Draw the graphs of the equations $x=3, x=5$ and $2x-y-4=0$. Also,find the area of the quadrilateral formed by these lines and the $x$-axis.

(8) Given equations of lines are $2x-y-4=0$,$x=3$,and $x=5$.
For the line $2x-y-4=0$,we find two points:

$x$	$0$	$2$
$y=2x-4$	$-4$	$0$

Plot the points $P(0, -4)$ and $Q(2, 0)$ and draw the line passing through them. Also,draw the vertical lines $x=3$ and $x=5$.
The quadrilateral formed by the lines $x=3$,$x=5$,the $x$-axis,and the line $2x-y-4=0$ is a trapezium with parallel sides at $x=3$ and $x=5$.
At $x=3$,$y = 2(3)-4 = 2$. So,point $D$ is $(3, 2)$.
At $x=5$,$y = 2(5)-4 = 6$. So,point $C$ is $(5, 6)$.
The parallel sides of the trapezium are $AD = 2$ units and $BC = 6$ units. The distance between them is $AB = 5-3 = 2$ units.
Area of trapezium $= \frac{1}{2} \times (\text{sum of parallel sides}) \times (\text{height})$
$= \frac{1}{2} \times (AD + BC) \times AB$
$= \frac{1}{2} \times (2 + 6) \times 2$
$= 8 \text{ sq units}$.
Thus,the area of the quadrilateral is $8 \text{ sq units}$.