
A.
6cm 
B.
5cm 
C.
4cm 
D.
3cm
Correct Answer: Option C
Explanation
Base of pyramid of a square of side 8cm vertex directly above the centre edge = (4sqrt{3})cm
From the diagram, the diagonal of one base is AC^{2} = 8^{2} + 8^{2}
Ac^{2} = 64 + 64 = 128
AC = (8sqrt{2})
but OC = (frac{1}{2})AC = 8(sqrt{frac{2}{2}}) = (4sqrt{2})cm
OE = h = height
h^{2} = ((4sqrt{3}))^{2}
16 x 2 – 16 x 2
48 – 32 = 16
h = (sqrt{16})
= 4