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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

3 votes
the answer is 16.36 meters
User Joma
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6 votes

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 = √(267.75)

h = 16.36 m (approx)

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




The Tetrahedron Superyacht designed by Jonathan Schwinge is a regular tetrahedron-example-1
User FlyingCat
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