If one side of a triangle is 7 with its perimeter equal to 18, and area equal to $$\sqrt{108}$$, then the other two sides are:

Given,

One side of the triangle (a) = 7

Let the other two sides of the triangle are b and c

Perimeter of the triangle = 18

$$=$$>  a+b+c = 18

$$=$$>  7+b+c = 18

$$=$$>  b+c = 11 .................(1)

Semi perimeter of the triangle (s) = $$\frac{18}{2}=9$$

Area of the triangle = $$\sqrt{108}$$

$$=$$>  $$\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}=\sqrt{108}$$

$$=$$>  $$\sqrt{9\left(9-7\right)\left(9-b\right)\left(9-c\right)}=\sqrt{108}$$

$$=$$>  $$9\left(2\right)\left(9-b\right)\left(9-c\right)=108$$

$$=$$>  $$\left(9-b\right)\left(9-c\right)=6$$

$$=$$>  $$81-9c-9b+bc=6$$

$$=$$>  $$81-9\left(b+c\right)+bc=6$$

$$=$$>  $$81-9\left(11\right)+bc=6$$

$$=$$>  $$81-99+bc=6$$

$$=$$>  $$bc=6+99-81$$

$$=$$>  $$bc=24$$

$$=$$>  $$c=\frac{24}{b}$$

Substituting $$c=\frac{24}{b}$$ in equation (1)

$$b+\frac{24}{b}=11$$

$$=$$>  $$b^2+24=11b$$

$$=$$>  $$b^2-11b+24=0$$

$$=$$>  $$b^2-8b-3b+24=0$$

$$=$$>  $$b\left(b-8\right)-3\left(b-8\right)=0$$

$$=$$>  $$\left(b-8\right)\left(b-3\right)=0$$

$$=$$>  $$b=8$$  or  $$b=3$$

when $$b=8$$,  $$c=\frac{24}{8}=3$$

when $$b=3$$,  $$c=\frac{24}{3}=8$$        

$$\therefore\ $$The other two sides of the triangle are 3 and 8

Hence, the correct answer is Option D