Find the area of hexagon PQRSTU (in $m^{2}$ ) in which QN is perpendicular to PS, RL is perpendicular to PS, TM is perpendicular to PS, UO is perpendicular to PS such that PO= 4 m, ON=2 m, NM= 6 m, ML=2 m, LS= 6 m, UO= 6 m, TM= 8 m, QN= 2 m and RL=4 m.
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Solution

Area of the hexagon PQRSTU = Area of $Δ$ PQN + Area of quad QRLN + Area of $Δ$ RSL + Area of $Δ$ TMS + Area of quad OMTU + Area of $Δ$ POU

=( $\frac{1}{2}$ $\times$ PN $\times$ QN) + { $\frac{1}{2}$ $\times$ (RL+QN) $\times$ NL} + ( $\frac{1}{2}$ $\times$ LS $\times$ RL) + ( $\frac{1}{2}$ $\times$ MS $\times$ TM) + { $\frac{1}{2}$ $\times$ (TM+UO) $\times$ OM} + ( $\frac{1}{2}$ $\times$ PO $\times$ UO)

= ( $\frac{1}{2}$ $\times$ 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 $m^{2}$

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