Question

Fie ABCA'B'C' un trunchi de piramida triunghiulara regulata care are AB=36 cm, A'B'=12 cm si inaltimea OO'=12 cm Calculati aria totala si volumul

Answer 1

Răspuns:

At = 360√3 + 288√10 cm²

V = 1872√3 cm³

Explicație pas cu pas:

A'D' ⊥ B'C'

A'D' = A'B'√3/2 = 12√3/2 = 6√3cm

O'D' = A'D'/3 = 2√3 cm

AD = AB√3/2 = 36√3/2 = 18√3 cm

OD = AD/3 = 18√3/3 = 6√3 cm

ducem D'F ⊥ OD

FD = OD - O'D' = 6√3 - 2√3 = 4 cm

In ΔD'DF aplicam Pitagora

D'D = √D'F²+FD² = √144+16 = √160 = 4√10 cm

D'D = 4√10 cm = a (apotema)

At = A(B) + A(b) + A(l)

A(b) = A'B'²√3/4 = 144√3/4 = 36√3 cm²

A(B) = AB²√3/4 = 1296√3/4 = 324√3 cm²

A(l) = (P(B) + P(b))×a/2 = (3×36 + 3×12)× 4√10/2

     = (108 + 36)×2√10

     = 288√10 cm²

At = 36√3 cm² + 324√3 cm² + 288√10 cm²

    = 360√3 + 288√10 cm²

V = h(AB + Ab + √AB×Ab)/3

   = 12×[324√3 + 36√3 + √(36√3×324√3)] / 3

   = 4×(360√3 + √34992)

   = 4×(360√3 + 108√3)

   = 4×468√3

   = 1872√3 cm³