The frustum can be considered as consisting of a square pyramid, less the
cut volume.
Reasons:
The given parameter of the frustum are;
The density of the frustum, ρ = 85 g/cm³
Height of pyramid, h = 9 cm
Side length of base = 10 cm
Height of frustum
By proportional shapes, the side length of the top of the frustum can be found as follows;
[tex]\dfrac{4 \, cm}{9 \, cm} = \dfrac{x}{10 \, cm}[/tex]
[tex]x = \dfrac{4 \, cm}{9 \, cm} \times 10 \, cm = 4\frac{4}{9} \, cm[/tex]
[tex]V = \dfrac{h}{3} \times \left(B_1 + B_2 + \sqrt{B_1 \times B_2} \right)[/tex]
B₁ = 10², B₂ = [tex]\left(4\frac{4}{9} \right)^2[/tex]
Therefore;
[tex]V = \dfrac{4}{3} \times \left(10^2 + \left(4\frac{4}{9} \right)^2 + \sqrt{10^2 \times \left(4\frac{4}{9} \right)^2} \right) \approx 218.93[/tex]
The volume of the stand, V ≈ 218.93 cm³
Mass = Volume × Density
∴ Mass of the stand, m = 218.93 cm³ × 85 g/cm³ = 18,609.05 grams = 18.60905 kg.
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