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Question Number 176192 by adhigenz last updated on 14/Sep/22

$$\mathrm{Suppose}\mathrm{ABCD}\mathrm{is}\mathrm{a}\mathrm{rectangle}.\mathrm{X}\mathrm{and}\mathrm{Y}\mathrm{are}\mathrm{points}\mathrm{on}\mathrm{BC}\mathrm{and}\mathrm{CD}\mathrm{respectively},$$$$\mathrm{such}\mathrm{that}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{ABX},\mathrm{CXY},\mathrm{and}\mathrm{AYD}\mathrm{are}3{\mathrm{cm}}^2,4{\mathrm{cm}}^2,\mathrm{and}5{\mathrm{cm}}^2\mathrm{respectively}.$$$$\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{AXY}.$$

Answered by som(math1967) last updated on 15/Sep/22

$$letAB=mBC=n$$$$\frac{1}{2}\times m\times BX=3$$$$BX=\frac{6}{m}$$$$CX=n-\frac{6}{m}=\frac{mn-6}{m}$$$$YC=\frac{8m}{mn-6}$$$$DY=m-\frac{8m}{mn-6}=m\left(\frac{mn-14}{mn-6}\right)$$$$ar.△AYD=5$$$$\frac{1}{2}\times n\times m\left(\frac{mn-14}{mn-6}\right)=5$$$$\frac{a^2-14a}{a-6}=10[mn=ar.ofrectangle=a]$$$$a^2-24a+60=0$$$$a=\frac{24\pm \sqrt{576-240}}{2}=\frac{24\pm 18.33}{2}$$$$\therefore a=21.2\left(approx\right)or2.8$$$$buta>5$$$$\therefore a=21.2$$$$\therefore area△AXY=21.2-(3+4+5)$$$$=21.2-12=9.2{cm}^2$$

Commented by adhigenz last updated on 15/Sep/22

$$\mathrm{many}\mathrm{thanks}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

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