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Sagot :
Given:
The perimeter of a rectangle, P=140 ft.
Let l be the length and w be the width of the rectangle.
Since the length is triple the width increasesd by 6, the length of the rectangle can be expressed as,
[tex]l=3w+6\text{ ----(1)}[/tex]Now, the perimeter of the rectangle can be expressed as,
[tex]\begin{gathered} P=2(l+w) \\ P=2(3w+6+w) \\ P=2(4w+6) \\ P=8w+12 \end{gathered}[/tex]Substitute P=140 in the above equation and solve for w.
[tex]\begin{gathered} 140=8w+12 \\ 8w=140-12 \\ 8w=128 \\ w=\frac{128}{8} \\ w=16\text{ ft} \end{gathered}[/tex]Now, put w=16 in equation (1) and find l.
[tex]\begin{gathered} l=3\cdot16+6 \\ =48+6 \\ =54\text{ ft} \end{gathered}[/tex]Therefore, the length of the rectangle is 54 ft and width of the rectangle is 16 ft.
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