SOLUTION: Half of the perimeter of a rectangle is 51 cm. If the difference between the length ang the width is 13 cm, find the area of the rectangle.

Half of the perimeter of a rectangle is 51 cm. If the 
difference between the length and the width is 13 cm, 
find the area of the rectangle.

First find the length L and the width W:

  __________
 |    L     |
W|          |W
 |__________|
      L

   P = PERIMETER = DISTANCE AROUND = TOP SIDE + 
                                      RIGHT SIDE + 
                                         BOTTOM SIDE + 
                                             LEFT SIDE = 
                          Length + Width + Length + Width = 
                                          2·Length + 2·Width = 
                                                         2L + 2W, so

P = 2L + 2W

>>...Half of the perimeter of a rectangle is 51 cm...<<

(1/2)P = 51

Multiply both sides by 2·P

2·(1/2)P = 2·51

       P = 102

Since P = 2L + 2W,

P = 2L + 2W = 102 

>>...the difference between the length and the width is 13 cm...<<

     L - W = 13

So you have this system of two equations in two unknowns:

2L + 2W = 102
 L -  W =  13 

Can you solve this?  If not post again to ask that.

Solution to that system of equations:  L = 32 cm, W = 19 cm

Now AREA = LENGTH × WIDTH = LW = (32 cm)(19 cm) = 608 cm²

Edwin