Question

Find the area of hexagon ABCDEF in which BL ⊥ AD, CM ⊥ AD, EN ⊥ AD and FP ⊥ AD such that AP = 6 cm, PL = 2 cm, LN = 8 cm, NM = 2 cm, MD = 3 cm, FP = 8 cm, EN = 12 cm, BL = 8 cm and CM = 6 cm.

Answer 1

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{hexagon}\mathrm{ABCDEF} \\ =\left(\mathrm{Area}\mathrm{of}∆\mathrm{AFP}\right)+\left(\mathrm{Area}\mathrm{of}\mathrm{trapezium}\mathrm{FENP}\right)+\left(\mathrm{Area}\mathrm{of}∆\mathrm{END}\right) \\ +\left(\mathrm{Area}\mathrm{of}∆\mathrm{CMD}\right)+\left(\mathrm{Area}\mathrm{of}\mathrm{trapezium}\mathrm{BCML}\right)+\left(\mathrm{Area}\mathrm{of}∆\mathrm{ALB}\right) \end{gathered}$$

$$\begin{gathered} =\left(\frac{1}{2}\times \mathrm{AP}\times \mathrm{FP}\right)+\left(\frac{1}{2}\times \left(\mathrm{FP}+\mathrm{EN}\right)\times \mathrm{PN}\right)+\left(\frac{1}{2}\times \mathrm{ND}\times \mathrm{EN}\right)+\left(\frac{1}{2}\times \mathrm{MD}\times \mathrm{CM}\right) \\ +\left(\frac{1}{2}\times \left(\mathrm{CM}+\mathrm{BL}\right)\times \mathrm{LM}\right)+\left(\frac{1}{2}\times \mathrm{AL}\times \mathrm{BL}\right) \\ =\left(\frac{1}{2}\times \mathrm{AP}\times \mathrm{FP}\right)+\left(\frac{1}{2}\times \left(\mathrm{FP}+\mathrm{EN}\right)\times \left(\mathrm{PL}+\mathrm{LN}\right)\right)+\left(\frac{1}{2}\times \left(\mathrm{NM}+\mathrm{MD}\right)\times \mathrm{EN}\right) \\ +\left(\frac{1}{2}\times \mathrm{MD}\times \mathrm{CM}\right)+\left(\frac{1}{2}\times \left(\mathrm{CM}+\mathrm{BL}\right)\times \left(\mathrm{LN}+\mathrm{NM}\right)\right)+\left(\frac{1}{2}\times \left(\mathrm{AP}+\mathrm{PL}\right)\times \mathrm{BL}\right) \\ =\left(\frac{1}{2}\times 6\times 8\right)+\left(\frac{1}{2}\times \left(8+12\right)\times \left(2+8\right)\right)+\left(\frac{1}{2}\times \left(2+3\right)\times 12\right)+\left(\frac{1}{2}\times 3\times 6\right) \\ +\left(\frac{1}{2}\times \left(6+8\right)\times \left(8+2\right)\right)+\left(\frac{1}{2}\times \left(6+2\right)\times 8\right){\mathrm{cm}}^2 \\ =\left(24+100+30+9+70+32\right){\mathrm{cm}}^2 \\ =265{\mathrm{cm}}^2 \end{gathered}$$

$\mathrm{Hence},\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{hexagon}\mathrm{is}265{\mathrm{cm}}^2.$