WHAT IS THE PERIMETER OF A SQUARE WHOSE VERTICLES ARE A(-4,-3), B(-5,1), C(-1,2), AND D(0,-2)?
AB=sqrt(((-5)-(-4))^2+(1-(-3))^2)=
sqrt((-1)^2+4^2)=sqrt(17)
The perimeter=4*AB
THANK YOU
You're welcome.
P=4.(17)^1/2
To find the perimeter of a square, we need to know the length of one side of the square. In this case, we have the vertices of the square, which are A(-4, -3), B(-5, 1), C(-1, 2), and D(0, -2).
We can calculate the length of a side of the square by using the distance formula between two points.
The distance formula between two points (x₁, y₁) and (x₂, y₂) is given by:
distance = √((x₂ - x₁)² + (y₂ - y₁)²)
Let's calculate the distances between each consecutive pair of vertices:
Distance between A and B:
distance_AB = √((-5 - (-4))² + (1 - (-3))²)
Distance between B and C:
distance_BC = √((-1 - (-5))² + (2 - 1)²)
Distance between C and D:
distance_CD = √((0 - (-1))² + (-2 - 2)²)
Distance between D and A:
distance_DA = √((-4 - 0)² + (-3 - (-2))²)
Now, we can add up the distances of all sides of the square to get the perimeter:
perimeter = distance_AB + distance_BC + distance_CD + distance_DA
Calculate each distance and add them up to find the perimeter of the square.