Question

All of the edges of a regular triangular pyramid have length 12. A frustum of the pyramid has height half of that of the pyramid. What is the:
1) Volume of the frustum?

2) Surface area of the frustum?

Answer 1

Answer:

1) The volume of the frustum is 27·(√3)·h

2) The surface area the frustum is 24·h + 54·√3

Step-by-step explanation:

1) A frustum of the regular triangular prism is a portion of the triangular prism

The length of the edge of the triangle prism = 12

The area of the triangular face, A = (√3/4)·a²

∴ A = (√3/4)·12² = 36·√3

When

The volume of the triangular prism, V = A·h = (36·√3)·h

By triangle proportionality theorem, the cross sectional area of the top half of the triangular prism =  (√3/4)·6² = 9·(√3)

The volume of the top half, V₂ = 9·(√3)·h

The volume of the frustum = V - V₂ = (36·√3)·h - 9·(√3)·h = 27·(√3)·h

The volume of the frustum = 27·(√3)·h

2) The surface area of the triangular side of the frustum is given as follows;

A = 36·√3- 9·(√3) = 27·(√3)

The surface area the frustum, $F_A$ = 2×27·(√3) + 2 ×6 ×h + 12×h = 24·h + 54·√3

The surface area the frustum = 24·h + 54·√3