The sides of a triangle are 3x + 4y,4x + 3y,a | vedclass

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(C) Let the sides be $a = 3x + 4y$,$b = 4x + 3y$,and $c = 5x + 5y$.Since $x, y > 0$,we compare the sides. Note that $c = 5x + 5y$ is the largest side

Vedclass · Accepted Answer

Obtuse angled

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(C) Let the sides be $a = 3x + 4y$,$b = 4x + 3y$,and $c = 5x + 5y$.Since $x, y > 0$,we compare the sides. Note that $c = 5x + 5y$ is the largest side because $5x + 5y > 3x + 4y$ and $5x + 5y > 4x + 3y$.For a triangle to be obtuse,the square of the largest side must be greater than the sum of the squares of the other two sides $(c^2 > a^2 + b^2)$.$a^2 + b^2 = (3x + 4y)^2 + (4x + 3y)^2 = (9x^2 + 24xy + 16y^2) + (16x^2 + 24xy + 9y^2) = 25x^2 + 48xy + 25y^2$.$c^2 = (5x + 5y)^2 = 25x^2 + 50xy + 25y^2$.Comparing $c^2$ and $a^2 + b^2$,we see that $25x^2 + 50xy + 25y^2 > 25x^2 + 48xy + 25y^2$ since $x, y > 0$.Thus,$c^2 > a^2 + b^2$,which implies the triangle is obtuse angled.

The sides of a triangle are $3x + 4y$,$4x + 3y$,and $5x + 5y$ units,where $x, y > 0$. The triangle is:

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