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the length of a rectagle is 5 in longer than its width. if the perimeter of the rectangle is 58 in, find its length and width

Sagot :

OllieFoxidn

Answer:

Length: 17 in, Width: 12 in

Step-by-step explanation:

Let l=length and w=width

l=w+5

2(w+5)+2w=58

4w=48

w=12, l=17

Answer:

  • Length = 17 inches

  • Width = 12 inches

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Step-by-step explanation:

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As it is given that, the length of a rectangle is 5 in longer than its width and the perimeter of the rectangle is 58 in and we are to find the length and width of the rectangle. So,

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Let us assume the width of the rectangle as x inches and therefore, the length will be (x + 5) inches .

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Now, According to the Question :

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[tex]{\longrightarrow \qquad { \pmb{\frak {2 ( Length + Breadth )= Perimeter_{(Rectangle)} }}}}[/tex]

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[tex]{\longrightarrow \qquad { {\sf{2 ( x + 5 + x )= 58 }}}}[/tex]

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[tex]{\longrightarrow \qquad { {\sf{2 ( 2x + 5 )= 58 }}}}[/tex]

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[tex]{\longrightarrow \qquad { {\sf{ 4x + 10= 58 }}}}[/tex]

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[tex]{\longrightarrow \qquad { {\sf{ 4x = 58 - 10}}}}[/tex]

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[tex]{\longrightarrow \qquad { {\sf{ 4x = 48}}}}[/tex]

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[tex]{\longrightarrow \qquad { {\sf{ x = \dfrac{48}{4} }}}}[/tex]

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[tex]{\longrightarrow \qquad{ \underline{ \boxed { \pmb{\mathfrak {x = 12}} }}} }\: \: \bigstar[/tex]

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Therefore,

  • The width of the rectangle is 12 inches .

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Now, We are to find the length of the rectangle:

[tex]{\longrightarrow \qquad{ { \frak{\pmb{Length = x + 5 }}}}}[/tex]

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[tex]{\longrightarrow \qquad{ { \frak{\pmb{Length = 12 + 5 }}}}}[/tex]

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[tex]{\longrightarrow \qquad{ { \frak{\pmb{Length = 17}}}}}[/tex]

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Therefore,

  • The length of the rectangle is 17 inches .

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