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=4分步解答梯形的面积:A=12(b1+b2)hA = \tfrac{1}{2}(b_1 + b_2) hA=21(b1+b2)h。代入 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)。化简:A=26A = 26A=26。答案A=26A = 26A=26想解其他题?打开 area 求解器 →相关例题/solve/geometry/area-rectangle-5-by-8/solve/geometry/area-triangle-base-10-height-6