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

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

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