Question

Square ABCD has vertices A(-8,-8), B(-8, 1), C(1, 1), and D(1, -8). Findthe length of one side of square ABCD. Round your answer to thenearest tenth if necessary.

Answer 1

Answer:

9 units.

Step-by-step explanation:

In a square, all sides have the same length. So in this question, the distance between any of these two points is the length of a side.

Distance between two points.

Point A(x1, y1)

Point B(x2,y2)

The distance is given by:

$D=\sqrt{(x2-x1)^2+(y2-y1)^2}$

In this question:

I am going to draw the square

The length of the side can be the distance between any of these points:

A and B, B and C, C and D or A and D.

I am going to use A and D.

A(-8,-8)

D(1,-8)

$D=\sqrt{(1-(-8))^2+(-8-(-8))^2}=\sqrt{(9)^2+(0)^2}=\sqrt{81}=9$

The length of one side is of 9 units.9 units.