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? (4 points)

No, the rectangle cannot have x = 20 and y = 11 because x + y ≠ 64
No, the rectangle cannot have x = 20 and y = 11 because x + y ≠ 32
Yes, the rectangle can have x = 20 and y = 11 because x + y is less than 64
Yes, the rectangle can have x = 20 and y = 11 because x + y is less than 32

Answer 1

d

Answer 2

What is the perimeter of a rectangle?

Answer 3

2 (w + h)

Answer 4

Great.

The problem is calling the length x and the width y.
According to the variables of the problem, then, the perimeter is

P = 2(x + y)

Ok?

Answer 5

We are told the length is 20 and the width is 11.
What would the perimeter be?
P = 2(x + y)
P = 2(20 + 11)
P = 2(31)
P = 62
With the length 20 and width 11, the perimeter is 62, not 64.
This means the perimeter cannot be 64.

Answer 6

Notice that when you calculate the perimeter, you add the length and width, and then you multiply that by 2.
That means the length + the width must equal half the perimeter.
In our case, the length + the width = 20 + 11 = 31
Since we are told the perimeter is 64, half the perimeter is 32.
31 is not half the perimeter, so 64 cannot be the perimeter.

Answer 7

So it would be B?

Answer 8

@mathstudent55

Answer 9

Or would it be A? @mathstudent55

Answer 10

b

Answer 11

The answer is B.