Length of the given rectangle is 8 units.
Option - D
Solution:
Given that
Length of a rectangle = [tex](2x-1)\text { units}[/tex]
Width of a rectangle = [tex](x-2)\text{ units}[/tex]
Area of rectangle = 20 square units
Need to calculate length of rectangle.
[tex]\text { Formula of area of rectangle }=\text { Length of a rectangle } \times \text { Width of a rectangle }[/tex]
Substituting given value of Area of rectangle and expressions for length and width of rectangle in formula, we get
[tex]20=(x-2)(2 x-1)[/tex]
On solving above equation for x we get
[tex]\begin{array}{l}{20=x(2 x-1)-2(2 x-1)} \\\\ {\Rightarrow 20=2 x^{2}-x-4 x+2} \\\\ {\Rightarrow 2 x^{2}-5 x+2-20=0} \\\\ {\Rightarrow 2 x^{2}-5 x-18=0}\end{array}[/tex]
On splitting the middle term in such a way that product of split terms comes as [tex]2x^2\times -18 = -36x^2[/tex] and summation comes as -5x, we get
[tex]\begin{array}{l}{\Rightarrow 2 x^{2}+4 x-9 x-18=0} \\\\ {=>2 x(x+2)-9(x+2)=0} \\\\ {\Rightarrow(2 x-9)(x+2)=0} \\\\ {\Rightarrow(2 x-9)=0 \text { or }(x+2)=0} \\\\ {\Rightarrow x=\frac{9}{2} \text { or } x=-2}\end{array}[/tex]
[tex]\begin{array}{l}{\text { When } x=\frac{9}{2}, \text { length }=(2 x-1) \text { units }=\left(2 \times \frac{9}{2}-1\right)=8 \text { units }} \\\\ {\text { When } x=-2, \text { length }=(2 x-1) \text { units }=(2 \times(-2)-1)=-5 \text { units }}\end{array}[/tex]
As length cannot be negative hence we can ignore negative value.
So length is 8 units.
