Match the polygons formed by the sets of points with their perimeters ( rounded to the nearest hundredth). me i don't understand this question. 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)

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 with explanation:

→→Using the Distance formula, 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+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+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+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+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 +V W+W X+XY+YU

      = 8+10+12+5+3

       =38 units

question is incomplete.

536

step-by-step explanation: