d the area of pentagon ABCDE in which BL ⊥ AC,

Area of pentagon $\mathrm{ABCDE}=($ Area of $\Delta \mathrm{AEN})+($ Area of trapezium EDMN $)+($ Area of $\Delta \mathrm{DMC})$ $+($ Area of $\Delta \mathrm{ACB})$

$=\left(\frac{1}{2} \times \mathrm{AN} \times \mathrm{EN}\right)+\left(\frac{1}{2} \times(\mathrm{EN}+\mathrm{DM}) \times \mathrm{NM}\right)+\left(\frac{1}{2} \times \mathrm{MC} \times \mathrm{DM}\right)+\left(\frac{1}{2} \times \mathrm{AC} \times \mathrm{BL}\right)$

$=\left(\frac{1}{2} \times \mathrm{AN} \times \mathrm{EN}\right)+\left(\frac{1}{2} \times(\mathrm{EN}+\mathrm{DM}) \times(\mathrm{AM}-\mathrm{AN})\right)+\left(\frac{1}{2} \times(\mathrm{AC}-\mathrm{AM}) \times \mathrm{DM}\right)+\left(\frac{1}{2} \times \mathrm{AC} \times \mathrm{BL}\right)$

$=\left[\left(\frac{1}{2} \times 6 \times 9\right)+\left(\frac{1}{2} \times(9+12) \times(14-6)\right)+\left(\frac{1}{2} \times(18-14) \times 12\right)+\left(\frac{1}{2} \times 18 \times 4\right)\right] \mathrm{cm}^{2}$

$=(27+84+24+36) \mathrm{cm}^{2}$

$=171 \mathrm{~cm}^{2}$

Hence, the area of the given pentagon is $171 \mathrm{~cm}^{2}$.