#### Answer

$4x^2+60x$

#### Work Step by Step

Step 1. Using the figure given in the exercise, the area of the pool is given by $A_1=20\times10=200\ m^2$
Step 2. The area of the bigger rectangle, including the pool and the path, is given by $A_2=(20+2x)(10+2x)=200+60x+4x^2\ m^2$
Step 3. The area of the path is the difference of the above; we have $A=A_2-A_1=4x^2+60x\ m^2$