Answer 1

Given that, the width of each section is same. Therefore,

IB = BJ = CK = CL = DM = DN = AO = AP

IL = IB + BC + CL

28 = IB + 20 + CL

IB + CL = 28 cm − 20 cm = 8 cm

IB = CL = 4 cm

Hence, IB = BJ = CK = CL = DM = DN = AO = AP = 4 cm

Area of section BEFC = Area of section DGHA

=`[1/2 (20 + 28)(4)] cm^2 = 96 cm^2`

Area of section ABEH = Area of section CDGF

⇒Area of section ABEH = Area of section CDGF

= 1216+244=80 cm²