Question

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.

Answer 1

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

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

Therefore,

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

Answer 2

Let width be x

• Length=x+8

`\\ \rm\longmapsto 2(x+x+8)=48`

`\\ \rm\longmapsto 2x+8=24`

`\\ \rm\longmapsto 2x=24-8=16`

`\\ \rm\longmapsto x=(16)/(2)=8`