keeping in mind that the area of a circle is πr².
the actual area will just be the area of the rectangular backyard plus the pool, namely (10*20) + (π15²), which gives us an actual area of 200 + 225π.
now, we know the model and actual are on a 1:20 ratio.
[tex]\bf \cfrac{model}{actual}\qquad 1:20\qquad \cfrac{1}{20}\qquad \qquad \cfrac{m}{200+225\pi }=\cfrac{1}{20}
\\\\\\
m=\cfrac{(200+225\pi )1}{20}\implies m=\cfrac{200+225\pi }{20}\implies m=\cfrac{200}{20}+\cfrac{225\pi }{2}
\\\\\\
m=10+112.5\pi [/tex]
