WHAT IS THE PERIMETER OF A SQUARE WHOSE VERTICLES ARE A(-4,-3), B(-5,1), C(-1,2), AND D(0,-2)?
Use the formula for distance between two points between the points of each of the four sides, and sum the distances.
Distance between two points P1(x1,y1) and P2(x2,y2):
D=sqrt((x2-x1)^2+(y2-y1)^2)
P.S.
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To find the perimeter of a square, you need to calculate the distance between its consecutive vertices and sum up those distances.
In this case, you have the coordinates of the four vertices of the square: A(-4,-3), B(-5,1), C(-1,2), and D(0,-2).
To find the distance between two points (x1, y1) and (x2, y2), you can use the distance formula, which is derived from the Pythagorean theorem:
Distance = √[(x2 - x1)^2 + (y2 - y1)^2]
Now, let's calculate each side of the square:
Distance AB:
x1 = -4, y1 = -3
x2 = -5, y2 = 1
Distance AB = √[(-5 - (-4))^2 + (1 - (-3))^2]
= √[(-5 + 4)^2 + (1 + 3)^2]
= √[(-1)^2 + 4^2]
= √[1 + 16]
= √17
Distance BC:
x1 = -5, y1 = 1
x2 = -1, y2 = 2
Distance BC = √[(-1 - (-5))^2 + (2 - 1)^2]
= √[(1 + 5)^2 + (2 - 1)^2]
= √[6^2 + 1^2]
= √[36 + 1]
= √37
Distance CD:
x1 = -1, y1 = 2
x2 = 0, y2 = -2
Distance CD = √[(0 - (-1))^2 + (-2 - 2)^2]
= √[(0 + 1)^2 + (-2 - 2)^2]
= √[1^2 + (-4)^2]
= √[1 + 16]
= √17
Distance DA:
x1 = 0, y1 = -2
x2 = -4, y2 = -3
Distance DA = √[(-4 - 0)^2 + (-3 - (-2))^2]
= √[(-4)^2 + (-3 + 2)^2]
= √[16 + 1]
= √17
Now, to find the perimeter of the square, you need to sum up all four distances:
Perimeter = AB + BC + CD + DA
= √17 + √37 + √17 + √17
Therefore, the perimeter of the square is √17 + √37 + √17 + √17 units.