What is the perimeter of a polygon with vertices at (−2, 1) , ​ (−2, 4) ​, (2, 7) , ​ (6, 4) ​, and (6, 1) ​?Enter your answer in the box. Do not round any side lengths.​____ units

Answer:24 unitsStep-by-step explanation:The polygon with 5 vertices A (-2,1), B (-2, 4) , C (2, 7), D (6, 4) and E (6, 1).where A [tex](x_{1} , y_{1}) = (-2, 1)[/tex],  B [tex](x_{2} , y_{2}) = (-2, 4)[/tex]C [tex](x_{3} , y_{3}) = (2, 7)[/tex] , D [tex](x_{4} , y_{4}) = (6, 4)[/tex]E [tex](x_{5} , y_{5}) = (6, 1)[/tex]Now using distance formula:Length of AB = [tex]\sqrt{\left(x_{2}- x_{1}\right)^{2}+ \left (y_{2} -y_{1}\right)^{2}}[/tex]= [tex]\sqrt{(-2+2)^{2} +(4 - 1)^{2} }[/tex]= [tex]\sqrt{0^{2} +3^{2} }[/tex]= [tex]\sqrt{3^{2} } = 3 unit[/tex]Length of BC =  [tex]\sqrt{\left(x_{3}- x_{2}\right)^{2}+ \left (y_{3} -y_{2}\right)^{2}}[/tex]= [tex]\sqrt{(2- (-2))^{2} +(7 - 4)^{2} }[/tex]= [tex]\sqrt{4^{2} +3^{2} }[/tex]=[tex]\sqrt{16 + 9}[/tex]= [tex]\sqrt{25^{2} } = 5 unit[/tex]Length of CD = [tex]\sqrt{\left(x_{4}- x_{3}\right)^{2}+ \left (y_{4} -y_{3}\right)^{2}}[/tex]= [tex]\sqrt{(6-2)^{2} +(4 - 7)^{2} }[/tex]= [tex]\sqrt{4^{2} +(-3)^{2} }[/tex]= [tex]\sqrt{16 + 9}[/tex]= [tex]\sqrt{25} = 5 unit[/tex]Length of DE = [tex]\sqrt{\left(x_{5}- x_{4}\right)^{2}+ \left (y_{5} -y_{4}\right)^{2}}[/tex]= [tex]\sqrt{(6 - 6)^{2} +(1 - 4)^{2} }[/tex]= [tex]\sqrt{0^{2} +(-3)^{2} }[/tex]=  [tex]\sqrt{0 + 9}[/tex]= 3 unitLength of EA =  [tex]\sqrt{\left(x_{5}- x_{1}\right)^{2}+ \left (y_{5} -y_{1}\right)^{2}}[/tex]= [tex]\sqrt{(6 - (- 2))^{2} +(1 - 1)^{2} }[/tex]= [tex]\sqrt{(6 + 2)^{2} +(0)^{2} }[/tex]= [tex]\sqrt{8^{2} +(0)^{2} }[/tex]= 8 unitPerimeter of polygon = Length of (AB + BC + CD + DE + EA)= (3 + 5 + 5 + 3 + 8)= 24 units