Problem:
Geometry-Quadrilaterals (InterMath)

The perimeter of a rectangle is tripled and the new perimeter is 66.

The original length is 3 more than the original width.

Find the original length and width.

 

Solution:

Let A= original rectangle
B= new rectangle= original rectangle tripled

Remember that perimeter of a rectangle is
P= 2w + 2l
P= (2 * width) + (2 * length)

So the Perimeter of A= 2x + 2( x+3)and the Perimeter of B= 3 x Perimeter A =66

= 3( 2x + 2(x+3))= (6x+ 6x +18)
= 12x + 18= 66

Now solve for x.


12x+18= 66
12x= 48
x=4= original width

Original length: Plug in the original width (4) into l = w +3
Original length = 7

The original perimeter:
Perimeter of A= 2(4) + 2(4 + 3)
Perimeter of A= 22

The original width of the original perimeter = 4


The original length of the original perimeter = 7

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