Question

The perimeter of a rectangle is 64 units. Can the length x of the rectangle can be 20 units when its width y is 11 units? A. No, the rectangle cannot have x = 20 and y = 11 because x + y ≠ 64 B. No, the rectangle cannot have x = 20 and y = 11 because x + y ≠ 32 C. Yes, the rectangle can have x = 20 and y = 11 because x + y is less than 64 D.Yes, the rectangle can have x = 20 and y = 11 because x + y is less than 32

Answer 1

P = 2(L + W)
P = 64
W = 11

64 = 2(L + 11)
64 = 2L + 22
64 - 22 = 2L
42 = 2L
21 = L

can length be 20....no, because x + y does not equal 32

Answer 2

Answer:

No, the rectangle cannot have x = 20 and y = 11 because x + y ≠ 32

Step-by-step explanation:

The perimeter of a rectangle is 64 units.

Let the length of the rectangle be = x

The width is 11 units.

The perimeter of rectangle is given as:

$P=2(L+W)$

=> $64=2(x+11)$

=> $64=2x+22$

=> $64-22=2x$

=> $42=2x$

x = 21 units.

So, we get the length as 21 units.

Therefore the answer will be - No, the rectangle cannot have x = 20 and y = 11 because x + y ≠ 32