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?
2 answers:
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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