Question

A piece of steel is in the form of a tetrahedron whose base is a right-angled triangle. The sides forming the right-angle are 5 cm and 6 cm long respectively. If the height of the tetrahedron is 7 cm, calculate its volume. If the tetrahedron is melted down, calculate the length of bar of diameter 20 mm that can be made from it, assuming no waste.

Answer 1

Known: AB = 5cm, BC = 6cm, h_tetrahedron = 7cm, d = 20mm

V_tetrahedron=1/3 S*h, where

S = S_ABC = AB*BC/2, and h - height, h = 7, then

$V_{tetrahedron} = \frac{AB*BC*h}{6}\\ V_{tetrahedron} = \frac{5\times6\times7}{6} = 35 cm^3, then\\ V_{tetrahedron} = 35000 mm^3$

$V_{bar} = V_{cylinder} = \pi R^2 h, where R = d/2, then\\ V_{bar} = \frac{\pi d^2h}{2^2}, hence\;\\ h = \frac{4V}{\pi d^2}, then \, h = \frac{4\times35000}{\pi\times20^2}=111.4mm$

Answer: the length of bar = 111.4mm or 11.14 cm