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. A cylinder is cut out of the middle of a hexagonal prism. The hexagon has an apothem with a length of 2 centimeters and base side lengths of 2.3 centimeters. The prism has a height of 2 centimeters. The cylinder has a diameter of 1.6 centimeters. The equation for the area of a regular hexagon = one-half (apothem) (perimeter). 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:

$V=V_p-V_cV=27.712 cm^3 - 4.019 cm^3V=23.693 cm^3V=23.6 cm^3$

Step-by-step explanation:

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=?

$V=V_p-V_c$

$\texttt {Volume of the prism}: V_p\\\\\texttt {Volume of the cylinder}: V_c$

Prism:

$V_p=Ab h$

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°

$\tan \frac{A}{2} =\frac{L}{2} / a$

$\tan \frac{60}{2} =\frac{L}{2} / 2cm$

$\tan 30 =\frac{L}{2} / 2cm$

$\frac{\sqrt{3} }{3} =\frac{\frac{L}{2} }{2}$

$2\frac{\sqrt{3} }{3} =\frac{L}{2}$

$L=4\frac{\sqrt{3} }{3}$

$Ab=n *L *\frac{a}{2}$

$Ab=6(4\frac{\sqrt{3} }{3} )(2cm)/2$

$=24\frac{\sqrt{3} }{3} cm^2\\\\=8\sqrt{3} cm^2$

$V_p=Ab h$

$=(8\sqrt{3} )cm^2(2cm)\\\\=16\sqrt{3} cm^3\\\\=16(1.732)cm^3\\\\=27.712cm^3$

Cylinder:

$V_c=\pi\frac{d^2}{4} L$

π=3.14

d=1.6 cm

Height of the cylinder: h=t=2 cm

$V_c=3.14\times\frac{1.6^2}{4} \times2\\\\=3.14\times\frac{2.56}{4} \times 2\\\\=2.0096\times2\\\\=4.019cm^3$

$V=V_p-V_cV=27.712 cm^3 - 4.019 cm^3V=23.693 cm^3V=23.6 cm^3$