[tex]\begin{gathered} Question\text{ 8} \\ \text{square} \\ \text{Area = 64yd}^2 \\ \text{square area = side}^2 \\ \text{side}^2=\text{64yd}^2 \\ \sqrt[]{\text{side}^2}=\sqrt[]{\text{64yd}^2} \\ side\text{ =8yd} \\ The\text{ square's side is 8yd} \\ \\ \text{Question 9} \\ \text{Triangle} \\ base=\text{ }10.5in \\ \text{height}=14in \\ \text{Triangle's area=}\frac{base\cdot height}{2} \\ \text{Triangle's area=}\frac{10.5\cdot14}{2} \\ \text{Triangle's area=}\frac{147}{2} \\ \text{Triangle's area=73.5} \\ \text{The triangle's area is 73.5in}^2 \end{gathered}[/tex]
