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Question Number 72579 by TawaTawa last updated on 30/Oct/19

Answered by mind is power last updated on 30/Oct/19

$$\mathrm{CF}.\mathrm{AD}=\mathrm{DC}.\mathrm{AE}=\mathrm{AB}.\mathrm{AE}$$$$\Rightarrow \mathrm{AD}=\frac{16.8}{10}=\frac{64}{5}=12.8\mathrm{cm}$$$$2)\mathrm{ar}\left(\mathrm{ABCD}\right)=\mathrm{DC}.\mathrm{AD}.\sin \left(\mathrm{\angle }\mathrm{CDA}\right)$$$$\mathrm{ar}\left(\mathrm{EFGH}\right)=\mathrm{ar}\left(\mathrm{ABCD}\right)-\mathrm{ar}\left(\mathrm{EBF}\right)-\mathrm{ar}\left(\mathrm{FCG}\right)-\mathrm{ar}\left(\mathrm{GDH}\right)-\mathrm{ar}\left(\mathrm{DHE}\right)$$$$=\mathrm{ar}\left(\mathrm{ABCD}\right)-4\mathrm{ar}\left(\mathrm{EBF}\right)$$$$\mathrm{ar}\left(\mathrm{EBF}\right)=\frac{1}{2}.\frac{\mathrm{AB}}{2}.\frac{\mathrm{BF}}{2}.\sin \left(\mathrm{\angle }\mathrm{EBF}\right)=\frac{\mathrm{DC}.\mathrm{AC}.\sin \left(\mathrm{\angle }\mathrm{CDA}\right)}{8}=\frac{\mathrm{ar}\left(\mathrm{ABCD}\right)}{8}$$$$\Rightarrow \mathrm{ar}\left(\mathrm{EFGH}\right)=\mathrm{ar}\left(\mathrm{ABCD}\right)-\frac{4}{8}\mathrm{ar}\left(\mathrm{ABCD}\right)=\frac{\mathrm{ar}\left(\mathrm{ABCD}\right)}{2}$$$$\mathrm{let}\mathrm{a}=\mathrm{\angle }\mathrm{ADC}$$$$3)\mathrm{ar}\left(\mathrm{APB}\right)=\mathrm{ar}\left(\mathrm{ABCD}\right)-\{\mathrm{ar}\left(\mathrm{ADP}\right)+\mathrm{ar}\left(\mathrm{PCD}\right)\}$$$$=\frac{\mathrm{AD}.\mathrm{DC}}{}\sin \left(\mathrm{a}\right)-\{\frac{\mathrm{AD}.\mathrm{DP}}{2}.\sin \left(\mathrm{a}\right)+\frac{\mathrm{PC}.\mathrm{AD}}{2}\sin \left(\mathrm{a}\right)\}=\mathrm{AD}.\mathrm{DCsin}\left(\mathrm{a}\right)-\frac{\mathrm{AD}.\mathrm{DCsin}\left(\mathrm{a}\right)}{2}$$$$=\frac{1}{2}.\mathrm{Ar}\left(\mathrm{ABCD}\right)=\mathrm{ar}\left(\mathrm{BQC}\right)$$$$$$

Commented by TawaTawa last updated on 30/Oct/19

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