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.

Final Answer:

x =  5.831 cm

Step-by-step explanation:

$$\begin{gathered} a=4\text{cm} \\ v=8\text{cm} \\ u=\sqrt{2}\cdot a=\sqrt{2}\cdot 4=4\sqrt{2}\text{cm}\doteq 5.6569\text{cm} \\ s=\sqrt{v^2+(u/2{)}^2}=\sqrt{8^2+(5.6569/2{)}^2}=6\sqrt{2}\text{cm}\doteq 8.4853\text{cm} \\ {s}_2=s/2=8.4853/2=3\sqrt{2}\text{cm}\doteq 4.2426\text{cm} \\ ACV=\arctan (\frac{v}{u/2})=\arctan (\frac{8}{5.6569/2})\doteq 1.231\text{rad} \\ {x}^2=u^2+s_2^2-2\cdot u\cdot s_2\cdot \cos (ACV) \\ x=\sqrt{u^2+s_2^2-2\cdot u\cdot s_2\cdot \cos (ACV)}=\sqrt{5.6569^2+4.2426^2-2\cdot 5.6569\cdot 4.2426\cdot \cos 1.231}=\sqrt{34}=5.831\text{cm} \end{gathered}$$

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