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Two adjacent sides of a triangle are $3x^2-5y^2$ and $7x^2-xy$. Find its perimeter.
Given: Two adjacent sides of a rectangle are $3x^2-5y^2$and $7x^2-xy$
To do: Find its perimeter.
Solution:
Perimeter of rectangle =$2(length + breadth)$
Perimeter =$2(3x^2-5y^2+7x^2-xy)$
Perimeter =$2(10x^2 -5y^2-xy)$
Perimeter =$20x^2-10y^2 -2xy$
Therefore , the perimeter is $20x^2-10y^2 -2xy$
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