AL = 10 cm; AM = 20 cm; AN = 50 cm; AO = 60 cm; AD = 90 cm given

LM = AM – AL = 20 – 10 = 10 cm

MN = AN – AM = 50 – 20 = 30 cm

OD = AD – AO = 90 – 60 = 30 cm

ON = AO – AN = 60 – 50 = 10 cm

DN = OD + ON = 30 + 10 = 40 cm

OM = MN + ON = 30 + 10 = 40 cm

LN = LM + MN = 10 + 30 = 40 cm

Area of given figure = area of triangle AMF + area of trapezium FMNE + area of triangle END + area of triangle ALB + area of trapezium LBCN + area of triangle DNC

Area of right angled triangle = \(\frac{1}{2}\)x base x altitude

Area of trapezium = \(\frac{1}{2}\)x (sum of parallel sides) x altitude

Area of given hexagon = \(\frac{1}{2}\)x AM x FM + \(\frac{1}{2}\) x (MF + OE) x OM + \(\frac{1}{2}\) x OD x OE + \(\frac{1}{2}\)x AL x BL + \(\frac{1}{2}\)x (BL + CN) x LN + \(\frac{1}{2}\)x DN x CN

Area of given hexagon = \(\frac{1}{2}\) x 20 x 20 + \(\frac{1}{2}\) x (20 + 60) x 40 + \(\frac{1}{2}\) x 30 x 60 + \(\frac{1}{2}\)x 10 x 30 + \(\frac{1}{2}\)x (30 + 40) x 40 + \(\frac{1}{2}\)x 40 x 40

Area of given hexagon = 200 + 1600 + 900 + 150 + 1400 + 800 = 5050 cm²

Therefore area of given hexagon is 5050 cm²