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The sides of a square are 3 cm long. One vertex of the
square is at (2,0) on a square coordinate grid marked in
centimeter units. Which of the following points could
also be a vertex of the square?
F. (−4, 0)
G. ( 0, 1)
H. ( 1,−1)
J. ( 4, 1)
K. ( 5, 0)


Sagot :

K (5.0) It's easy just use 2plus3and that's it.

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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