Question

The width of a rectangle is 4 cm. What should the length of the other side be so that the area of the rectangle is greater than the perimeter of a square with a side of 6 cm?

Answer 1

Answer:

6

Step-by-step explanation:

because if you get the perimeter what is 6+6+4+4 that equal to 20cm then the area would be 24 cm cubed.

Answer 2

Answer:

The length of the other side should be larger than 6cm so that the area of the rectangle is greater than the perimeter of a square with a side of 6 cm.

Step-by-step explanation:

The perimeter of a square with a side of s cm is given by the following formula:

$P = 4s$

So, with a side of 6 cm, we have that;

$P = 4s = 4(6) = 24$cm

The area of a rectangle, with width w and length l is given by the following formula

$A = lw$

In this problem, we have that:

The width of a rectangle is 4 cm. This means that $w = 4$

What should the length of the other side be so that the area of the rectangle is greater than the perimeter of a square with a side of 6 cm?

We want to have:

$A > 24$

So

$lw > 24$

$4l > 24$

$l > 6$

The length of the other side should be larger than 6cm so that the area of the rectangle is greater than the perimeter of a square with a side of 6 cm.