05.05 HC) Figure ABCD has vertices A(−2, 3), B(4, 3), C(4, −2), and D(−2, 0). What is the area of Figure ABCD?

Answer:The area of ABCD is 24 units²Step-by-step explanation:* Lets explain how to solve the problem- All the point in a vertical line have the same x-coordinates- The length of the vertical line is y2 - y1- All the point in a horizontal line have the same y-coordinates- The length of the horizontal line is x2 - x1- The horizontal and the vertical lines are perpendicular to each other- The trapezoid has two parallel bases not equal in length and the other  two sides are nonparallel sides- The area of the trapezoid = 1/2 (sum of the two // bases) × height* Lets solve the problem∵ ABCD is a quadrilateral∵ A = (-2 , 3) , B = (4 , 3) , C = (4 , -2) , D = (-2 , 0)∵ Side AD has same x-coordinates in A and D (-2)∴ AD is vertical side∴ AD = 3 - 0 = 3∵ Side BC has same x-coordinates in B and C (4)∴ BC is vertical side∴ BC = 3 - (-2) = 3 + 2 = 5∵ AD and BC are vertical lines∴ AD // BC∵ Side AB has same y-coordinates in A and B (3)∴ AB is horizontal side∴ AB = 4 - (-2) = 4 + 2 = 6∵ The horizontal and the vertical lines are perpendicular to each other∴ AB is perpendicular on AD and BC∵ The side CD is not vertical or horizontal∴ ABCD has only two parallel sides AD and BC ∵ AD ≠ BC∴ ABCD is a trapezoid∵ The two parallel bases are AD and BC∵ Its height is AB∵ AD = 3 , BC = 5 , AB = 6∴ Its area = 1/2 (3 + 5) × 6 = 1/2 (8) × 6 = 4 × 6 = 24 units²* The area of ABCD is 24 units²