Please help!! Square ABCD has vertices A(-2,-3), B(4, -1), C(2,5), and D(-4,3). Find the length of a side

of the square. Round your answer to the nearest tenth if necessary.

The length of a side of the square is 6.3 units.

Step-by-step explanation:

When given vertices for a given shape, the length of the side is calculated using the formula:

√(x2 - x1)² + (y2 - y1)²

When given vertices (x1 , y1) and (x2 , y2)

Square ABCD has vertices A(-2,-3), B(4, -1), C(2,5), and D(-4,3). Find the length of a side

Side AB : A(-2,-3), B(4, -1)

√(x2 - x1)² + (y2 - y1)²

= √(4 -(-2))² + (-1 -(-3))²

= √ 6² + 2²

= √36 + 4

= √40

= 6.3245553203

≈ 6.3 units

B(4, -1), C(2,5),

√(x2 - x1)² + (y2 - y1)²

= √ (2- 4)² + (5- (-1))²

= √-2² + 6²

= √4 + 36

= √40

= 6.3245553203

≈ 6.3 units

C(2,5), D(-4,3).

√(x2 - x1)² + (y2 - y1)²

= √(-4 - 2)² + (3 - 5)²

= √-6² + -2²

= √36 + 4

= √40

= 6.3245553203

≈ 6.3 units

A(-2,-3), D(-4,3).

√(x2 - x1)² + (y2 - y1)²

= √(-4 -(-2))² + (3 - (-3))²

= √-2² + 6²

= √4 + 36

= √40

= 6.3245553203

≈ 6.3 units

what is the question here

step-by-step explanation: