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

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 cmApothem of the hexagon: a=2 cmAssuming the thickness of the steel hex nut: t=2 cmVolume of metal in the hex nut: V=?V=Vp-VcVolume of the prism: VpVolume of the cylinder: Vc
Prism:Vp=Ab hAb=n L a / 2Number of the sides: n=6Side of the hexagon: LHeight of the prism: h=t=2 cmCentral angle in the hexagon: A=360°/nA=360°/6A=60°tan (A/2)=(L/2) / atan (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/22 sqrt(3)/3 cm = L/2Solving for L:2 (2 sqrt(3)/3 cm)=L4 sqrt(3)/3 cm = LL=4 sqrt(3)/3 cmAb=n L a / 2Ab=6 (4 sqrt(3)/3 cm)(2 cm) / 2Ab=24 sqrt(3)/3 cm^2Ab=8 sqrt(3) cm^2Vp=Ab hVp=(8 sqrt(3) cm^2)(2 cm)Vp=16 sqrt(3) cm^3Vp=16 (1.732) cm^3(1) Vp=27.712 cm^3
Cylinder:Vc=(π d^2/4) hπ=3.14d=1.6 cmHeight of the cylinder: h=t=2 cmVc=[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-VcV=27.712 cm^3 - 4.019 cm^3V=23.693 cm^3V=23.6 cm^3