Height of pyramid

The pyramid ABCDV has edge lengths: AB = 4, AV = 7. What is its height?

Correct answer:

h =  6.4031

Step-by-step explanation:

$$\begin{gathered} a=4 \\ s=7 \\ u=\sqrt{2}\cdot a=\sqrt{2}\cdot 4=4\sqrt{2}\doteq 5.6569 \\ {h}^2=s^2-(u\mathrm{/}2{)}^2 \\ h=\sqrt{s^2-(u\mathrm{/}2{)}^2}=\sqrt{7^2-(5.6569\mathrm{/}2{)}^2}=\sqrt{41}=6.4031 \end{gathered}$$

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