Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

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

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

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

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

$${let}\:{AB}={m}\:\:{BC}={n} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}}×{m}×{BX}=\mathrm{3} \\ $$$$\:{BX}=\frac{\mathrm{6}}{{m}} \\ $$$${CX}={n}−\frac{\mathrm{6}}{{m}}=\frac{{mn}−\mathrm{6}}{{m}} \\ $$$$\:{YC}=\frac{\mathrm{8}{m}}{{mn}−\mathrm{6}} \\ $$$${DY}={m}−\frac{\mathrm{8}{m}}{{mn}−\mathrm{6}}={m}\left(\frac{{mn}−\mathrm{14}}{{mn}−\mathrm{6}}\right) \\ $$$${ar}.\bigtriangleup{AYD}=\mathrm{5} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×{n}×{m}\left(\frac{{mn}−\mathrm{14}}{{mn}−\mathrm{6}}\right)=\mathrm{5} \\ $$$$\frac{{a}^{\mathrm{2}} −\mathrm{14}{a}}{{a}−\mathrm{6}}=\mathrm{10}\:\:\:\left[{mn}={ar}.{of}\:{rectangle}={a}\right] \\ $$$${a}^{\mathrm{2}} −\mathrm{24}{a}+\mathrm{60}=\mathrm{0} \\ $$$$\:{a}=\frac{\mathrm{24}\pm\sqrt{\mathrm{576}−\mathrm{240}}}{\mathrm{2}}=\frac{\mathrm{24}\pm\mathrm{18}.\mathrm{33}}{\mathrm{2}} \\ $$$$\therefore{a}=\mathrm{21}.\mathrm{2}\:\left({approx}\right)\:{or}\mathrm{2}.\mathrm{8} \\ $$$${but}\:{a}>\mathrm{5} \\ $$$$\therefore{a}=\mathrm{21}.\mathrm{2} \\ $$$$\therefore\:{area}\:\bigtriangleup{AXY}=\mathrm{21}.\mathrm{2}−\left(\mathrm{3}+\mathrm{4}+\mathrm{5}\right) \\ $$$$\:\:\:\:=\mathrm{21}.\mathrm{2}−\mathrm{12}=\mathrm{9}.\mathrm{2}{cm}^{\mathrm{2}} \\ $$

Commented by adhigenz last updated on 15/Sep/22

many thanks sir. God bless you.

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com