A square pyramid has wiring PQRS and vertex O. Each whet has length equal to 20. Summate the shortest loftiness withal the outer surface of the pyramid from P to T, the midpoint of OR.
As usual, watch the video for a solution.
Shortest path virtually pyramid
Or alimony reading.
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Answer To Shortest Path Virtually Pyramid
(Pretty much all posts are transcribed quickly without I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).
There are two paths we need to check: one path withal the wiring of the pyramid and the other withal the faces. We will summate the length of each path and take the shorter length.
We will summate the length of each path by unfolding the pyramid into a net. Then the path from P to T will be a straight line segment and calculated easily.
Path withal square base
We know RT = 10, and each triangular squatter is an equilateral triangle with each wile equal to 60 degrees. Thus the horizontal loftiness from T to the triangular wiring is 10 sin (60°) = 5√3 and the other leg of the triangle is 10 cos (60°) = 5.
Each side of the square wiring has length equal to 20, so the legs of the triangle with hypotenuse PT have lengths of 20 – 5 = 15 and 20 5√3. Thus we have:
length from P to T
= √(152 (20 5√3)2)
= √(225 400 200√3 75)
= 10√(7 2√3)
≈ 32.348
Path withal triangular faces
The net of this pyramid looks like this:
We know PO = 20 and OT = 10, and the wile POT = 60° 60° = 120°. Thus we can find PT by Al-Kashi’s law of cosines.
length from P to T
= √(202 102 – 2(20)(10) cos (120°))
= √(400 100 200)
= 10√7
≈ 26.458
This path has a shorter length, and thus this is the loftiness from P to T. The wordplay is 10√7 ≈ 26.458.
Source
Math StackExchange
https://math.stackexchange.com/questions/3289765/shortest-distance-around-a-pyramid
