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Answer: The required point that could also be a vertex of the square is K(5, 0).
Step-by-step explanation: Given that the sides of a square are 3 cm long and one vertex of the square is at (2,0) on a square coordinate grid marked in centimeter units.
We are to select the co-ordinates of the point that could also be a vertex of the square.
To be a vertex of the given square, the distance between the point and the vertex at (2, 0) must be 3 cm.
Now, we will be suing the distance formula to calculate the lengths of the segment from the point to the vertex (2, 0).
If the point is F(-4, 0), then the length of the line segment will be
[tex]\ell=\sqrt{(-4-2)^2+(0-0)^2}=\sqrt{6^2+0^2}=\sqrt{6^2}=6~\textup{cm}\neq 3~\textup{cm}.[/tex]
If the point is G(0, 1), then the length of the line segment will be
[tex]\ell=\sqrt{(0-2)^2+(1-0)^2}=\sqrt{2^2+1^2}=\sqrt{4+1}=\sqrt5~\textup{cm}\neq 3~\textup{cm}.[/tex]
If the point is H(1, -1), then the length of the line segment will be
[tex]\ell=\sqrt{(1-2)^2+(-1-0)^2}=\sqrt{1^2+1^2}=\sqrt{1+1}=\sqrt2~\textup{cm}\neq 3~\textup{cm}.[/tex]
If the point is J(4, 1), then the length of the line segment will be
[tex]\ell=\sqrt{(4-2)^2+(1-0)^2}=\sqrt{2^2+1^2}=\sqrt{4+1}=\sqrt5~\textup{cm}\neq 3~\textup{cm}.[/tex]
If the point is K(5, 0), then the length of the line segment will be
[tex]\ell=\sqrt{(5-2)^2+(0-0)^2}=\sqrt{3^2+0^2}=\sqrt{3^2}=3~\textup{cm}.[/tex]
Thus, the required point that could also be a vertex of the square is K(5, 0).
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