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The perimeter of a triangle is 8y^{2} – 9y + 4 and its two sides are 3y^{2} – 5y and 4y^{2} + 12. Find its third side.

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#### Solution

Perimeter of the triangle = Sum of three sides

= 8y^{2} – 9y + 4

Sum of two sides = 3y^{2} – 5y + 4y^{2} + 12

= 7y^{2} − 5y + 12

∴ (8y^{2 } – 9y + 4) – (7y^{2} – 5y + 12)

= 8y^{2} – 9y + 4 – 7y^{2} + 5y – 12

= y^{2} – 4y – 8

Hence third side = y^{2} – 4y – 8

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