Question

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 1

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

$h = \sqrt{267.75}$

h = 16.36 m (approx)

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

Answer 2

The answer is 16.36 meters