The tetrahedron superyacht designed by jonathan schwinge is a regular tetrahedron with edges 20 m long. the distance r from one corner to the center of the base is 11.5 m. what is the vertical height, h, of the tetrahedron?

Answer

Find out the  the vertical height (h)of the tetrahedron .

To prove

By using the pythagorus theorem

Hypotenuse² = Perpendicular² + Base²

As given

The Tetrahedron Superyacht designed by Jonathan Schwinge is a regular tetrahedron with edges 20 m long.

The distance r from one corner to the center of the base is 11.5 m.

As shown in the diagram given below.

AB² = AD² + DB²

As given in the question

AD = h , AB = 20 m , DB = 11.5 m

Put in the above

(20)² = h² + (11.5)²

h² =  (20)² - (11.5)²

h² = 400 - 132.25

h² = 267.75

h = 16.36 m (approx)

Therefore the vertical height,h ,of the tetrahedron is 16.36 m (approx) .

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