Question

Let ABCD be a tetrahedron with AB = 2, BC = 3, AC = 4, and AD = BD = CD =32/15

What is the volume of ABCD?

Answers

  1. The volume of the tetrahedron ABCD as described in the task content in which case; AD = BD = CD =32/15 is; 1.84 cubic units.

    What is the volume of the tetrahedron described in the task content where; AD = BD = CD =32/15?

    It follows from the task content that the dimensions of the tetrahedron are such that;
    AB = 2, BC = 3, AC = 4.
    Hence, by determining the area of the base triangle by means of the formula;
    A = √(s(s-a)+(s-b)+(s-c)) where; s = (a+b+c)/2
    Area, A = 2.9.
    The height of the base triangle whose midpoint is the equi-center can be evaluated as follows; 2(2.9)/3 = 1.93.
    Half that length then forms a right triangle with the side; CD = 32/15.
    Hence, the height of the tetrahedron can be evaluated by means of Pythagoras theorem as;
    height = √(2.13)² – (0.97)²
    height = √3.62.
    Ultimately, height, h = 1.90.
    Therefore, the volume of the tetrahedron in discuss is;
    Volume, V = (1/3) × 2.9 × 1.9
    Volume = 1.84 cubic units.
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