Question

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.

• A. 142
• B. 144
• C. 156
• D. 147

Answer 1

The correct option is A

142

Area of the hexagon PQRSTU = Area of $\Delta$PQN + Area of quad QRLN + Area of $\Delta$RSL + Area of $\Delta$TMS + Area of quad OMTU + Area of $\Delta$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 $m^2$