[tex] {\qquad\qquad\huge\underline{{\sf Answer}}} [/tex]
As per the given information ~
The given triangles are similar, so their corresponding sides should be in same ratio :
[tex]\qquad \sf \dashrightarrow \: \dfrac{2x}{4} = \cfrac{5}{x - 3} [/tex]
[tex]\qquad \sf \dashrightarrow \: \dfrac{x}{2} = \cfrac{5}{x - 3} [/tex]
[tex]\qquad \sf \dashrightarrow \: x(x - 3) = 10[/tex]
[tex]\qquad \sf \dashrightarrow \: {x}^{2} - 3x = 10 [/tex]
[tex]\qquad \sf \dashrightarrow \: {x}^{2} - 3x - 10 = 0[/tex]
[tex]\qquad \sf \dashrightarrow \: {x}^{2} - 5x + 2x - 10 = 0[/tex]
[tex]\qquad \sf \dashrightarrow \: x(x - 5) + 2(x - 5) = 0[/tex]
[tex]\qquad \sf \dashrightarrow \: (x + 2)(x - 5) = 0[/tex]
so, x = 5 or -2
but since side length can't be negative, we will take x = 5 neglecting the negative value (-2)
So, side lengths of unknown sides are :
[tex] \qquad \sf \dashrightarrow \: {2x = 2 × 5 = 10 units} [/tex]
and
[tex]\qquad \sf \dashrightarrow \: x - 3 = 5 - 3 = 2 \: \: units[/tex]
