If the points of the vertices are (-3,1_,(4,2),(9,-3),(2,-4) then the perimeter of the figure is 28.28 units.
Given points of the vertices of the figure having four sides be (-3,1),(4,2),(9,-3),(2,-4).
We have to find the perimeter of the figure.
Perimeter is the sum of lengths of all the sides of a figure.
First give some name to the points so that it will be easy to understand the sides like A(-3,1),B(4,2),C(9,-3),D(2,-4).
Perimeter of the figure =AB+BC+CD+DA
AB=[tex]\sqrt{(2-1)^{2} +(4+3)^{2} }[/tex]
=[tex]\sqrt{1+49}[/tex]
=[tex]\sqrt{50}[/tex] units
BC=[tex]\sqrt{(9-4)^{2} +(-3-2)^{2} }[/tex]
=[tex]\sqrt{25+25}[/tex]
=[tex]\sqrt{50}[/tex]units
CD=[tex]\sqrt{(2-9)^{2} +(-4+3)^{2} }[/tex]
=[tex]\sqrt{49+1}[/tex]
=[tex]\sqrt{50}[/tex] units
DA=[tex]\sqrt{(2+3)^{2} +(-4-1)^{2} }[/tex]
=[tex]\sqrt{25+25}[/tex]
=[tex]\sqrt{50}[/tex] units
Perimeter=[tex]\sqrt{50} +\sqrt{50} +\sqrt{50} +\sqrt{50}[/tex]
=4[tex]\sqrt{50}[/tex]
=4*5*[tex]\sqrt{2}[/tex]
=20[tex]\sqrt{2}[/tex]
=20*1.411
=28.284
After rounding off we will get 28.28 units.
Hence if the points of the vertices are (-3,1_,(4,2),(9,-3),(2,-4) then the perimeter of the figure is 28.28 units.
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