Please answer only if you know for sure :) Comes from pre-calc btw
2 answers:
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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, = 2×27·(√3) + 2 ×6 ×h + 12×h = 24·h + 54·√3
The surface area the frustum = 24·h + 54·√3