contestada

the length of a rectagle is 5 in longer than its width. if the perimeter of the rectangle is 56 in, find its length and width

Respuesta :

Answer:

Width: 11.5 inches; Length: 16.5 inches

Step-by-step explanation:

Let l=length and w=width

l=w+5

2(w+5)+2w=56

4w=46

w=11.5, l=16.5

Answer:

  • Length = 16.5 inches

  • Width = 11.5 inches

Step-by-step 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 56 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 :

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

[tex]{\longrightarrow \qquad { {\sf{2 ( x + 5 + x )= 56 }}}}[/tex]

[tex]{\longrightarrow \qquad { {\sf{2 ( 2x + 5 )= 56 }}}}[/tex]

[tex]{\longrightarrow \qquad { {\sf{ 4x + 10= 56 }}}}[/tex]

[tex]{\longrightarrow \qquad { {\sf{ 4x = 56 - 10}}}}[/tex]

[tex]{\longrightarrow \qquad { {\sf{ 4x = 46}}}}[/tex]

[tex]{\longrightarrow \qquad { {\sf{ x = \dfrac{46}{4} }}}}[/tex]

[tex]{\longrightarrow \qquad{ \underline{ \boxed { \pmb{\mathfrak {x = 11.5}} }}} }\: \: \bigstar[/tex]

Therefore,

  • The width of the rectangle is 11.5 inches .

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

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

[tex]{\longrightarrow \qquad{ { \frak{\pmb{Length = 11.5 + 5 }}}}}[/tex]

[tex]{\longrightarrow \qquad{\underline{\boxed { \frak{\pmb{Length = 16.5}}}}}} \: \: \bigstar[/tex]

Therefore,

  • The length of the rectangle is 16.5 inches .