✏️RECTANGLE
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[tex]\underline{\mathbb{PROBLEM:}}[/tex]
- A rectangle has a perimeter of 86 feet. The width is 10 feet less than the length. Find the dimensions of the rectangle.
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[tex]\underline{\mathbb{ANSWER:}}[/tex]
[tex]\qquad \Large \rm» \: \: \green{ Length = 26.5 \: feet }[/tex]
[tex]\qquad \Large\rm» \: \: \green{Width = 16.5 \: feet }[/tex]
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[tex]\underline{\mathbb{SOLUTION:}}[/tex]
- Represent l and w as the length and width of the rectangle respectively. Formulated equations of the given statement.
- [tex] \begin{cases}86 = 2(l + w)& \red{(eq. \: 1)} \\ w = l - 10& \red{(eq. \: 2)} \end{cases}[/tex]
- Substitute w from the second equation to the first equation in terms of l.
- [tex] \begin{cases}86 = 2(l + l - 10) \\ w = l - 10 \end{cases}[/tex]
- [tex] \begin{cases}86 = 2(2l - 10) \\ w = l - 10 \end{cases}[/tex]
- [tex] \begin{cases}86 = 4l - 20 \\ w = l - 10 \end{cases}[/tex]
- [tex] \begin{cases}86 + 20 = 4l \\ w = l - 10 \end{cases}[/tex]
- [tex] \begin{cases}106 = 4l \\ w = l - 10 \end{cases}[/tex]
- [tex] \begin{cases} \begin{gathered} \frac{106}{4} = \frac{ \cancel4l}{ \cancel4} \end{gathered}\\ w = l - 10 \end{cases}[/tex]
- [tex] \begin{cases}l = 26.5 \\ w = l - 10 \end{cases}[/tex]
- The length of the rectangle is 26.5 feet, then substitute it to the first equation to find the width of the rectangle.
- [tex] \begin{cases}l = 26.5 \\ w = 26.5 - 10 \end{cases}[/tex]
- [tex] \begin{cases}l = 26.5 \\ w = 16.5 \end{cases}[/tex]
[tex]\therefore[/tex] The length of the rectangle is 26.5 feet while the width of the rectangle is 16.5 feet
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