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