Question

A polygon has its vertices at the following points.

Answer 1

Answer:

isosceles trapezoid

Step-by-step explanation:

1) find the distance of the points

$\sqrt{(x2-x1)^2 + (y2 -y1)^2}$

AB = $\sqrt{(4-2)^2 + (7-5)^2} = \sqrt{4 + 4} = \sqrt{8} = 2 \sqrt{2}$

BC = |7-4| = 3

CD = $\sqrt{(9-7)^2 + (5-7)^2}= \sqrt{4+4} = \sqrt{8} = 2 \sqrt{2}$

AD = |9-2| = 7

2) equation of the line that passes threw BC

y = 7

3) equation of the line that passes threw AD

y = 5

conclusion

the quadrilateral has two parallel sides and two congruent sides, so it is a isosceles trapezoid