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

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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