The required trapezium is as given in figure.

Draw perpendiculars DP and CQ on AB.

Let AP = x cm.

Note that Δ APD ≅ ΔBQC.

Therefore,

QB = x cm.

**Also,**

**By Pythagoras theorem,**

DP = QC = \(\sqrt{100-x^2}.\)

**Let A be the area of the trapezium.**

**Then,**

A = A(x) = \(\frac{1}{2}\)

**(sum of parallel sides) x (height)**

**Now,**

A'(x) = 0 gives

2x^{2} + 10x - 100 = 0,

i.e., x = 5 and x = - 10

**Since x represents distance, it cannot be negative.**

**So, x = 5.**

**Thus, **

**Area of trapezium is maximum at x = 5 and the maximum area is given by**