ProblemA=12(b1+b2)h, b1=5, b2=8, h=4A = \tfrac{1}{2}(b_1 + b_2) h,\ b_1=5,\ b_2=8,\ h=4A=21(b1+b2)h, b1=5, b2=8, h=4Step-by-step solutionDiện tích hình thang: A=12(b1+b2)hA = \tfrac{1}{2}(b_1 + b_2) hA=21(b1+b2)h.Thế b1=5b_1 = 5b1=5, b2=8b_2 = 8b2=8, h=4h = 4h=4: A=12(5+8)(4)=12(13)(4)A = \tfrac{1}{2}(5 + 8)(4) = \tfrac{1}{2}(13)(4)A=21(5+8)(4)=21(13)(4).Rút gọn: A=26A = 26A=26.AnswerA=26A = 26A=26Want to solve a different problem? Open the area solver →Related worked examples/solve/geometry/area-rectangle-5-by-8/solve/geometry/area-triangle-base-10-height-6