# The length of a rectangle is two units more than four times the width. What is the length if the perimeter is of 20 units?

justaguide | Certified Educator

Let us represent the width of the rectangle by W. The length is equal to two units more than four times the length or 2 + 4*W.

Now the perimeter is given as 20.

The perimeter is also equal to 2*W + 2*L

So 2*W + 2*L =20

=>2W + 2(2 + 4W) = 20

=> 2W + 4 + 8W = 20

=> 10 W = 16

=> W = 16/10

The length is 2 + 4W = 2 + 4*16/10 = 8.4

Therefore the required length is 8.4.

neela | Student

If the width is x, then the length is 2 more than 4times width. So length is 4x+2.

The perimeter p is 2(length+width) = 2(4x+2+x) = 2(5x+2).

But the actual perimeter = 20.

=> 2(5x+2) = 20

=> (5x+2) = 20/2 = 10.

=> 5x+2 = 10.

=> 5x = 10-2 = 8

=>x = 8/5 = 1.6.

=> 4x+2 = 1.6*4+2 = 8.4 units.

So width = 1.6 units, and length = 8.4 units.

giorgiana1976 | Student

We have to specify that the length of a rectangle is bigger than the width.

We'll put the width of the rectangle to be a units and the length  be b inches.

We know, from enunciation, that the width is 2 units more than 4 times it's length and we'll write the constraint mathematically:

a - 2 = 4b

We'll subtract 4b and add 2 both sides:

a - 4b = 2 (1)

The perimeter of the rectangle is 20 units.

We'll write the perimeter of the rectangle:

P = 2(a+b)

20 = 2(a+b)

We'll divide by 2:

10 = a + b

We'll use the symmetric property:

a + b = 10 (2)

a - 4b + 4a + 4b = 2 + 40

We'l eliminate and combine like terms:

5a = 42

We'll divide by 5:

a = 42/5

a = 8.4 units

8.4 + b = 10

b = 10 - 8.4

b  =  1.6 units

So, the width of the rectangle is of 8.4 units and the lengths of the rectangle is of 1.6 units.

Since the width cannot be larger than the length, we'll change and we'll put the length of 8.4 units and the width of 1.6 units.