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

Area of pentagon $\mathrm{ABCDE}=($ Area of $\Delta \mathrm{ABC})+($ Area of $\Delta \mathrm{ACD})+($ Area of $\Delta \mathrm{ADE})$

$=\left(\frac{1}{2} \times \mathrm{AC} \times \mathrm{BL}\right)+\left(\frac{1}{2} \times \mathrm{AD} \times \mathrm{CM}\right)+\left(\frac{1}{2} \times \mathrm{AD} \times \mathrm{EM}\right)$

$=\left[\left(\frac{1}{2} \times 10 \times 3\right)+\left(\frac{1}{2} \times 12 \times 7\right)+\left(\frac{1}{2} \times 12 \times 5\right)\right] \mathrm{cm}^{2}$

$=(15+42+30) \mathrm{cm}^{2}$

$=87 \mathrm{~cm}^{2}$

Hence, the area of the pentagon is $87 \mathrm{~cm}^{2}$.