A rectangle is twice as long as it is wide and has the same perimeter as a square whose area is 36 square feet larger than that of the rectangle. what are the dimensions of both the rectangle and the square?

You want the dimensions of a rectangle and square such that they have the same perimeter, but the square has an area of 36 more square feet. The rectangle is twice as long as wide.

Setup

Let x represent the side length of the square. Then its area is x², and the area of the rectangle is (x² -36).

This area is the product of length and width. The expression factors as ...

rectangle area = x² -36 = (x +6)(x -6)

Solution

If we assume these factors are the dimensions of the rectangle, then the longer dimension is twice the shorter one:

(x +6) = 2(x -6)

18 = x . . . . . . . . . add 12-x to both sides

This is the side length of the square. The rectangle dimensions are ...

x+6 = 18+6 = 24 . . . . feet long

x -6 = 18 -6 = 12 . . . . feet wide

The rectangle is 12 feet wide and 24 feet long. The square is 18 feet on a side.

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Additional comment

The perimeter in each case is 4x = 2((x+6) +(x -6)) = 72 ft.

The area of the rectangle is (12 ft)(24 ft) = 288 ft². The area of the square is (18 ft)² = 324 ft², a value that is 36 ft² more than the rectangle area.

Alternate solution

Using x for the width of the rectangle, the length is 2x and its area is x(2x) = 2x². The perimeter is 2(x +2x) = 6x, so the side length of the square is (6x)/4 = 3/2x and its area is (3/2x)² = 9/4x².

The difference is 36 square feet, so we have ...

9/4x² -2x² = 36 = x²/4 ⇒ x = √(36·4) = 6·2 = 12 . . . . width of rectangle