\(40 = 2^3 \times 5\)

which is \(8 \times 5\)

Now, if you consider that 8+5=13 and the perimeter is \(13 \times 2= 26\)

The perimeter is at the very least 26 or above.

20 is impossible

So A is wrong and all the other choices are.

Regards

As mentioned by others. The key to solving this question quickly is that given an area for a rectangle the perimeter is always minimized when the rectangle is a square. This will happen when length and width are the same. Since lw=40 this means l=rad(40) and w=rad(40) and the smallest perimeter possible is 4rad(40) which is around 25.3 if you plug it in the calculator. So 20 is the only one below that.

Everything about 25.3 will be valid since we can let the width be literally any positive number and adjust the length to make it area 40, therefore the perimeter itself is unbounded above.

Question)
If the area of a rectangle is 40, which of the following could be the perimeter of the rectangle? Indicate all such perimeters.
❑ 20
❑ 40
❑ 200
❑ 400
❑ 2,000
❑ 4,000

Solution:
Since the area of the rectangle is 40, possible cases are:

#1. L x W = 40 x 1 => Perimeter = 2(40 + 1) = 82
#2. L x W = 20 x 2 => Perimeter = 2(20 + 2) = 44
#3. L x W = 8 x 5 => Perimeter = 2(8 + 5) = 26

We can observe that closer the values of L and W, smaller the perimeter of the rectangle.
Thus, the minimum perimeter would be when L = W
=> L = W = \(\sqrt{40}\) = 6.32
=> Perimeter = 2(6.32 + 6.32) = 25.3

Thus, the minimum perimeter is 25.3

Thus, any value of the perimeter higher than 25.3 is possible.

For example, if you are thinking how the perimeter can have such large values as mentioned in the last two options:
We can assume the length as 2000 and then the width becomes 40/2000 = 0.02
=> Perimeter = 2(2000 + 0.02) = 4000.04


Answer: B, C, D, E, F

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