Use the trapezium rule with intervals of width 0,5 to find an approximation for integral_1^(2.5) (1)/(1+ln x) dx, giving your answer correct to 2 decimal places.

Point C lies on the x-axis, so its y-coordinate is 0. Let $C = (x_C, 0)$.
The coordinates of A are $(1,4)$ and B are $(6,5)$.
The slope of line AB is $m_{AB} = \frac{5-4}{6-1} = \frac{1}{5}$.
Angle ABC is a right angle, so line AB is perpendicular to line BC.
The product of their slopes must be -1: $m_{AB} \cdot m_{BC} = -1$.
The slope of line BC is $m_{BC} = \frac{0-5}{x_C-6} = \frac{-5}{x_C-6}$.

So, the coordinates of C are $(7,0)$.

Line CD is parallel to line BA. The slope of BA is the same as the slope of AB.
So, $m_{CD} = m_{AB} = \frac{1}{5}$.

Using point $C(7,0)$ and slope $m_{CD} = \frac{1}{5}$:

Multiplying by 5 to clear the fraction:

The equation of the line CD is $\boxed{x - 5y - 7 = 0}$.

Line AD is parallel to the y-axis, which means it is a vertical line. Since A is $(1,4)$, the x-coordinate of D must be 1. So, $D = (1, y_D)$.
Point D lies on the line CD, whose equation is $y = \frac{1}{5}x - \frac{7}{5}$. Substitute $x=1$ into this equation:

So, the coordinates of D are $(1, -\frac{6}{5})$.

The vertices are $A(1,4)$, $B(6,5)$, $C(7,0)$, $D(1, -\frac{6}{5})$.
We know that AB is parallel to CD (from part a, $m_{AB} = m_{CD} = \frac{1}{5}$).
We also know that angle ABC is a right angle, which means BC is perpendicular to AB.
Therefore, ABCD is a right trapezium (or right trapezoid) with parallel sides AB and CD, and height BC.

Length of AB (base 1):

Length of CD (base 2):

Length of BC (height):

The area of the quadrilateral ABCD is $\boxed{28.6 \text{ units}^2}$.