For this type of problems we use the formula for the volume of a pyramid:
[tex]\begin{gathered} V=\text{ }\frac{1}{3}A_bh \\ A_b\text{ is the area of the base} \\ h\text{ is the height of the pyramid} \end{gathered}[/tex]
Substituting h=12 yd and knowing that the area of a square is side*side we get that:
[tex]\begin{gathered} A_b=\text{ 10yd }\cdot10yd=100yd^2 \\ V=\frac{1}{3}100yd^212yd=100yd^24yd=400yd^3 \end{gathered}[/tex]
