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The width of a rectangle is 8 inches less than the length. The perimeter is 48 inches. Find the length and the
width.
The length is
* inches and the width is
* inches.

The width of a rectangle is 8 inches less than the length The perimeter is 48 inches Find the length and the width The length is inches and the width is inches class=

Respuesta :

Let width be x

  • Length=x+8

[tex]\\ \rm\longmapsto 2(x+x+8)=48[/tex]

[tex]\\ \rm\longmapsto 2x+8=24[/tex]

[tex]\\ \rm\longmapsto 2x=24-8=16[/tex]

[tex]\\ \rm\longmapsto x=\dfrac{16}{2}=8[/tex]

manissaha129

Answer:

Let the required width of the rectangle be x inches and the length of the rectangle be (x+8) inches

Perimeter of the rectangle is 2 ( length + width )

According to the above problem, we get

[tex]2[(x + 8) + x] = 48 \\ 2(2x + 8) = 48 \\ 4x + 16 = 48 \\ 4x = 48 - 16 \\ 4x = 32 \\ x = \frac{32}{4} \\ \boxed{ x = 8}[/tex]

Therefore,

The length of the rectangle is 16 inches and the width of the rectangle 8 inches.

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