Arectangular garden must have a perimeter of 160 feet and an area of at least 1300 square feet. describe the possible lengths of the garden. round your answers to the nearest foot.

9514 1404 393

  y = 21

Step-by-step explanation:

In straightforward fashion, we can find the midpoints of the two segments, and set the distance between them equal to 10.

  M = midpoint of AB

  M = (A+B)/2 = ((-1, 0) +(-3, 6))/2 = (-4, 6)/2 = (-2, 3)

  N = midpoint of CD

  N = (C+D)/2 = ((3, 1) +(5, y))/2 = (8, 1+y)/2 = (4, (1+y)/2)

Then the distance MN is given by the distance formula.

  |N-M| = 10 = √((4 -(-2))² +((1+y)/2 -3)²) . . . . . distance formula*

  100 = 36 +((1+y)/2 -3)² . . . . . square both sides

  64 = ((1 +y)/2 -3)² . . . . . . . . . subtract 36

  8 = (1 +y)/2 -3 . . . . . . . . . . . . (positive) square root

  11 = 1+y/2 . . . . . . . . add 3

  22 = 1+y . . . . . . . . . multiply by 2

  21 = y . . . . . . . . . . . subtract 1

* The distance formula is an application of the Pythagorean theorem. Basically, the differences of x- and y-coordinates are taken to be the legs of a right triangle. They length of the hypotenuse is the distance between the two points.

  d² = (x2 -x1)² +(y2 -y1)² . . . . . . . from the Pythagorean theorem

  d = √((x2 -x1)² +(y2 -y1)²) . . . . . the usual distance formula