Question

find the length (distance) of one side of square abcd.

Explanation:

Step1: Recall the distance formula

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). Let's take two adjacent vertices of the square, say \(A(2,5)\) and \(B(6,0)\).

Step2: Substitute the coordinates into the formula

Here, \(x_1 = 2\), \(y_1 = 5\), \(x_2 = 6\), \(y_2 = 0\). Substituting into the formula:
\[

$$\begin{align*} d&=\sqrt{(6 - 2)^2+(0 - 5)^2}\\ &=\sqrt{4^2+(- 5)^2}\\ &=\sqrt{16 + 25}\\ &=\sqrt{41} \end{align*}$$

\]
We can also check with another pair, e.g., \(A(2,5)\) and \(D(-3,1)\):
\[

$$\begin{align*} d&=\sqrt{(-3 - 2)^2+(1 - 5)^2}\\ &=\sqrt{(-5)^2+(-4)^2}\\ &=\sqrt{25 + 16}\\ &=\sqrt{41} \end{align*}$$

\]

Answer:

\(\sqrt{41}\)