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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

User Tim Rae
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2 Answers

20 votes
20 votes

Answer:

Length: 17 in, Width: 12 in

Explanation:

Let l=length and w=width

l=w+5

2(w+5)+2w=58

4w=48

w=12, l=17

User Ching Chong
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8 votes
8 votes

Answer:

  • Length = 17 inches

  • Width = 12 inches

Explanation:

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,

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

Now, According to the Question :


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


{\longrightarrow \qquad { {\sf{2 ( x + 5 + x )= 58 }}}}


{\longrightarrow \qquad { {\sf{2 ( 2x + 5 )= 58 }}}}


{\longrightarrow \qquad { {\sf{ 4x + 10= 58 }}}}


{\longrightarrow \qquad { {\sf{ 4x = 58 - 10}}}}


{\longrightarrow \qquad { {\sf{ 4x = 48}}}}


{\longrightarrow \qquad { {\sf{ x = (48)/(4) }}}}


{\longrightarrow \qquad{ \underline{ \boxed { \pmb{\mathfrak {x = 12}} }}} }\: \: \bigstar

Therefore,

  • The width of the rectangle is 12 inches .

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


{\longrightarrow \qquad{ { \frak{\pmb{Length = x + 5 }}}}}


{\longrightarrow \qquad{ { \frak{\pmb{Length = 12 + 5 }}}}}


{\longrightarrow \qquad{ { \frak{\pmb{Length = 17}}}}}

Therefore,

  • The length of the rectangle is 17 inches .

User Braydie
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