If the vertices of a triangle are (0, 4),(4, | vedclass

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(A) Let the vertices be $A(0, 4)$,$B(4, 1)$,and $C(7, 5)$.The distance $AB = \sqrt{(4-0)^2 + (1-4)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25

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$5(2 + \sqrt{2})$

Vedclass · Answer

(A) Let the vertices be $A(0, 4)$,$B(4, 1)$,and $C(7, 5)$.The distance $AB = \sqrt{(4-0)^2 + (1-4)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.The distance $BC = \sqrt{(7-4)^2 + (5-1)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.The distance $AC = \sqrt{(7-0)^2 + (5-4)^2} = \sqrt{7^2 + 1^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2}$.The perimeter of the triangle is $AB + BC + AC = 5 + 5 + 5\sqrt{2} = 10 + 5\sqrt{2} = 5(2 + \sqrt{2})$.

If the vertices of a triangle are $(0, 4)$,$(4, 1)$,and $(7, 5)$,find its perimeter.

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