the base of a right pyramid is an equilateral triangle of side 6cm.if height ishalf of its slant height find h

Answer 1

Draw a line from the middle of an edge of the base into the center, and another from one of the angles to the center.

That would be a 30-60-90 triangle where the longer side is 6/2 = 3. So the shorter side of that triangle would be 3/sqrt(3) = sqrt(3).

Now you have a new right triangle where the base is the line we just drew, the hypotenuse is the slant height, and the height of the pyramid is one half the hypotenuse.

You can either use Pyth Thm, or just realize that whenever one side of a right triangle is 1/2 the hypotenuse, that also is a 30-60-90 triangle. So the longer side is sqrt(3), which we just calculated, and the shorter side, which is the height of the pyramid, would be sqrt(3)/sqrt(3), or 1.

Answer 2

So we need to know the length from the corner to the center of the base of the pyramid which is exactly half the distance from the corner to the side of the pyramid.

We know that with an equilateral triangle, when we bisect the angle, a right triangle is formed and its opposite side length is half. So using this and pythgorean equation, we can find the length from the corner to the other side of the side of the base of the pyramid. So 3^2 + x^2 = 6^2 thus the length is SQRT(27).

Thus the length from the corner to the center is SQRT(27)/2. Again using pythogrean (SQRT(27)/2)^2 + h^2 = (2h)^2. Simplifying we get h = 3/2
so the height is