Question Number 176192 by adhigenz last updated on 14/Sep/22
$$\mathrm{Suppose}\:\mathrm{ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{rectangle}.\:\mathrm{X}\:\mathrm{and}\:\mathrm{Y}\:\mathrm{are}\:\mathrm{points}\:\mathrm{on}\:\mathrm{BC}\:\mathrm{and}\:\mathrm{CD}\:\mathrm{respectively}, \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{ABX},\:\mathrm{CXY},\:\mathrm{and}\:\mathrm{AYD}\:\mathrm{are}\:\mathrm{3}\:\mathrm{cm}^{\mathrm{2}} ,\:\mathrm{4}\:\mathrm{cm}^{\mathrm{2}} ,\:\mathrm{and}\:\mathrm{5}\:\mathrm{cm}^{\mathrm{2}} \:\mathrm{respectively}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{AXY}. \\ $$
Answered by som(math1967) last updated on 15/Sep/22
![let AB=m BC=n (1/2)×m×BX=3 BX=(6/m) CX=n−(6/m)=((mn−6)/m) YC=((8m)/(mn−6)) DY=m−((8m)/(mn−6))=m(((mn−14)/(mn−6))) ar.△AYD=5 (1/2)×n×m(((mn−14)/(mn−6)))=5 ((a^2 −14a)/(a−6))=10 [mn=ar.of rectangle=a] a^2 −24a+60=0 a=((24±(√(576−240)))/2)=((24±18.33)/2) ∴a=21.2 (approx) or2.8 but a>5 ∴a=21.2 ∴ area △AXY=21.2−(3+4+5) =21.2−12=9.2cm^2](https://www.tinkutara.com/question/Q176204.png)
$${let}\:{AB}={m}\:\:{BC}={n} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}}×{m}×{BX}=\mathrm{3} \\ $$$$\:{BX}=\frac{\mathrm{6}}{{m}} \\ $$$${CX}={n}−\frac{\mathrm{6}}{{m}}=\frac{{mn}−\mathrm{6}}{{m}} \\ $$$$\:{YC}=\frac{\mathrm{8}{m}}{{mn}−\mathrm{6}} \\ $$$${DY}={m}−\frac{\mathrm{8}{m}}{{mn}−\mathrm{6}}={m}\left(\frac{{mn}−\mathrm{14}}{{mn}−\mathrm{6}}\right) \\ $$$${ar}.\bigtriangleup{AYD}=\mathrm{5} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×{n}×{m}\left(\frac{{mn}−\mathrm{14}}{{mn}−\mathrm{6}}\right)=\mathrm{5} \\ $$$$\frac{{a}^{\mathrm{2}} −\mathrm{14}{a}}{{a}−\mathrm{6}}=\mathrm{10}\:\:\:\left[{mn}={ar}.{of}\:{rectangle}={a}\right] \\ $$$${a}^{\mathrm{2}} −\mathrm{24}{a}+\mathrm{60}=\mathrm{0} \\ $$$$\:{a}=\frac{\mathrm{24}\pm\sqrt{\mathrm{576}−\mathrm{240}}}{\mathrm{2}}=\frac{\mathrm{24}\pm\mathrm{18}.\mathrm{33}}{\mathrm{2}} \\ $$$$\therefore{a}=\mathrm{21}.\mathrm{2}\:\left({approx}\right)\:{or}\mathrm{2}.\mathrm{8} \\ $$$${but}\:{a}>\mathrm{5} \\ $$$$\therefore{a}=\mathrm{21}.\mathrm{2} \\ $$$$\therefore\:{area}\:\bigtriangleup{AXY}=\mathrm{21}.\mathrm{2}−\left(\mathrm{3}+\mathrm{4}+\mathrm{5}\right) \\ $$$$\:\:\:\:=\mathrm{21}.\mathrm{2}−\mathrm{12}=\mathrm{9}.\mathrm{2}{cm}^{\mathrm{2}} \\ $$
Commented by adhigenz last updated on 15/Sep/22
$$\mathrm{many}\:\mathrm{thanks}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$