The volume of metal in the hex nut to the nearest tenth is 27.6 cm^3.
Calculations and Parameters:
Given that:
- 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
The 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
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