Question

The perimeter of a rectangle can be found using the equation P = 2L + 2W, where P is the perimeter, L is the length, and W is the width of the rectangle. Can the perimeter of the rectangle be 64 units when its width is 11 units and its length is 20 units?
A.No. If the length is 20 and the width is 11, the perimeter is P = 20 + 22 = 42, not 64.
B.No. If the length is 20 and the width is 11, the perimeter is P = 40 + 22 = 62, not 64.
C.Yes. If the perimeter is 64 units and the width is 11 units, then P + W is greater than D.40.
Yes. If the perimeter is 64 units and the width is 11 units, then P + W is less than 40.

Answer 1

Answer:

B:  No

Step-by-step explanation:

P = 2L + 2W.  If L = 20 and W = 11, then P = 2(20) + 2(11), or 62.  This matches Answer B.

Answer 2

Answer:  The correct option is

(B) No. If the length is 20 and the width is 11, the perimeter is P = 40 + 22 = 62, not 64.

Step-by-step explanation:  Given that the perimeter of a rectangle can be found using the equation P = 2L + 2W, where P is the perimeter, L is the length, and W is the width of the rectangle.

We are to check whether the perimeter of the rectangle can be 64 units when its width is 11 units and its length is 20 units.

According to the given information, we have

Length, L = 20 units,  width, W = 11 units  and  Perimeter, P = 64 units.

We have

$P\\\\=2L+2W\\\\=2\times20+2\times11\\\\=40+22\\\\=62\neq64.$

Therefore, if length is 20 units and width is 11 units, then perimeter P = 40 + 22 = 62 units, not 64.

Thus, the answer is NO.

Thus, option (B) is CORRECT.