Option A: ![O(0,0), S(0,a), T(a,a), W(a,0)](https://tex.z-dn.net/?f=O%280%2C0%29%2C%20S%280%2Ca%29%2C%20T%28a%2Ca%29%2C%20W%28a%2C0%29)
Option D: ![O(0,0), S(a,0), T(a,a), W(0,a)](https://tex.z-dn.net/?f=O%280%2C0%29%2C%20S%28a%2C0%29%2C%20T%28a%2Ca%29%2C%20W%280%2Ca%29)
Step-by-step explanation:
Option A: ![O(0,0), S(0,a), T(a,a), W(a,0)](https://tex.z-dn.net/?f=O%280%2C0%29%2C%20S%280%2Ca%29%2C%20T%28a%2Ca%29%2C%20W%28a%2C0%29)
To find the sides of a square, let us use the distance formula,
![d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%5Cleft%28x_%7B2%7D-x_%7B1%7D%5Cright%29%5E%7B2%7D%2B%5Cleft%28y_%7B2%7D-y_%7B1%7D%5Cright%29%5E%7B2%7D%7D)
Now, we shall find the length of the square,
![\begin{array}{l}{\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } S T=\sqrt{(a-0)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } T W=\sqrt{(a-a)^{2}+(0-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7B%5Ctext%20%7B%20Length%20%7D%20O%20S%3D%5Csqrt%7B%280-0%29%5E%7B2%7D%2B%28a-0%29%5E%7B2%7D%7D%3D%5Csqrt%7Ba%5E%7B2%7D%7D%3Da%7D%20%5C%5C%7B%5Ctext%20%7B%20Length%20%7D%20S%20T%3D%5Csqrt%7B%28a-0%29%5E%7B2%7D%2B%28a-a%29%5E%7B2%7D%7D%3D%5Csqrt%7Ba%5E%7B2%7D%7D%3Da%7D%20%5C%5C%7B%5Ctext%20%7B%20Length%20%7D%20T%20W%3D%5Csqrt%7B%28a-a%29%5E%7B2%7D%2B%280-a%29%5E%7B2%7D%7D%3D%5Csqrt%7Ba%5E%7B2%7D%7D%3Da%7D%20%5C%5C%7B%5Ctext%20%7B%20Length%20%7D%20O%20W%3D%5Csqrt%7B%28a-0%29%5E%7B2%7D%2B%280-0%29%5E%7B2%7D%7D%3D%5Csqrt%7Ba%5E%7B2%7D%7D%3Da%7D%5Cend%7Barray%7D)
Thus, the square with vertices
has sides of length a.
Option B: ![O(0,0), S(0,a), T(2a,2a), W(a,0)](https://tex.z-dn.net/?f=O%280%2C0%29%2C%20S%280%2Ca%29%2C%20T%282a%2C2a%29%2C%20W%28a%2C0%29)
Now, we shall find the length of the square,
![\begin{aligned}&\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text {Length } S T=\sqrt{(2 a-0)^{2}+(2 a-a)^{2}}=\sqrt{5 a^{2}}=a \sqrt{5}\\&\text {Length } T W=\sqrt{(a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{2 a^{2}}=a \sqrt{2}\\&\text {Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26%5Ctext%20%7B%20Length%20%7D%20O%20S%3D%5Csqrt%7B%280-0%29%5E%7B2%7D%2B%28a-0%29%5E%7B2%7D%7D%3D%5Csqrt%7Ba%5E%7B2%7D%7D%3Da%5C%5C%26%5Ctext%20%7BLength%20%7D%20S%20T%3D%5Csqrt%7B%282%20a-0%29%5E%7B2%7D%2B%282%20a-a%29%5E%7B2%7D%7D%3D%5Csqrt%7B5%20a%5E%7B2%7D%7D%3Da%20%5Csqrt%7B5%7D%5C%5C%26%5Ctext%20%7BLength%20%7D%20T%20W%3D%5Csqrt%7B%28a-2%20a%29%5E%7B2%7D%2B%280-2%20a%29%5E%7B2%7D%7D%3D%5Csqrt%7B2%20a%5E%7B2%7D%7D%3Da%20%5Csqrt%7B2%7D%5C%5C%26%5Ctext%20%7BLength%20%7D%20O%20W%3D%5Csqrt%7B%28a-0%29%5E%7B2%7D%2B%280-0%29%5E%7B2%7D%7D%3D%5Csqrt%7Ba%5E%7B2%7D%7D%3Da%5Cend%7Baligned%7D)
This is not a square because the lengths are not equal.
Option C: ![O(0,0), S(0,2a), T(2a,2a), W(2a,0)](https://tex.z-dn.net/?f=O%280%2C0%29%2C%20S%280%2C2a%29%2C%20T%282a%2C2a%29%2C%20W%282a%2C0%29)
Now, we shall find the length of the square,
![\begin{array}{l}{\text { Length OS }=\sqrt{(0-0)^{2}+(2 a-0)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } S T=\sqrt{(2 a-0)^{2}+(2 a-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } T W=\sqrt{(2 a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } O W=\sqrt{(2 a-0)^{2}+(0-0)^{2}}=\sqrt{4 a^{2}}=2 a}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7B%5Ctext%20%7B%20Length%20OS%20%7D%3D%5Csqrt%7B%280-0%29%5E%7B2%7D%2B%282%20a-0%29%5E%7B2%7D%7D%3D%5Csqrt%7B4%20a%5E%7B2%7D%7D%3D2%20a%7D%20%5C%5C%7B%5Ctext%20%7B%20Length%20%7D%20S%20T%3D%5Csqrt%7B%282%20a-0%29%5E%7B2%7D%2B%282%20a-2%20a%29%5E%7B2%7D%7D%3D%5Csqrt%7B4%20a%5E%7B2%7D%7D%3D2%20a%7D%20%5C%5C%7B%5Ctext%20%7B%20Length%20%7D%20T%20W%3D%5Csqrt%7B%282%20a-2%20a%29%5E%7B2%7D%2B%280-2%20a%29%5E%7B2%7D%7D%3D%5Csqrt%7B4%20a%5E%7B2%7D%7D%3D2%20a%7D%20%5C%5C%7B%5Ctext%20%7B%20Length%20%7D%20O%20W%3D%5Csqrt%7B%282%20a-0%29%5E%7B2%7D%2B%280-0%29%5E%7B2%7D%7D%3D%5Csqrt%7B4%20a%5E%7B2%7D%7D%3D2%20a%7D%5Cend%7Barray%7D)
Thus, the square with vertices
has sides of length 2a.
Option D: ![O(0,0), S(a,0), T(a,a), W(0,a)](https://tex.z-dn.net/?f=O%280%2C0%29%2C%20S%28a%2C0%29%2C%20T%28a%2Ca%29%2C%20W%280%2Ca%29)
Now, we shall find the length of the square,
![\begin{aligned}&\text { Length OS }=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } S T=\sqrt{(a-a)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } T W=\sqrt{(0-a)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } O W=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26%5Ctext%20%7B%20Length%20OS%20%7D%3D%5Csqrt%7B%28a-0%29%5E%7B2%7D%2B%280-0%29%5E%7B2%7D%7D%3D%5Csqrt%7Ba%5E%7B2%7D%7D%3Da%5C%5C%26%5Ctext%20%7B%20Length%20%7D%20S%20T%3D%5Csqrt%7B%28a-a%29%5E%7B2%7D%2B%28a-0%29%5E%7B2%7D%7D%3D%5Csqrt%7Ba%5E%7B2%7D%7D%3Da%5C%5C%26%5Ctext%20%7B%20Length%20%7D%20T%20W%3D%5Csqrt%7B%280-a%29%5E%7B2%7D%2B%28a-a%29%5E%7B2%7D%7D%3D%5Csqrt%7Ba%5E%7B2%7D%7D%3Da%5C%5C%26%5Ctext%20%7B%20Length%20%7D%20O%20W%3D%5Csqrt%7B%280-0%29%5E%7B2%7D%2B%28a-0%29%5E%7B2%7D%7D%3D%5Csqrt%7Ba%5E%7B2%7D%7D%3Da%5Cend%7Baligned%7D)
Thus, the square with vertices
has sides of length a.
Thus, the correct answers are option a and option d.