show that the altitudes of equivilent root3/2 of a side

Wouldn't be easier just to post the problem as stated? The language above is gibberish. How about

show that the altitudes of an equilateral triangle are √3/2 of a side

Draw the altitude.
Now you have a right triangle, with legs h, s/2 and hypotenuse s.

Using the Pythagorean Theorem,

h^2 + (s/2)^2 = s^2
h^2 + s^2/4 = s^2
h^2 = 3/4 s^2
h = √3/2 s

To show that the altitudes of an equilateral triangle are √3/2 times the length of a side, we can use the formula for the area of an equilateral triangle.

The area of an equilateral triangle is given by the formula:

Area = (s^2 * √3) / 4

where s is the length of a side.

Now, let's consider the altitude of the triangle, which is the line segment drawn from one vertex of the triangle to the opposite side, perpendicular to that side.

When the altitude is drawn, it creates two right-angled triangles. Each of these right-angled triangles can be thought of as half of a rectangle (with the altitude being the longer side and s/2 being the shorter side) within the triangle.

Now, let's focus on one of these right-angled triangles to find the length of the altitude.

We have a right-angled triangle with one leg measuring s/2 and the hypotenuse being s (the length of a side of the equilateral triangle). Using the Pythagorean theorem, we can find the length of the altitude (h).

Using the Pythagorean theorem:

(s/2)^2 + h^2 = s^2

Simplifying this equation:

s^2/4 + h^2 = s^2

h^2 = s^2 - s^2/4

h^2 = (4s^2 - s^2)/4

h^2 = 3s^2/4

Taking the square root of both sides:

h = √(3s^2)/2

h = (s√3)/2

Therefore, the length of the altitude of an equilateral triangle is (s√3)/2, which is √3/2 times the length of a side.