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
2x2 + 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