Question

nd the surface area of the pyramid. The faces of the triangular pyramid are equilateral triangles.
40 mm
40 mm
34.6 mm
Question content area bottom
Part 1
The surface area is
  
2292
mm cubed .
mm cubed .
mm squared .
mm.
​(Type an integer or a​ decimal.)

Answer 1

I’m missing the full picture/which edges are which. Could you upload the diagram or say exactly which three edges are 40 mm and which is 34.6 mm (for example: “the three edges from the apex to the base vertices are 40, 40, 34.6” or “the three lateral edges are 40 and the base edges are 34.6,” etc.)?

How I’ll solve it once you confirm:
- Find the area of each triangular face, then add them.
- If a face is equilateral with side s: area = (√3/4) s^2.
- If a face is not equilateral (three side lengths known): use Heron’s formula
A = sqrt[p(p−a)(p−b)(p−c)] where p = (a+b+c)/2.

Some quick example results so you can check which matches your diagram:
- If the pyramid is a regular tetrahedron with every edge 40 mm: SA = 4·(√3/4·40^2) = 1600·√3 ≈ 2771.3 mm^2.
- If every edge were 34.6 mm: SA = 1197.16·√3 ≈ 2073.0 mm^2.
- If the three lateral faces are isosceles triangles with sides 40, 40, 34.6 (area ≈ 624.0 mm^2 each) and the base is equilateral 34.6 (area ≈ 518.8 mm^2), total SA ≈ 2390.8 mm^2.

You mentioned 2292 mm^2 — that doesn’t match any of the common interpretations above. Tell me which edges correspond to the 40 and 34.6 labels and I’ll compute the exact surface area.