Answer:
[tex]SA=5\ yd^2[/tex]
Step-by-step explanation:
we know that
The surface area of the square pyramid is equal to the area of the square base plus the area of its four triangular faces
so
[tex]SA=b^2+4[\frac{1}{2}(b)(h)][/tex]
we have
[tex]b=1\ yd\\h=2\ yd[/tex]
substitute
[tex]SA=1^2+4[\frac{1}{2}(1)(2)][/tex]
[tex]SA=5\ yd^2[/tex]