) A área lateral de um tronco de cone é 40 cm². Os raios das b...

1- A) 1,67cm

2- D) 616π cm²

Explicação passo-a-passo:

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1) Al = π.g. (R + r)

Al = π.g. (R + r)40 = 3 . g . (6 + 2)

Al = π.g. (R + r)40 = 3 . g . (6 + 2)40 = 3.g.8

Al = π.g. (R + r)40 = 3 . g . (6 + 2)40 = 3.g.840 = 24.g

Al = π.g. (R + r)40 = 3 . g . (6 + 2)40 = 3.g.840 = 24.g40/24 = g

Al = π.g. (R + r)40 = 3 . g . (6 + 2)40 = 3.g.840 = 24.g40/24 = gG = 1,67 cm

2) g² = (R – r)² + h²

g² = (R – r)² + h²g² = (16 – 10)² + 82

g² = (R – r)² + h²g² = (16 – 10)² + 82g² = 36 + 64

g² = (R – r)² + h²g² = (16 – 10)² + 82g² = 36 + 64g = √100

g² = (R – r)² + h²g² = (16 – 10)² + 82g² = 36 + 64g = √100g = 10

Al = π.g. (R + r)

Al = π.g. (R + r)Al = π .10. (10 + 16)

Al = π.g. (R + r)Al = π .10. (10 + 16)Al = 260π cm²

Ab = π.r²

Ab = π.r²Ab = π.10²

Ab = π.r²Ab = π.10²Ab = 100π cm²

AB = π.R²

AB = π.R²AB = π.16²

AB = π.R²AB = π.16²AB = 256π cm²

At = AB + Ab + Al

At = AB + Ab + AlAt = 256π + 100π + 260π

At = AB + Ab + AlAt = 256π + 100π + 260πAt = 616π cm²