Geometry help PLEASE!?

Question

1). The vertices of a triangle are located at X(-2, -1), Y(2,5) and Z(4,3). What is the perimeter of this triangle? Round to the nearest tenth.

2). What is the perimeter of a square whose vertices are A(-4,-3), B(-5,1), C(-1,2), and D(0,-2)?

can anyone explain these two questions to me...i have no idea how to slove them...maybe i need to use distance formula?

Answer 1

these are all magnitude problems

length of each segment = sqrt(y2-y1)^2 + (x2-x1)^2)

Answer 2

1.
Using Pythagoras, to find the distance between any two vertices.
XY = Sqrt[(2--2)^2+(5--1)^2]
XY = sqrt [4^2 + 6^2]
XY = sqrt[16+36]
XY = sqrt 52
XY = 7.2

YZ= sqrt[(4-2)^2+(3-5)^2]
YZ = sqrt[6^2 + (-2)^2]
YZ = sqrt[36 + 4]
YZ = sqrt40
YZ = 6.3

XZ = sqrt[(4--2)^2+(3--1)^2]
XZ = sqrt[6^2 + 4^2]
XZ = sqrt[36 + 16]
XZ = sqrt52
XZ = 7.2

Therefore perimeter is
XY + YZ + XZ = 7.2 + 6.3 + 7.2 = 20.7 ( 20 7/10)

2.
Again use Pythagoras to find the distance between any two vertices. Also note, because it is a square, only one length needs to be found. Since it is a square then the three other sides are also the same length.
Once the length is calculated multiply by 4 to find the perimeter.
AB = sqrt[(-5--4)^2 + (1--3)^2]
AB = sqrt[(-1)^2 + 4^2)]
AB = sqrt[1 + 17]
AB = sqrt18
AB = 4.24264
Hence perimeter is 4.24262 x 4 = 16.97056 = 17.0 (1dp)