What is the measure of the third side of a triangle given that its two sides are $a^{2} - 2 a + 1$ and $3 a^{2} - 5 a + 3$ and has a perimeter $6 a^{2} - 4 a + 9$ ?

2 a^{2} - 3 a - 5

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2 a^{2} + 3 a - 5

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2 a^{2} + 3 a + 5

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2 a^{2} - 3 a + 5

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Solution

The correct option is B $2 a^{2} + 3 a + 5$
Let the third side of the triangle be $x .$

The perimeter of a triangle is equal to the sum of its sides so,
$\begin{matrix} x + (a^{2} - 2 a + 1) + (3 a^{2} - 5 a + 3) & = 6 a^{2} - 4 a + 9 x + (a^{2} + 3 a^{2}) + (- 2 a - 5 a) + (1 + 3) & = 6 a^{2} - 4 a + 9 x + 4 a^{2} - 7 a + 4 & = 6 a^{2} - 4 a + 9 x & = (6 a^{2} - 4 a^{2}) + (- 4 a + 7 a) + (9 - 4) = 2 a^{2} + 3 a + 5 \end{matrix}$

Hence, option $C$ is correct.

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