The area of the rectangle is 63.91 m and the width is 8.3 how many times the perimeter of the rectangle

The area of the rectangle is 63.91 m and the width is 8.3 how many times the perimeter of the rectangle is greater than the perimeter of the square, the area of which is 25 dm.

Firstly, note that the area of the rectangle is found by the formula:

S = a * b,

where S is its area,
a and b are its length and width.

It follows that the length of the rectangle (a) should be sought by the formula:

a = S: b.

Then:

a = S: b = 63.91: 8.3 = 7.7 (m).

Now let’s calculate the perimeter of a rectangle with sides of 7.7 m and 8.3 m:

P = 2 * (7.7 + 8.3) = 32 (m).

Find the side of the square:

a sq. = √25 = 5 (dm).

Its perimeter:

R sq. = 4 * 5 = 20 (dm).

Then:

P / P sq. = 32 m: 20 dm = 16.

Answer: 16 times.