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A steel hex nut has two regular hexagonal bases and a cylindrical hole with a diameter of 1.6 centimeters through the middle. The apothem of the hexagon is 2 centimeters. What is the volume of metal in the hex nut, to the nearest tenth? Use 3.14 for π. 21.1 cm3 23.6 cm3 27.6 cm3 31.6 cm3

  • Professor1994:

    Answer:

    Subtracting the volume of the cylinder from the volume of the prism, the volume of metal in the hex nut to the nearest tenth is 23.6 cm^3 (second option)

    Step-by-step explanation:

    Diameter of the cylinder: d=1.6 cm

    Apothem of the hexagon: a=2 cm

    Assuming the thickness of the steel hex nut: t=2 cm

    Volume of metal in the hex nut: V=?

    V=Vp-Vc

    Volume of the prism: Vp

    Volume of the cylinder: Vc


    Prism:

    Vp=Ab h

    Ab=n L a / 2

    Number of the sides: n=6

    Side of the hexagon: L

    Height of the prism: h=t=2 cm

    Central angle in the hexagon: A=360°/n

    A=360°/6

    A=60°

    tan (A/2)=(L/2) / a

    tan (60°/2)=(L/2) / (2 cm)

    tan 30° = (L/2) / (2 cm)

    sqrt(3)/3=(L/2) / (2 cm)

    Solving for L/2:

    (2 cm) sqrt(3)/3 = L/2

    2 sqrt(3)/3 cm = L/2

    Solving for L:

    2 (2 sqrt(3)/3 cm)=L

    4 sqrt(3)/3 cm = L

    L=4 sqrt(3)/3 cm

    Ab=n L a / 2

    Ab=6 (4 sqrt(3)/3 cm)(2 cm) / 2

    Ab=24 sqrt(3)/3 cm^2

    Ab=8 sqrt(3) cm^2

    Vp=Ab h

    Vp=(8 sqrt(3) cm^2)(2 cm)

    Vp=16 sqrt(3) cm^3

    Vp=16 (1.732) cm^3

    (1) Vp=27.712 cm^3


    Cylinder:

    Vc=(π d^2/4) h

    π=3.14

    d=1.6 cm

    Height of the cylinder: h=t=2 cm

    Vc=[3.14 (1.6 cm)^2 / 4] (2 cm)

    Vc=[3.14 (2.56 cm^2) / 4] (2 cm)

    Vc=(2.0096 cm^2) (2 cm)

    Vc=4.019 cm^3


    V=Vp-Vc

    V=27.712 cm^3 - 4.019 cm^3

    V=23.693 cm^3

    V=23.6 cm^3

  • kmanley36622:

    Answer:

    person above is correct!

    Step-by-step explanation:

    edge. 2022 :)

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