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Question Number 15904 by mrW1 last updated on 15/Jun/17

Commented by mrW1 last updated on 15/Jun/17

$$\mathrm{A}\mathrm{convex}\mathrm{quadrilateral}\mathrm{ABCD}\mathrm{is}$$$$\mathrm{divided}\mathrm{into}4\mathrm{triangle}\mathrm{by}\mathrm{its}\mathrm{two}$$$$\mathrm{diagonals}.\mathrm{The}\mathrm{centroids}\mathrm{of}\mathrm{these}$$$$\mathrm{triangles}\mathrm{form}\mathrm{a}\mathrm{new}\mathrm{quadrilaterial}.$$$$\mathrm{Find}\mathrm{out}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{new}\mathrm{quadri}-$$$$\mathrm{lateral}\mathrm{in}\mathrm{relation}\mathrm{to}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}$$$$\mathrm{original}\mathrm{quadrilateral}.$$

Answered by ajfour last updated on 15/Jun/17

$$letAO=a,BO=b,CO=c,DO=d$$$$and\alpha =\mathrm{\angle }AOB,\mathrm{\angle }BOC=\pi -\alpha$$$$\mathrm{\angle }COD=\alpha ,\mathrm{\angle }DOA=\pi -\alpha$$$${Area}_{ABCD}=\frac{1}{2}[ab\sin \alpha +$$$$bc\sin (\pi -\alpha )+cd\sin \alpha +da\sin (\pi -\alpha )]$$$${Area}_{ABCD}=\frac{1}{2}\left(\sin \alpha \right)(ab+bc+cd+da)$$$$\Rightarrow {Area}_{ABCD}=\frac{1}{2}(a+c)(b+d)\sin \alpha$$$$\mathrm{\Delta }AOBand\mathrm{\Delta }COBhavethe$$$$sameheight(withtheirbases$$$$alongAC)=b\sin \alpha andtheir$$$$centroidsareEandFrespectively$$$$withheight=\frac{b}{3}\sin \alpha .$$$$soEFisparalleltoAC.$$$$similarlyEH,FG//BD.$$$$wecanconcludeEFGHisa$$$$parallelogram.$$$$letitsbasebealongACoflength$$$$=\frac{1}{3}(a+c)$$$$itsheight\mathrm{\perp }toAC=\frac{1}{3}(b+d)\sin \alpha$$$${Area}_{EFGH}=\frac{1}{9}(a+c)(b+d)\sin \alpha$$$$=\frac{2}{9}\times \frac{1}{2}(a+c)(b+d)\sin \alpha$$$${\mathit{A}\mathit{r}\mathit{e}\mathit{a}}_{\mathit{E}\mathit{F}\mathit{G}\mathit{H}}=\frac{2}{9}\left({\mathit{A}\mathit{r}\mathit{e}\mathit{a}}_{\mathit{A}\mathit{B}\mathit{C}\mathit{D}}\right).$$

Commented by mrW1 last updated on 15/Jun/17

$$\mathrm{very}\mathrm{correct},\mathrm{thanks}!$$

Commented by ajfour last updated on 15/Jun/17

$$thanksSir,Godblessyou.$$

Answered by mrW1 last updated on 15/Jun/17

$$\mathrm{An}\mathrm{other}\mathrm{way}\mathrm{to}\mathrm{solve}:$$$$\because \mathrm{OE}=\frac{2}{3}\mathrm{OP}$$$$\therefore {\mathrm{A}}_{\mathrm{\Delta }\mathrm{OEA}}=\frac{2}{3}{\mathrm{A}}_{\mathrm{\Delta }\mathrm{OPA}}$$$$\because \mathrm{OT}=\frac{1}{3}\mathrm{OA}$$$$\therefore {\mathrm{A}}_{\mathrm{\Delta }\mathrm{OET}}=\frac{1}{3}{\mathrm{A}}_{\mathrm{\Delta }\mathrm{OEA}}=\frac{1}{3}\times \frac{2}{3}{\mathrm{A}}_{\mathrm{\Delta }\mathrm{OPA}}=\frac{2}{9}{\mathrm{A}}_{\mathrm{\Delta }\mathrm{OPA}}$$$$\mathrm{similarly}$$$${\mathrm{A}}_{\mathrm{\Delta }\mathrm{OES}}=\frac{2}{9}{\mathrm{A}}_{\mathrm{\Delta }\mathrm{OPB}}$$$$......$$$${\mathrm{A}}_{\mathrm{EFGH}}={\mathrm{A}}_{\mathrm{\Delta }\mathrm{OET}}+{\mathrm{A}}_{\mathrm{\Delta }\mathrm{OES}}+.......$$$$=\frac{2}{9}{\mathrm{A}}_{\mathrm{\Delta }\mathrm{OPA}}+\frac{2}{9}{\mathrm{A}}_{\mathrm{\Delta }\mathrm{OPB}}+......$$$$=\frac{2}{9}{\mathrm{A}}_{\mathrm{ABCD}}$$

Commented by mrW1 last updated on 15/Jun/17

Commented by ajfour last updated on 15/Jun/17

$$brillianttechnique,Sir.$$$$thankyousir.$$

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