Height of pyramid

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

Correct result:

h =  6.403

Solution:

a=4 s=7  u=2⋅ a=2⋅ 4≐4 2≐5.6569  h2=s2−(u/2)2  h=s2−(u/2)2=72−(5.6569/2)2=41=6.403a=4 \ \\ s=7 \ \\ \ \\ u=\sqrt{ 2 } \cdot \ a=\sqrt{ 2 } \cdot \ 4 \doteq 4 \ \sqrt{ 2 } \doteq 5.6569 \ \\ \ \\ h^2=s^2 - (u/2)^2 \ \\ \ \\ h=\sqrt{ s^2 - (u/2)^2 }=\sqrt{ 7^2 - (5.6569/2)^2 }=\sqrt{ 41 }=6.403

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