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:

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.

Answer 2

Answer:

Length of the other side should be grater then 6 cm.

Step-by-step explanation:

Given information:

The perimeter of a square of side s is:

$P=4s$

So, perimeter if side is 6 cm.

$P=4\times6\\P=24 cm.$

Now, the area of rectangle

$A=l\times w$

The width is given as 4 cm.

For grater area the other side should be grater

Hence, area should be grater

$A>24$

So,

$l\times w>24\\4\times l>24\\l>6 cm.$

Hence, length of the other side should be grater then 6 cm.

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