The perimeter of the kite, using the formula to calculate the distance between two points, is 2√53+10 units
Solution:
Perimeter of kite WXYZ: P=?
The kite is a polygon, and the perimeter of a polygon is the sum of the lengths of its sides. In this case:
P=WX+XY+YZ+WZ
1. WX is an horizontal segment, then its length is the difference of the abscissas of their end points:
WX=Xx-Xw
W=(-3,3)=(Xw,Yw)→Xw=-3, Yw=3
X=(2,3)=(Xx,Yx)→Xx=2, Yx=3
Replacing the values:
WX=Xx-Xw
WX=2-(-3)
WX=2+3
WX=5 units
2. XY is an oblique segment, then to determine its length we must use the formula to calculate the distance between two points:
If P1=(x1,y1) and P2=(x2,y2)→P1P2=√[(x2-x1)^2+(y2-y1)^2]
XY=√[(Xy-Xx)^2+(Yy-Yx)^2]
X=(2,3)=(Xx,Yx)→Xx=2, Yx=3
Y=(4,-4)=(Xy,Yy)→Xy=4, Yy=-4
Replacing the values:
XY=√[(Xy-Xx)^2+(Yy-Yx)^2]
XY=√[(4-2)^2+(-4-3)^2]
XY=√[(2)^2+(-7)^2]
XY=√(4+49)
XY=√53 units
3. YZ is an oblique segment, then to determine its length we must use the formula to calculate the distance between two points:
If P1=(x1,y1) and P2=(x2,y2)→P1P2=√[(x2-x1)^2+(y2-y1)^2]
YZ=√[(Xz-Xy)^2+(Yz-Yy)^2]
Y=(4,-4)=(Xy,Yy)→Xy=4, Yy=-4
Z=(-3,-2)=(Xz,Yz)→Xz=-3, Yz=-2
Replacing the values:
YZ=√[(Xz-Xy)^2+(Yz-Yy)^2]
YZ=√[(-3-4)^2+(-2-(-4))^2]
YZ=√[(-7)^2+(-2+4)^2]
YZ=√[(-7)^2+(2)^2]
YZ=√(49+4)
YZ=√53 units
4. WZ is a vertical segment, then its length is the difference of the ordinates of their end points:
WZ=Yw-Yz
W=(-3,3)=(Xw,Yw)→Xw=-3, Yw=3
Z=(-3,-2)=(Xz,Yz)→Xz=-3, Yz=-2
Replacing the values:
WZ=Yw-Yz
WZ=3-(-2)
WZ=3+2
WZ=5 units
Replacing the values in the formula of perimeter:
P=WX+XY+YZ+WZ
P=5 units+√53 units+√53 units+5 units
Adding similar terms:
P=2√53+10 units