Question

A steel hex nut has two regular hexagonal faces and a hole with a diameter of 0.4 cm through the middle. Its mass is 3.03 grams.

A regular hexagonal prism is shown. A cylinder is cut out of the middle of the prism. The hexagon has an apothem with length 0.5 centimeters. The base edges have lengths of 0.6 centimeters and the height of the prism is 0.5 centimeters. The cylinder has a diameter of 0.4 centimeters.
Complete the steps to calculate the density of the steel.

Calculate the volume of the prism. Recall that the area of a hexagon is One-half times the apothem times the perimeter. V =
cm3
Calculate the volume of the cylinder. Round to the nearest hundredth. V =
cm3
Find the volume of the composite figure.
V =
cm3
Calculate the density by dividing the mass by the volume. d =
g/cm3

Answer 1

Area of hexagon = 1/2 × apothem × perimeter = 1/2 × 0.5 × (6×0.6) = 0.9 cm².
Volume of prism = base area × height = 0.9 × 0.5 = 0.45 cm³.

Volume of cylinder = πr²h = π(0.2)²(0.5) = 0.02π ≈ 0.06 cm³ (rounded to nearest hundredth).

Volume of composite = prism − cylinder = 0.45 − 0.06 = 0.39 cm³.

Density = mass / volume = 3.03 g / 0.39 cm³ ≈ 7.77 g/cm³.

(If you use the unrounded cylinder volume 0.02π ≈ 0.06283 cm³, the composite volume ≈ 0.38717 cm³ and density ≈ 7.83 g/cm³.)