What is the area of arectangle whose perimeter is 40 and the length is 4 more than 4 time the width?

To calculate the area of the rectangle, we'll have to know
the values of the lengths of the sides of the
rectangle.

Let's note the length as a and the width as
b.

From enunciation we know
that:

a = 4 + 4b
(1)

From enunciation we also know that the
perimeter is 40 units.

The formula for perimeter of a
rectangle is:

P =
2(a+b)

40 =
2(a+b)

We'll divide by 2 and we'll use the symmetric
property:

a+b = 20
(2)

To calculate a and b we have to solve
the system of equations (1) and (2).

We'll substitute (1)
in (2):

4 + 4b + b = 20

We'll
subtract 4:

5b = 20-4

5b =
16

We'll divide by
5:

b = 16/5
units

We'll substitute b in
(2):

a + 16/5 = 20

We'll
subtract 16/5:

a = 20 -
16/5

a =
84/5

Now, we'll calculate the area of the
rectangle:

A = a*b

A =
84*16/5*5

A = 53.76 square
units