We are given –
- The perimeter is 40 feet and the length of a rectangle is 4 more than its width.
- Let the length be " l " and the width be " w ".
As we know that –
- Rectangle is a four sided polygonal figure, i.e, a quadrilateral, in which the opposite sides are of equal length and are parallel. Also its diagonals bisects each other.
[tex].\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \huge \boxed{ \bf{ \: \: \: \: \: \: }} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tiny\sf{A \: rectangle} [/tex]
- Perimeter = 2 (l + w) units
- Area = lb sq.units
According to the question –
- Width = w feet
- Length = w+4 feet
[tex]\qquad[/tex][tex] \pink{\bf \longrightarrow Perimeter _{(Rectangular) } = 2( Length + Breadth) }[/tex]
[tex]\qquad[/tex][tex] \bf \longrightarrow Perimeter _{(Rectangular) } = 2 (l + w)[/tex]
[tex]\qquad[/tex][tex] \sf \longrightarrow 40 = 2( w+ 4 +w)[/tex]
[tex]\qquad[/tex][tex] \sf \longrightarrow 40 = 2 ( 2w +4)[/tex]
[tex]\qquad[/tex][tex] \sf \longrightarrow 40 = 4w +8[/tex]
[tex]\qquad[/tex][tex] \sf \longrightarrow 4w = 40-8[/tex]
[tex]\qquad[/tex][tex] \sf \longrightarrow 4w = 32[/tex]
[tex]\qquad[/tex][tex]\sf \longrightarrow w = \cancel{\dfrac{32}{4}}[/tex]
[tex]\qquad[/tex][tex] \pink{\bf\longrightarrow w = 8 \:feet}[/tex]
- Width of perimeter is = 8 feet
Substituting the value of w–
[tex]\qquad[/tex][tex]\purple{\bf \longrightarrow Length\: of \: rectangle = (w+4)\: feet }[/tex]
[tex]\qquad[/tex][tex]\bf \longrightarrow Length\: of \: rectangle = 8+ 4 \:feet [/tex]
[tex]\qquad[/tex][tex]\purple{\bf \longrightarrow Length\: of \: rectangle = 12\: feet}[/tex]
- Henceforth, The width is 8 ft and the length is 12 ft.
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