數學figure問題

FH,BH and BF are face diagonals, their lengths are (√2)(4) = 4√2 cm

A line connect mid point of BF and H is a height. Its length can be calculated as [(1/2)FB]√3 = [(1/2)(4√2)]√3 = 2 (√2)(√3)

Area = (1/2)(BF)(height) = (1/2)(4√2)[2 (√2)(√3)] = 8√3 cm² or 13.86 cm²

2014-03-09 16:21:31 補充：
a)　　Let height of the circular cone be h.

　　　10² = h² + 6²

　　　h = 8 cm

Volume = (1/3)[π(6)²](8) + (1/2)(4/3)π6³ = 96π + 144π = 240π cm³ = 754.0 cm³

2014-03-09 16:25:24 補充：
b)
　　　Triangular part : (1/2)(base)(height = (1/2)(12)(8) = 48 cm²

　　　Semi circle : (1/2)πr² = (1/2)π6² = 18π cm²

　　　Total = (48 + 18π) cm² = 104.55 cm²

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