Answer:
[tex]SA=672\ yd^2[/tex]
The net in the attached figure
Step-by-step explanation:
we know that
The surface area of the square pyramid using a net, is equal to the area of a square plus the area of its four lateral triangular faces
so
[tex]SA=16^2+4[\frac{1}{2}(16)(13)][/tex]
[tex]SA=256+416=672\ yd^2[/tex]
