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 (shown in the picture below) is 11.5 m.
image of a tetrahedron - shaped superyacht
what is the vertical height h of the tetrahedron?
round your answer to the nearest tenth of a meter.
( h approx square ) m
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Explanation:

Step1: Identify the right triangle

The vertical height \( h \), the edge of the tetrahedron (20 m), and the distance from the corner to the center of the base (11.5 m) form a right triangle. So we can use the Pythagorean theorem \( a^2 + b^2 = c^2 \), where \( c = 20 \) (the hypotenuse), \( b = 11.5 \), and \( a = h \). Rearranging for \( h \), we get \( h=\sqrt{c^{2}-b^{2}} \).

Step2: Substitute the values

Substitute \( c = 20 \) and \( b = 11.5 \) into the formula:
\( h=\sqrt{20^{2}-11.5^{2}}=\sqrt{400 - 132.25}=\sqrt{267.75} \)

Step3: Calculate the square root

Calculate \( \sqrt{267.75}\approx16.4 \) (rounded to the nearest tenth).

Answer:

\( 16.4 \)