Find the perimeter of the original rectangle if all measurements are whole numbers.

Question

The area of a rectangle is 120m^2. If its length is decreased by 8 meters and its width is increased by 4m then the area does not change. Find the perimeter of the original rectangle if all measurements are whole numbers.

Answer 1

1. A = L*W = 120   {Eqn 1}
2. A = (L-8)*(W+4) = 120
3. (L-8)*(W+4) = LW
4. LW + 4L - 8W -32 = LW
5. 4L -8W - 32 = 0
6. L - 2W - 8 = 0
7. L = 2W + 8
8. A = (2W+8)W = 120
9. 2W² + 8W = 120
10. W² + 4W = 60
11. W² + 4W - 60 = 0
12. (W + 10)(W - 6) = 0
13. W = -10 or W = +6, but W cannot be a negative number since it is a length.  So, W must equal 6
14. If W = 6, then L =20 (by Eqn 1)
15. The perimeter is therefore (2*6) + (2*20) = 12 + 40 = 52m
16. Doublecheck the answer.  After changing the dimensions to form the new rectangle, the new W is 10 and new L is 12.  This has an area of 120m².  So, it all checks out.

Answer 2

xy = 120

(x - 8)(y + 4) = 120

xy - 8y + 4x - 32 = 120

120 - 8y + 4x - 32 =120

-8y + 4x -32 = 0

-8y + 4x = 32

-2y + x = 8

-2y + 120/y = 8

-2y² + 120 = 8y

-2y² - 8y + 120 = 0

y² + 4y - 60 = 0

(x + 10)(x - 6) = 0

 

The original is 20 x 6.