Question

The sides of a triangle are 3x²-y²,4x²-7xy+4y² and-3x²+7xy+8y².Find it's perimeter​

Answer 1

Answer :

• The perimeter of the triangle = 4x² + 11y².

Explanation :

Given :-

The sides of a triangle are :

• 3x² – y², 4x² – 7xy + 4y² & –3x² + 7xy + 8y².

To Find :-

• The perimeter of the triangle.

Solution :-

We know that,

Perimeter of the triangle = a + b + c

Where,

• a = 3x² – y².
• b = 4x² – 7xy + 4y².
• c = –3x² + 7xy + 8y².

$\implies ({3x}^{2} - {y}^{2}) + ( {4x}^{2} - 7xy + {4y}^{2} ) + ( { - 3x}^{2} + 7xy + {8y}^{2} ) \\ \\ \implies {3x}^{2} - {y}^{2} + {4x}^{2} - 7xy + {4y}^{2} - {3x}^{2} + 7xy + {8y}^{2} \\ \\ \implies {3x}^{2} + {4x}^{2} - {3x}^{2} - {y}^{2} + {4y}^{2} + {8y}^{2} - 7xy + 7xy \\ \\ \implies {4x}^{2} + {11y}^{2} - 0 \\ \\ \implies {4x}^{2} + {11y}^{2}$

Hence, the perimeter of the triangle is 4x² + 11y².

Answer 2

Answer:

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