Choose the correct option: 

If PO = 4 cm, ON = 2 cm, NM = 6 cm, ML = 2 cm, LS = 6 cm, UO = 6 cm, TM = 8 cm, QN = 2 cm, and RL = 4 cm, then find the area of the following hexagon.

Given, PO = 4 cm, ON = 2 cm, NM = 6 cm, ML = 2 cm, LS = 6 cm, UO = 6 cm, TM = 8 cm, QN = 2 cm, and RL = 4 cm

PN = PO+ON = 4+2 = 6 cm

NL= NM+ML = 6+2 = 8 cm

MS = ML+LS = 2+6 = 8 cm

OM = ON+MN = 2+6 = 8 cm

$\begin{array}{rcl}\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{hexagon} \mathrm{PQRSTU}& =& \mathrm{Area} \mathrm{of} ∆\mathrm{PQN} + \mathrm{Area} \mathrm{of} \mathrm{quad} \mathrm{QRLN}\\ & & + \mathrm{Area} \mathrm{of} ∆\mathrm{RSL} + \mathrm{Area} \mathrm{of} ∆\mathrm{TLS} \\ & & + \mathrm{Area} \mathrm{of} \mathrm{quad} \mathrm{OMTU} + \mathrm{Area} \mathrm{of} ∆\mathrm{POU}\end{array}$

$\begin{array}{rcl} & =& (\frac{1}{2}\times \mathrm{PN}\times \mathrm{QN})+\frac{1}{2}\times (\mathrm{RL}+\mathrm{QN})\times \mathrm{NL}\\ & & +(\frac{1}{2}\times \mathrm{LS}\times \mathrm{RL})+(\frac{1}{2}\times \mathrm{MS}\times \mathrm{TM})\\ & & +\frac{1}{2}\times (\mathrm{TM}+\mathrm{UO})\times \mathrm{OM}+(\frac{1}{2}\times \mathrm{PO}\times \mathrm{UO})\end{array}$

$\begin{array}{rcl} & =& (\frac{1}{2}\times 6\times 2)+\frac{1}{2}\times (4+2)\times 8\\ & & +(\frac{1}{2}\times 6\times 4)+(\frac{1}{2}\times 8\times 8)\\ & & +\frac{1}{2}\times (8+6)\times 8+(\frac{1}{2}\times 4\times 6)\\ & =& 6+24+12+32+56+12\\ & =& 142 {\mathrm{cm}}^2\end{array}$

Therefore, the area of the polygon PQRSTU = 142 cm²