Question

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 1

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

Answer 2

It's B, 23.6 cm^3  you're welcome