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

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neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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

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