## Prealgebra (7th Edition)

$9\frac{3}{4}\ yd^2$
The area of a triangle is 1\2 base times height. $A=\frac{1}{2}(6\frac{1}{2}yd)(3yd)=\frac{1}{2}(\frac{13}{2}yd)(3yd)=\frac{39}{4}yd^2=9\frac{3}{4}yd^2$