Question

The perimeter of triangle is 7x² - 17xy + 15y² + 8 and two side are 3x² - 6xy + 4y² - 5 and 2x² - 9xy + 6. Find the third side if triangle.​

Answer 1

Answer:

Since, the sum of lengths of any two sides in a triangle should be greater than the length of the third side.

It is given that two sides of triangle are 12cm and 15cm.

Therefore, the third side should be less than 12+15=27cm.

And also the third side cannot be less than the difference of the two sides.

Therefore, the third side has to be more than 15−12=3cm

Therefore, the third side should be the length more than 3cm and less than 27cm.

therefore, 3cm< 3rd side <27cm

Answer 2

Answer :-

•  The third side of the triangle is (4x² - 2xy + 5y² - 2a² + 7)

       __________

Given that:

 Perimeter of triangle is  7x² - 17xy + 15y² + 8

 First Side of triangle is 3x² - 6xy + 4y² - 5  

 Second Side of the triangle is  2x² - 9xy + 6.

Need to find:

 Third side of that triangle

❍ Let's say that, the third side of the triangle is 'A'

As we know that,

• Perimeter of a triangle  = Sum of all sides

       __________

T H E R E F O R E,

$\sf : \; \implies (3x^{2} - 6xy + 4y^{2} - 5) + (2a^{2} - 9xy + 6y^{2} + 6) + A = 7x^{2} - 17xy + 15y^{2} + 8$

$\sf : \; \implies 3x^{2} - 6xy + 4y^{2} - 5 + 2a^{2} - 9xy + 6y^{2} + 6 + A = 7x^{2} - 17xy + 15y^{2} + 8$

$\sf : \; \implies 3x^{2} - 6xy - 9xy + 4y^{2} + 6y^{2} + 2a^{2} + 6 - 5 + A = 7x^{2} - 17xy + 15y^{2} + 8$

$\sf : \; \implies 3x^{2} - 15xy + 10y^{2} + 2a^{2} + 1 + A = 7x^{2} - 17xy + 15y^{2} + 8$

$\sf : \; \implies A = (7x^{2} - 17xy + 15y^{2} + 8) - (3x^{2} - 15xy + 10y^{2} + 2a^{2} + 1)$

$\sf : \; \implies A = 7x^{2} - 17xy + 15y^{2} + 8 - 3x^{2} + 15xy - 10y^{2} - 2a^{2} - 1$

$\sf : \; \implies A = 7x^{2} - 3x^{2} - 17xy + 15xy + 15y^{2} - 10y^{2} - 2a^{2} + 8 - 1$

$\boxed{\bf{ : \; \implies A = 4x^{2} - 2xy + 5y^{2} - 2a^{2} + 7}} \; \bigstar$

       __________

• Henceforth, the third side of the triangle is (4x² - 2xy + 5y² - 2a² + 7)

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