it is possible that a rectangular plot of land has perimeter 360m and 38500m^2. if yes, find its sides.​

2 thoughts on “it is possible that a rectangular plot of land has perimeter 360m and 38500m^2. if yes, find its sides.​”

1. Answer :-

   Given :-

   • Perimeter = 360 m
   • Area = 38500 m²

   To Find :-

   • Sides ( if possible )

   Solution :-

   → Perimeter = 2 ( Length + Breadth )

   → 360 = 2 ( L + B )

   → L + B = 360 / 2

   → L + B = 180

   → Area = Length × Breadth

   → 38500 = L × B

   Now we have 2 equation :-

   1. L + B = 180
   2. L × B = 38500

   Now, squaring ( L + B ) :-

   → ( L + B )² = L² + B² + 2LB

   → (180)² = L² + B² + 2 × 38500

   → 32400 = L² + B² + 77000

   → L² + B² = 32400 – 77000

   → L² + B² = – 44600

   We know that, square of real number is always positive. So, sum of square of two numbers will also be always positive.

   Here, we are getting the value of square of two numbers as negative.

   Hence, there is no possible value of Length and Breadth.

   Reply

2. Step-by-step explanation:

   Let us try. Assume that length and breadth of a rectangular plot are X and Y respectively. Therefore

   2X + 2Y = 360(equation 1) and XY = 8500(equation 2)

   2X + 2Y = 360, 2X = 360 — 2Y, X = 180 — Y and Y = 180 — X

   XY = 8500, X(180 — X) = 8500, 180X — X^2 = 8500, X^2 — 180X + 8500 = 0

   X^2 -180X + 8500 — 400 = — 400

   X^2 — 180X +8100 = — 400, (X — 90)×(X — 90) = — 400

   (X — 90)×(X — 90) = (+/—)(20 × 20)

   X — 90 = (+/—)(20)

   X = (90+20) or (90-20)

   X = 110 or 70

   Same can be applied for Y

   So 2(110+70) = 360 perimeter is found.

   But 110 × 70 = 7700 area is not found.

   So a rectangle with an area of 8500 can not have perimeter of 360 or a rectangle with perimeter 360 can not have an area of 8500. So area and perimeter do not match and sides can not be found.

   Reply