The backyard is a rectangle, so the area is given by the product of the dimensions:
[tex] A_r = 10\cdot 8 = 80\ [/tex] squared yards
The pool is a circle, so the area is given by [tex] \pi r^2 [/tex]. The radius is half the diameter, so we have
[tex] A_c = \pi(2.5)^2 \approx 19.63 [/tex]
So, the remaining area is
[tex] A = A_r-A_c = 80-19.63 = 60.37 [/tex]
which you can round to 60.4