The perimeter of a triangle is 8 + 13a + 7a² and two of its sides are 2a² + 3a + 2 and 3a² - 4a - 1. Find the third side of the triangle.

Given:

Perimeter of a triangle = 8 + 13a + 7a²

Side 1 (S1) = 2a² + 3a + 2

Side 2 (S2) = 3a² - 4a - 1

Side 3 (S3) = ?

We know the formula,

Perimeter of a triangle = S1 + S2 + S3

⟹ S3 = Perimeter of a triangle - (S1 + S2)

Substituting the values above, we get:

S3 = (8 + 13a + 7a²) - ((2a² + 3a + 2) + (3a² - 4a - 1))

First, let's find S1 + S2:

$\begin{array}{rcccc} 2 & + & 3a & + & 2a^2 \\ -1 & - & 4a & + & 3a^2 \\ \hline 1 & - & a & + & 5a^2 \\ \hline \end{array}$

Sum (S1 + S2) = 1 - a + 5a²

Now, we have S3 = (8 + 13a + 7a²) - (1 - a + 5a²)

Subtract Sum (S1 + S2) from perimeter of a triangle:

$\begin{array}{rcccc} 8 & + & 13a & + & 7a^2 \\ +1 & - & a & + & 5a^2 \\ -\phantom{1} & + & & - \\ \hline 7 & + & 14a & + & 2a^2 \\ \hline \end{array}$

S3 = 7 + 14a + 2a²

∴ The third side of the triangle is 7 + 14a + 2a².