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The perimeter of a rectangle is 320 feet. Find the length and width if the length is an odd integer and the width is 5 times the next consecutive odd integer.

The Perimeter Of A Rectangle Is 320 Feet Find The Length And Width If The Length Is An Odd Integer And The Width Is 5 Times The Next Consecutive Odd Integer class=

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The length of rectangle is 25 feet and the width  of rectangle is 135 feet.

According to the question,

The perimeter of a rectangle is 320 feet. The length and width if the length is an odd integer and the width is 5 times the next consecutive odd integer.

Odd integers that follow each other and they differ by 2. If x is an odd integer, then x + 2, x + 4 and x + 6 are consecutive odd integers. The perimeter of a rectangle is 320 feet. Let the length of a rectangle is x. The next consecutive odd integer is x+2.width = 5(x+2)

    =5x+10.

Let, Length - l:Width - w; l = 5w and the perimeter of a rectangle is 320 feet. Formula for perimeter of rectangle is 2(Length + Breadth). Perimeter of a rectangle is the sum of the length of all sides of the rectangle. The perimeter of a rectangle formula is,

P = 2(l + b)

In words, it is equal to the sum of two times the length and two times the breadth of the rectangle. Perimeter for any figure is defined as the length of its boundary.

2(length + width) = 320

Length+ width=160

x+5x+10=160

6x=150

x=25

Length = 25 feet

width = 5(25+2)=135 feet.

Hence, the length of rectangle is 25 feet and the width of rectangle is 135 feet..

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