Therefore, QB = x cm Also, by
Pythagoras theorem, DP = QC = 
2
100
−x Let A be the area of the trapezium Then
A ≡ A(x) = 1
2 (sum of parallel sides) (height)
=
(
2)
1 (2
10 10)
100
2
x
x
+
+
−
=
(
2)
(
10)
100
x
x
+
−
or
A′(x) =
(
)
2
( 2 )2
(
10)
100
2 100
x
x
x
x
−
+
+
−
−
=
2
2
2
10
100
100
x
x
x
−
−
+
−
Now
A′(x) = 0 gives 2x2 + 10x – 100 = 0, i