Question

Please help faster and don't give irrelevant answers​

Answer 1

Answer:

Consider ΔBEC, draw a perpendicular from point E to F on BC

EF is the height of the triangle BEC

Area of a triangle = √s(s-a)(s-b)(s-c) = √45(20)(9)(16)

  = √3×3×5×5×2×2×3×3×2×2×2×2 = 3×5×2×3×2×2 = 360cm²

Area of a triangle = 1/2 bh = 360

                    1/2 36h = 360

                    h = 360×2/36 = 20cm

EF = DC = AB(because EF is the distance between AD and BC)

AB=DC=20cm

Area of a rectangle = lb = (36)(20) = 720cm²

Answer 2

Answer:

Answer:

Consider ΔBEC, draw a perpendicular from point E to F on BC

EF is the height of the triangle BEC

Area of a triangle = √s(s-a)(s-b)(s-c) = √45(20)(9)(16)

  = √3×3×5×5×2×2×3×3×2×2×2×2 = 3×5×2×3×2×2 = 360cm²

Area of a triangle = 1/2 bh = 360

                    1/2 36h = 360

                    h = 360×2/36 = 20cm

EF = DC = AB(because EF is the distance between AD and BC)

AB=DC=20cm

Area of a rectangle = lb = (36)(20) = 720cm²