Answer:
[tex]Perimeter\,\,of \,\,triangle =\frac{3x^2+7x+21}{3x^2}[/tex] so, Option D is correct.
Step-by-step explanation:
Perimeter of triangle = Sum of all sides
Perimeter of triangle = a+ b+ c
a= 1/3x, b= x+2/x and c = 7/x^2
[tex]Perimeter\,\,of\,\,triangle =\frac{1}{3x}+\frac{x+2}{x}+\frac{7}{x^2}[/tex]
LCM of 3x, x and x^2 : 3x^2
[tex]Perimeter\,\,of\,\,triangle =\frac{x+3x(x+2)+7(3)}{3x^2}\\Perimeter\,\,of \,\,triangle =\frac{x+3x^2+6x+21}{3x^2}\\Perimeter\,\,of\,\,triangle =\frac{3x^2+7x+21}{3x^2}[/tex]
So, [tex]Perimeter\,\,of\,\,triangle =\frac{3x^2+7x+21}{3x^2}[/tex] so, Option D is correct.
