the perimeter of a triangle is 7x sq-17xy+5y sq+8 and two of its sides are 4x sq-7xy+4y sq-3 and 5+6y sq - 8xy +x .find the third side of the triangle

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Solution

Answer :

Given : Perimeter of triangle = 7 x ² - 17 xy + 5 y ² + 8
And
Length of two sides are 4 x ² -7 xy + 4 y ² - 3 and 5 + 6 y ² - 8 xy + x
Let length of third side = l

We know perimeter of triangle = Sum of all three sides ,So

4 x ² -7 xy + 4 y ² - 3 + 5 + 6 y ² - 8 xy + x + l = 7 x ² - 17 xy + 5 y ² + 8

4 x ² -15 xy + 10 y ² + 2 + x + l = 7 x ² - 17 xy + 5 y ² + 8

l = 7 x ² - 17 xy + 5 y ² + 8 - ( 4 x ² -15 xy + 10 y ² + 2 + x )

l = 7 x ² - 17 xy + 5 y ² + 8 - 4 x ² +15 xy - 10 y ² - 2 - x

l = 3 x ² - 2 xy - 5 y ² + 6 - x

So,

Length of third side = 3 x ² - 2 xy - 5 y ² + 6 - x ( Ans )

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