The area of a rectangle r is 48. if its sides are whole numbers, then its perimeter cannot be

See below

Step-by-step explanation:

If its area is 48, and we know that the area of a rectangle is calculated by:

b*h = area

Where b is the base and h the height, we know that:

b*h = 48

So, b and h are whole numbers such that when multiplied gives as 48. Lets find which whole numbers give us this combination. Lets star with 1 and see if there is a whole number that when multiplied by 1 results in 48, then lets go with 2, then with 3, and so on until 48. Doing so we have:

1*48 = 48

2 * 24 = 48

3*16 = 48

4*12 = 48

6*8 = 48

Those are the combinations we have.

Now, we know that the perimeter is calculated by the sum of all sides. As in every rectangle we will have two identical pairs, this is, two sides with measure b (base) and two with measure h (height), the perimeter is:

perimeter = b + b + h + h = 2b + 2h

So, lets see for every of the 5 combination its respective perimeter:

1, 48 - -> 2*1 + 48*2 = 98

2, 24 - -> 2*2 + 2*24 = 52

3, 16 - -> 3*2 + 2*16 = 38

4, 12 - -> 4*2 + 12*2 = 32

6, 8 - -> 6*2 + 8*2 = 28

So, for every combination of sides the perimeters are listed above. Here we see that the perimeter cannot be, for example, an odd number. If your question has options, see if one of these options fits. Do not hesitate in contact if you have any doubt.