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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x = 1035 = MXXXV
Step-by-step explanation:
<u>Roman Numbers </u>
It's an ancient number system developed in Rome where numbers are represented by the letters {I,V,X,L,C,D,M}. The value of each are, respectively {1,5,10,50,100,500,1000}. If some letter is repeated, then it's value must be multiplied by the number of times the letter is present. For example, XXVII has two X's and two I's, so the conversion is 10+10+5+1+1=27.
If a less-valued letter appears before a greater one, it subtracts. For example CDI=-100+500+1=401
Our numbers are
MMMMMMMMCDXV=8*1000-100+500+10+5=8415
MDCCCLXIII = 1000+500+300+50+10+3=1863
DLXI = 500+50+10+1=561
XXVII = 20+5+2=27
We now operate in our numerical system
Converting back to roman