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

*print*Print*list*Cite

### 2 Answers

Given the perimeter of the rectangle = 40.

The length is 4 times width.

Let l be the length. and w be the width. Then l = 4w.

Therefore area of the perimeter of the rectangle = 40 =2 l+2w = 2*4w+2w

40 = 10w.

40/10 = w.

w = 4

Therefore width = 8 and length = 4w = 4*4 = 16.

Therefore area of the rectangle = lw = 16*4 = 64 sq units.

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**