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?

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. Read more on volume of a tetrahedron; https://brainly.com/question/14604466 #SPJ1 Reply

volumeof thetetrahedronABCD 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?

tetrahedronare such that;s = (a+b+c)/2A = 2.9.heightof thebase trianglewhose midpoint is the equi-center can be evaluated as follows; 2(2.9)/3 = 1.93.h = 1.90.volumeof thetetrahedronin discuss is;1.84 cubic units.volumeof atetrahedron;