1-24 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} 1\ut{acre}=&40\ut{perch}\times4\ut{perch}=&160\ut{perch^2}\\ 1\ut{rood}=&40\ut{perch}\times1\ut{perch}=&40\ut{perch^2}\\ 1\ut{perch}=&16.5\ut{ft} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \Ans=&2.80\ut{acre}+13.0\ut{perch}\times3.00\ut{perch}\\ =&2.80\ut{acre}\cdot\frac{160\ut{perch^2}}{1\ut{acre}}+39\ut{perch^2}\\ =&487\ut{perch^2}\\ =&487\ut{perch^2}\cdot\frac{1\ut{rood}}{40\ut{perch^2}}\\ =&12.175\ut{rood}\\ \approx&12.2\ut{rood} \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \Ans=&487\ut{perch^2}\\ =&487\ut{perch^2}\cdot\(\frac{16.5\ut{ft}}{1\ut{perch}}\cdot\frac{12\ut{in}}{1\ut{ft}}\cdot\frac{2.54\ut{cm}}{1\ut{in}}\cdot\frac{1\ut{m}}{100\ut{cm}}\)^2\\ =&\frac{76985120223}{6250000}\ut{m^2}\\ =&12317.61923568\ut{m^2}\\ \approx&1.23\times10^4\ut{m^2} \end{aligned} $$