# Find the area of hexagon ABCDEF in which BL ⊥ AD,

Question:

Find the area of hexagon ABCDEF in which BL ⊥ ADCM ⊥ ADEN ⊥ 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.

Solution:

Area of hexagon ABCDEF

\begin{aligned}=&(\text { Area of } \Delta \text { AFP })+(\text { Area of trapezium FENP })+(\text { Area of } \Delta \text { END }) \\ &+\text { (Area of } \Delta \text { CMD) }+(\text { Area of trapezium BCML })+(\text { Area of } \Delta \text { ALB }) \end{aligned}

\begin{aligned}=&\left(\frac{1}{2} \times \mathrm{AP} \times \mathrm{FP}\right)+\left(\frac{1}{2} \times(\mathrm{FP}+\mathrm{EN}) \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(\mathrm{CM}+\mathrm{BL}) \times \mathrm{LM}\right)+\left(\frac{1}{2} \times \mathrm{AL} \times \mathrm{BL}\right) \end{aligned}

\begin{aligned}=&\left(\frac{1}{2} \times \mathrm{AP} \times \mathrm{FP}\right)+\left(\frac{1}{2} \times(\mathrm{FP}+\mathrm{EN}) \times(\mathrm{PL}+\mathrm{LN})\right)+\left(\frac{1}{2} \times(\mathrm{NM}+\mathrm{MD}) \times \mathrm{EN}\right) \\ &+\left(\frac{1}{2} \times \mathrm{MD} \times \mathrm{CM}\right)+\left(\frac{1}{2} \times(\mathrm{CM}+\mathrm{BL}) \times(\mathrm{LN}+\mathrm{NM})\right)+\left(\frac{1}{2} \times(\mathrm{AP}+\mathrm{PL}) \times \mathrm{BL}\right) \end{aligned}

\begin{aligned}=&\left(\frac{1}{2} \times 6 \times 8\right)+\left(\frac{1}{2} \times(8+12) \times(2+8)\right)+\left(\frac{1}{2} \times(2+3) \times 12\right)+\left(\frac{1}{2} \times 3 \times 6\right) \\ &+\left(\frac{1}{2} \times(6+8) \times(8+2)\right)+\left(\frac{1}{2} \times(6+2) \times 8\right) \mathrm{cm}^{2} \end{aligned}

$=(24+100+30+9+70+32) \mathrm{cm}^{2}$

$=265 \mathrm{~cm}^{2}$

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