Answer:
Perimeter = 13.27 unit
Step-by-step explanation:
We have to find the perimeter of the shape given in the figure.
Coordinates of J are ( -3, 4 ) and K are (-1, 6)
Then JK = [tex]\sqrt{(x-x')^{2}+(y-y')^{2}}[/tex]
= [tex]\sqrt{(-3+1)^2+(6-4)^2}[/tex]
= [tex]\sqrt{4+4}[/tex]
= [tex]2\sqrt{2}[/tex]
Coordinates of K are (-1,6) and for ∠ are (1, 1)
KL = [tex]\sqrt{(1+1)^2+(1-6)^2}[/tex]
= [tex]\sqrt{2^2+5^2}[/tex]
= [tex]\sqrt{29}[/tex]
Coordinates of M are (-1, 3) and of ∠ are (1, 1)
∠M = [tex]\sqrt{(1+1)^2+(1-3)^2}[/tex]
= [tex]\sqrt{4+4}[/tex]
= [tex]2\sqrt{2}[/tex]
Coordinates of M are (-1, 3) and J are (-3, 4)
MJ = [tex]\sqrt{(-1+3)^2+(3-4)^2}[/tex]
= [tex]\sqrt{4+1}[/tex]
= [tex]\sqrt{5}[/tex]
Then perimeter = JK + KL + LM + MJ
= [tex]2\sqrt{2}[/tex] + [tex]\sqrt{29}[/tex] + [tex]2\sqrt{2}[/tex] + [tex]\sqrt{5}[/tex]
= [tex]4\sqrt{2}[/tex] + [tex]\sqrt{29}[/tex] + [tex]\sqrt{5}[/tex]
= 5.66 + 5.38 + 2.24 = 13.27
