Distance of points

A regular quadrilateral pyramid ABCDV is given, in which edge AB = a = 4 cm and height v = 8 cm. Let S be the center of the CV. Find the distance of points A and S.

Correct result:

x =  5.831 cm

Solution:

a=4 cm v=8 cm  u=2 a=2 4=4 2 cm5.6569 cm  s=v2+(u/2)2=82+(5.6569/2)2=6 2 cm8.4853 cm s2=s/2=8.4853/2=3 2 cm4.2426 cm  ACV=arctan(vu/2)=arctan(85.6569/2)1.231 rad  x2=u2+s222 u s2 cos(ACV)  x=u2+s222 u s2 cos(ACV)=5.65692+4.242622 5.6569 4.2426 cos1.231=34=5.831 cm



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See also our trigonometric triangle calculator.

 
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