The perimeter of a triangle is $14 a^{2} + 20 a + 13$ two of its sides are $3 a^{2} + 5 a + 1$ and $a^{2} + 10 a - 6$ . Find its third side.

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Solution

As we know that the perimeter of a triangle is the sum of all the sides.
Suppose that the third side of the triangle is $X$ then,
According to the given data in the question-
$14 a^{2} + 20 a + 13 = 3 a^{2} + 5 a + 1 + a^{2} + 10 a - 6 + X$
$14 a^{2} + 20 a + 13 = 4 a^{2} + 15 a - 5 + X$
$X = 14 a^{2} + 20 a + 13 - (4 a^{2} + 15 a - 5)$
$X = 14 a^{2} + 20 a + 13 - 4 a^{2} - 15 a + 5$
$X = 10 a^{2} + 5 a + 18$
Hence, the third side will be $10 a^{2} + 5 a + 18$ .

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