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Completion requirements
A rectangle has the dimensions shown in the diagram.
The perimeter of the rectangle is \(24\) units. Determine the length of the sides of the rectangle.
NPVs: \(r \ne 0, \thinspace 2\)
LCD: \(r\left(r - 2\right)\)
The two possible dimensions are:
Because the dimensions cannot be negative, \(r = \frac{5}{6}\) can be ignored as a possible answer. The dimensions of the rectangle are \(5\) units by \(7\) units.
The perimeter of the rectangle is \(24\) units. Determine the length of the sides of the rectangle.
\[2\left( {\frac{{15}}{r}} \right) + 2\left( {\frac{7}{{r - 2}}} \right) = 24\]
NPVs: \(r \ne 0, \thinspace 2\)
LCD: \(r\left(r - 2\right)\)
\[\begin{align}
2\left[ {\frac{{15}}{{{\color{red}\cancel{\color{#444}{r}}}}}} \right]{\color{red}\cancel{\color{#444}{r}}}\left( {r - 2} \right) + 2\left[ {\frac{7}{{{\color{red}\cancel{\color{#444}{r - 2}}}}}} \right]r {\color{red}\cancel{\color{#444}{\left(r - 2\right)}}}&= \left[ {24} \right]r\left( {r - 2} \right) \\
30\left( {r - 2} \right) + 14r &= 24r^2 - 48r \\
30r - 60 + 14r &= 24r^2 - 48r \\
0 &= 24r^2 - 92r + 60 \\
0 &= 4\left( {6r^2 - 23r + 15} \right) \\
0 &= 4\left( {6r - 5} \right)\left( {r - 3} \right) \\
r &= \frac{5}{6} \thinspace {\rm{and}} \thinspace r = 3 \\
\end{align}\]
2\left[ {\frac{{15}}{{{\color{red}\cancel{\color{#444}{r}}}}}} \right]{\color{red}\cancel{\color{#444}{r}}}\left( {r - 2} \right) + 2\left[ {\frac{7}{{{\color{red}\cancel{\color{#444}{r - 2}}}}}} \right]r {\color{red}\cancel{\color{#444}{\left(r - 2\right)}}}&= \left[ {24} \right]r\left( {r - 2} \right) \\
30\left( {r - 2} \right) + 14r &= 24r^2 - 48r \\
30r - 60 + 14r &= 24r^2 - 48r \\
0 &= 24r^2 - 92r + 60 \\
0 &= 4\left( {6r^2 - 23r + 15} \right) \\
0 &= 4\left( {6r - 5} \right)\left( {r - 3} \right) \\
r &= \frac{5}{6} \thinspace {\rm{and}} \thinspace r = 3 \\
\end{align}\]
The two possible dimensions are:
\[\frac{{15}}{r} = \frac{{15}}{3} = 5 \thinspace {\rm{and}} \thinspace \frac{7}{{r - 2}} = \frac{7}{{3 - 2}} = 7 \]
OR
\[ \frac{{15}}{r} = \frac{{15}}{{\frac{5}{6}}} = 18 \thinspace {\rm{and}} \thinspace \frac{7}{{r - 2}} = \frac{7}{{\frac{5}{6} - 2}} = - 6 \]
OR
\[ \frac{{15}}{r} = \frac{{15}}{{\frac{5}{6}}} = 18 \thinspace {\rm{and}} \thinspace \frac{7}{{r - 2}} = \frac{7}{{\frac{5}{6} - 2}} = - 6 \]
Because the dimensions cannot be negative, \(r = \frac{5}{6}\) can be ignored as a possible answer. The dimensions of the rectangle are \(5\) units by \(7\) units.