Question

Match the polygons formed by the sets of points with their perimeters ( rounded to the nearest hundredth). Please Help me I don't understand this question. Thank you
A( 1, 1), B(6,13), C(8,13), D(16,-2) E(1, -2 )
38 units
K(4,2), L(8,2), M(12,5), N(6,5), O(4,4)
50 units
F(14,-10), G(16, -10), H(19,-6), I (14,-2), J(11,-6)
25.4 units
P(7,2), Q (12,2), R(12,6), S(7,10), T(4,6)
19.24 units
U(4, -1), V(12, -1), W(20,-7), X(8, -7), Y(4,-4)

Answer 1

38 units-U(4, -1), V(12, -1), W(20,-7), X(8, -7), Y(4,-4)

25.4 units-P(7,2), Q (12,2), R(12,6), S(7,10), T(4,6)

50 units-A( 1, 1), B(6,13), C(8,13), D(16,-2) E(1, -2 )

19.24 units-K(4,2), L(8,2), M(12,5), N(6,5), O(4,4)

Answer 2

Answer with explanation:

→→Using the Distance formula,$\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2$ finding the distance between two points that is between two vertices

The Points which are vertices of Polygons are

1.→ A( 1, 1), B(6,13), C(8,13), D(16,-2) E(1, -2 )

$AB=\sqrt{(6-1)^2+(13-1)^2}\\\\AB=\sqrt{5^2+12^2}\\\\ AB=\sqrt{25+144}\\\\ AB=\sqrt{169}\\\\ AB=13$

$BC=\sqrt{(8-6)^2+(13-13)^2}\\\\ BC=\sqrt{2^2}\\\\ BC=2$

$CD=\sqrt{(16-8)^2+(-2-13)^2}\\\\ CD=\sqrt{8^2+15^2}\\\\CD=\sqrt{64+225}\\\\CD=\sqrt{289}\\\\CD=17\\\\DE=\sqrt{(16-1)^2+(-2+2)^2}\\\\DE=\sqrt{15^2}\\\\DE=15 \\\\AE=\sqrt{(1-1)^2+(-2-1)^2}\\\\ AE=\sqrt{3^2}\\\\ AE=3$

→→AB+BC+CD+DE+EA= 13+2+17+15+3

         =50 units

2. K(4,2), L(8,2), M(12,5), N(6,5), O(4,4)

$KL=\sqrt{(8-4)^2+(2-2)^2}=\sqrt{4^2}=4\\\\ LM=\sqrt{(8-12)^2+(2-5)^2}=\sqrt{4^2+3^2}=\sqrt{5^2}=5\\\\MN=\sqrt{(12-6)^2+(5-5)^2}=\sqrt{6^2}=6\\\\NO=\sqrt{(6-4)^2+(5-4)^2}=\sqrt{2^2+1^2}=\sqrt{5}\\\\KO=\sqrt{(4-4)^2+(4-2)^2}=\sqrt{2^2}=2$

→KL+LM+MN+NO+OK=4+5+6+√5+2

                                 =17+2.24

                                 = 19.24 units

3. F(14,-10), G(16, -10), H(19,-6), I (14,-2), J(11,-6)

$FG=\sqrt{(16-14)^2+(-10+10)^2}=\sqrt{2^2}=2\\\\ GH=\sqrt{(19-16)^2+(-6+10)^2}=\sqrt{4^2+3^2}=\sqrt{5^2}=5\\\\HI=\sqrt{(19-14)^2+(-6+2)^2}=\sqrt{5^2+4^2}=\sqrt{25+16}=\sqrt{41}\\\\IJ=\sqrt{(14-11)^2+(-2+6)^2}=\sqrt{3^2+4^2}=\sqrt{5^2}=5\\\\JF=\sqrt{(14-11)^2+(-10+6)^2}=\sqrt{3^2+4^2}=\sqrt{5^2}=5$

→FG+GH+HI+IJ+JF=2+5+√41+5+5=19+6.40=25.40

4.→ P(7,2), Q (12,2), R(12,6), S(7,10), T(4,6)

 $PQ=\sqrt{(12-7)^2+(2-2)^2}=\sqrt{5^2}=5\\\\ QR=\sqrt{(12-12)^2+(-6+2)^2}=\sqrt{4^2+0^2}=\sqrt{4^2}=4\\\\RS=\sqrt{(12-7)^2+(-6+10)^2}=\sqrt{5^2+4^2}=\sqrt{25+16}=\sqrt{41}\\\\ST=\sqrt{(7-4)^2+(-10+6)^2}=\sqrt{3^2+4^2}=\sqrt{5^2}=5\\\\PT=\sqrt{(7-4)^2+(-2+6)^2}=\sqrt{3^2+4^2}=\sqrt{5^2}=5$

PQ+QR+RS+ST+TP=5+4+√41+5+5

                              =19+6.40

                              = 25.40 units

5.→U(4, -1), V(12, -1), W(20,-7), X(8, -7), Y(4,-4)

$UV=\sqrt{(12-4)^2+(-1+1)^2}=\sqrt{8^2}=8\\\\ VW=\sqrt{(20-12)^2+(-7+1)^2}=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10\\\\WX=\sqrt{(20-8)^2+(-7+7)^2}=\sqrt{12^2+0^2}=12\\\\XY=\sqrt{(8-4)^2+(-7+4)^2}=\sqrt{3^2+4^2}=\sqrt{5^2}=5\\\\YU=\sqrt{(4-4)^2+(-4+1)^2}=\sqrt{3^2+0^2}=\sqrt{3^2}=3$

UV +V W+W X+XY+YU

      = 8+10+12+5+3

       =38 units