A rectangle has a width of x centimeters and a perimeter of 8x centimeters a square has sides of length 1/4 that of the length of the rectangle.

A. Find the length of the rectangle

B. Find the perimeter of the square

C. Find how many cm greater the rectangle's perimeter if x=4

D. Find how many square cm greater the rectangle's area is than the square's area if x=4

This is the concept of algebra, to solve the questions we proceed as follows:

a] Length of the rctangle

The width of rectangle is x cm

Perimeter of the rectangle is 8x

P=2 (L+W)

plugging in our values we get:

8x=2 (L+x)

8x=2L+2x

2L=8x-2x

2L=6x

hence

L=6x/2=3x

The length is 3x cm

B] The length of the square is 1/4 times the length of rectangle. The length of the square will be:

L=1/4*3x=3/4x cm

The perimeter of the square will be:

P=2 (L+W)

=2 (3/4x+3/4x)

=3x cm

C] Given that x=4, and the length of rectangle is 3x cm, then the actual length is:

substitute x in 3x we get

3x

= (3*4)

=12 cm

The length of the rectangle is 12 cm

D]

Given that x=4, to calculate the area, we need the actual lengths of the rectangle and square.

Length of rectangle=3x=12 cm

width of the rectangle=x=4cm

area of rectangle will be:

Area=length*width

Area=12*4=48 cm²

Next:

Length of the square will be:

length=1/4*12=3 cm

but in square length=width

hence

width=3 cm

Area of the square will be:

area=3*3=9 cm²

The difference in the areas will be:

48-9

=39 cm²

Therefore the area of rectangle is 39 cm² greater than the area of the square.