Question

can you find the height of the base to the nearest tenth when the surface area is 100in, the base is 6in, and the slant height is 8in of a regular triangular pyramid?

Answer 1

I need to clarify what you mean by "height of the base." Do you mean
- the altitude of the triangular base (the height of the equilateral triangle), or
- the vertical height of the pyramid (apex to base plane)?

Also, I assume "surface area is 100 in" means 100 in^2 (total surface area) and "base is 6 in" means the base is an equilateral triangle with side length 6 in.

Two useful results:

1) If you want the base triangle's altitude:
- altitude = (√3/2)·s = (√3/2)·6 = 3√3 ≈ 5.2 in (to the nearest tenth).

2) If you want the pyramid's vertical height:
- for a regular triangular pyramid the slant height l relates to the pyramid height h by l^2 = h^2 + (inradius of base)^2.
- inradius of an equilateral triangle with side 6 is s·√3/6 = √3 ≈ 1.732.
- with l = 8: h = √(8^2 − (√3)^2) = √(64 − 3) = √61 ≈ 7.8 in (to the nearest tenth).

Note: with s = 6 and l = 8 the total surface area is
base area + lateral area = (√3/4)·36 + 3·(1/2)·6·8 = 9√3 + 72 ≈ 87.6 in^2,
so the given total surface area 100 in^2 is inconsistent with s = 6 and l = 8.

If you actually mean total surface area = 100 in^2 and side s = 6, the slant height would need to be
l = (100 − 9√3)/9 ≈ 9.4 in, giving a pyramid height ≈ 9.2 in.

Tell me which "height" you want (base altitude or pyramid height) and whether SA = 100 is fixed; I can proceed from there.