theorem)
AD² + (5 ft)² = (10 ft)²
AD² = (100 - 25) ft²
AD² = 75 ft²
AD = √75 ft
AD = 5√3 ft

We can calculate the height of an equilateral triangle with the following formula:
h = (√3/2)*a
where h = height of the triangle
a = length of each side of the triangle

Using this formula, we have:
h = (√3/2)*10
h = 10√3/2
h = 5√3
h = 8.66 feet (or 2.64 metres)

Therefore, the height of the triangle is 8.66 feet (or 2.64 metres).

Hope this helps.
Thanks for reading.
Cheers!

The altitude (whose length is the height) is the side of a 30-60-90 triangle
the hypotenuse is 10
the altitude is opposite the 60 degree angle
sin =opp/hyp
sin(60) = opp/10
opp = 10*sin(60) = 10*sqrt(3) /2 = 5*sqrt(3)

altitude = height = 5*sqrt(3) = approx. 8.660254038 feet
so exactly 5*sqrt(3) feet
or
approx. 8.660254038 feet

==== Method 2
the altitude (length = height) bisects the 3rd side of the triangle
SO the altitude is one leg of a triangle with hypotenuse 10 and other side 10/2 =5
h^2 + 5^2 = 10^2
h^2 + 25 = 100
h^2 = 100-25 = 75
h = sqrt(75)= sqrt(25)*sqrt(3) = 5 *sqrt(3) feet or approx.
approx. 8.660254038 feet

sin 60⁰ = x / 10
x = 10 sin 60⁰
x = 5 √3 ft is the height of triangle

If you drop a perpendicular from its apex, it will be the height. It will intersect the base in its centre.

You now have two right angled triangles with hypotenuse 10 and base 5.

Use Pythagoras:
10^2 = 5^2 + h^2
100 = 25 + h^2
h^2 = 100 - 25
h^2 = 75
h = sq rt 75
h = 8.66 ft

Let h = height.
5^2 + h^2 = 10^2
25 + h^2 = 100
h^2 = 75
h = √75 = 5√3ft

The height divides the equilateral triangle into 2 equal right triangles and it can be calculated as follows:
h = sqrt(10^2 - 5^2)
h = sqrt(75)
h = 5sqrt(3) ft

5 sqrt 3 feet