Question

If the length of three sides of a trapezium, other than the base is equal to 10 cm each, then find the area of trapezium when it is maximum.

Answer 1

Answer:

Maximum area (A) = 15√75 cm² (or) 75√3 cm²

Step-by-step explanation:

Let ABCD be the given trapezium with

          AD = DC = BC = 10 cm

Draw DP ⊥ AB and CQ ⊥ AB

and let AP = x cm

            QB = x cm

Now,

      Area of trapezium

                        = 1/2 [10 + (10+2x) ] √100 - x²

                  A = (x+10)√100-x²

Let,             S = (x+10)².(100 - x²)

      ⇒ dS / dx = - 2x (x +10)² + 2(x + 10)(100 - x²)

                    = 2(x + 10)².(-x + 10 - x)

                     = 2(x + 10)².(10-2x)

          dS/dx = 0 ⇒ x = 5

dS/dx = -4(x+10)² + 4(x+10)(10-2x)

              = - 900

∴ Maximum area (A) 15√75 cm² (or) 75√3 cm²

Good luck !!

Answer 2

Final answer x=5 and area is 75root3