The perimeter of a triangle with vertices (0, | vedclass

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(A) To find the perimeter of a triangle,we calculate the sum of the lengths of all its sides. Let the vertices of the triangle be $A(0,4)$,$O(0,0)$,an

Vedclass · Accepted Answer

$12$

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(A) To find the perimeter of a triangle,we calculate the sum of the lengths of all its sides. Let the vertices of the triangle be $A(0,4)$,$O(0,0)$,and $B(3,0)$.The perimeter of $	riangle AOB = 	ext{Sum of the lengths of all its sides} = d(AO) + d(OB) + d(AB)$.The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.$1$. Distance $d(AO) = \sqrt{(0-0)^2 + (0-4)^2} = \sqrt{0 + 16} = 4$.$2$. Distance $d(OB) = \sqrt{(3-0)^2 + (0-0)^2} = \sqrt{9 + 0} = 3$.$3$. Distance $d(AB) = \sqrt{(3-0)^2 + (0-4)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.Perimeter $= 4 + 3 + 5 = 12$.Thus,the required perimeter of the triangle is $12$.

The perimeter of a triangle with vertices $(0,4), (0,0)$ and $(3,0)$ is

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