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:

3 and 8

One side of triangle = 7 cm

Perimeter of triangle = 18 cm

Area of triangle = $\sqrt{108}$

And we know that,

Area of a triangle = \(\sqrt {s(s-a)(s-b)(s-c) }\)

according to the question,

18 = (7 + b + c)

= (b + c) = 11

= c = 11 - b 

S = \(\frac{18}{2}\) = 9

\(\sqrt {108}\) = \(\sqrt {[(9)(9 - 7)(9 - b)(9 - c)] }\)

Squaring both sides,

108 = 18 × (9 - b) × (9 - c)

Put the value of c in above equation,

6 = (9 - b)[9 - (11 - b)]

= 6 = (9 - b)(b - 2)

= 6 = 9b + 2b - b² - 18

= b² - 11b + 24 = 0

=  b² - 8b - 3b + 24 = 0

= b(b - 8) - 3(b - 8) = 0

= (b - 8)(b - 3) = 0

= b = 8, 3

Take side b = 8

Then, c = 11 - 8 = 3

So, the other two sides are = 3 and 8