[tex]Fencing\ problems\ are\ what\ we\ can\ conclude\ in\ a\ \\ real\ life\ perimeters\ problems. \\ \\ In\ order\ to\ solve\ this,\ let's\ get\ first\ the\ formula \\ of\ the\ perimeter\ of\ the\ rectangle. \\ \\ ^{Formula:} \\ _{Perimeter}=2(l)+2(w) \\ \\ ^{Given:} \\ _{length}=9 \frac{5}{6}\ yards \\ \\ _{width}=5 \frac{1}{4}\ yards [/tex]
[tex]\bold{Equation:} \\ \\ Perimeter=2(l)+2(w) \\ \\ Perimeter=2(\ 9\frac{5}{6}\ yards )+2(\ 5\frac{1}{4}\ yards) \\ \\ Solve\ first\ the\ total\ length\ or\ \underline{TL} \\ \\ TL=2(\ 9 \frac{5}{6}\ yards) \\ \\ TL= \frac{2}{1}\cdot\ 9 \frac{5}{6}\ yards \\ \\ TL= \frac{2}{1}\ \cdot\ [ \frac{54+5}{6}\ yards= \frac{59}{6}\ yards] \\ \\ TL= \frac{^{1}\not{2}}{1}\ \cdot\ \frac{59}{\not6_{3}}\to \frac{59}{3} \\ \\ \boxed{\bold{TL=\ 19 \frac{2}{3}\ yards}} [/tex]
[tex]Solving\ for\ total\ width\ or\ \underline{TW} \\ \\ TW=2\ (5\ \frac{1}{4}\ yards) \\ \\ TW= \frac{2}{1}\ \cdot \ 5\ \frac{1}{4}\ yards \\ \\ TW= \frac{2}{1}\ \cdot\ \ [ \frac{20+1}{4}\ yards= \frac{21}{4}\ yards] \\ \\ TW= \frac{^{1}\not2}{1}\ \cdot\ \frac{21}{\not4_{2}}\to \frac{21}{2}\ yards \\ \\ \boxed{\bold{TW=10 \frac{1}{2}\ yards}} [/tex]
[tex]Final\ solution: \\ Perimeter=2(l)+2(w) \\ \\ Perimeter=19 \frac{2}{3}\ yards+\ 10 \frac{1}{2}\ yards \\ \\ Perimeter= \frac{59}{3}\ yards\ +\ \frac{21}{2}\ yards\ \ \ \ \ |\ ^{Change\ the\ following\ fractions\ to} \\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ ^{improper\ fractions.} \\ \\ Get\ their\ common\ denominator: \\ \\ \frac{19}{3}\ yards \to\ \frac{118}{6}\ yards \\ \\ \frac{21}{2}\ yards \to\ \frac{63}{6}\ yards [/tex]
[tex]Perimeter= \frac{118}{6}\ yards +\ \frac{63}{6}\ yards \\ \\ Perimeter= \frac{181}{6}\ yards \\ \\ \boxed{\boxed{\bold{Perimeter=\ 30\ \frac{1}{6}\ yards}}} \\ \\ \\ \\ Hope\ it\ Helps :) \\ Domini [/tex]
