Question-12933

Question Number 12933 by chux last updated on 07/May/17

Commented by ajfour last updated on 08/May/17

$${Touch}\:{and}\:{Draw} \\ $$

Commented by chux last updated on 07/May/17

this question was solved but was explained. please i need detailed explanation to the solution. Thanks for always helping.

$$\mathrm{this}\:\mathrm{question}\:\mathrm{was}\:\mathrm{solved}\:\mathrm{but}\:\mathrm{was} \\ $$$$\mathrm{explained}.\:\mathrm{please}\:\mathrm{i}\:\mathrm{need}\:\mathrm{detailed} \\ $$$$\mathrm{explanation}\:\mathrm{to}\:\mathrm{the}\:\mathrm{solution}. \\ $$$$ \\ $$$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{always}\:\mathrm{helping}. \\ $$

Commented by ajfour last updated on 07/May/17

Commented by ajfour last updated on 07/May/17

$$\mathrm{24}×\mathrm{5}=\mathrm{120}\:. \\ $$

Commented by ajfour last updated on 07/May/17

△EGD & △ECF have the same area , so Area of trapez(ABCD) = Area of rectangle(ABFG) . further mid-point of altitude FG is at E. EF=12 , BF=(√(13^2 −12^2 )) =5 . Area (ABCD)=Area(ABFG) =AB×BF = 24×5 =120 .

$$\bigtriangleup{EGD}\:\&\:\bigtriangleup{ECF}\:{have}\:{the}\:{same} \\ $$$${area}\:,\:{so}\:{Area}\:{of}\:{trapez}\left({ABCD}\right) \\ $$$$\:\:\:\:\:\:\:\:=\:{Area}\:{of}\:{rectangle}\left({ABFG}\right)\:. \\ $$$${further}\:{mid}-{point}\:{of}\:{altitude}\:{FG}\: \\ $$$${is}\:{at}\:{E}.\: \\ $$$${EF}=\mathrm{12}\:,\:\:{BF}=\sqrt{\mathrm{13}^{\mathrm{2}} −\mathrm{12}^{\mathrm{2}} }\:=\mathrm{5}\:. \\ $$$${Area}\:\left({ABCD}\right)={Area}\left({ABFG}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={AB}×{BF} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{24}×\mathrm{5}\:=\mathrm{120}\:. \\ $$

Commented by chux last updated on 07/May/17

thanks alot.... i′m so sorry but its still not clear.

$$\mathrm{thanks}\:\mathrm{alot}….\:\mathrm{i}'\mathrm{m}\:\mathrm{so}\:\mathrm{sorry}\:\mathrm{but}\:\mathrm{its}\: \\ $$$$\mathrm{still}\:\mathrm{not}\:\mathrm{clear}. \\ $$

Commented by chux last updated on 07/May/17

$$\mathrm{ok}…..\:\mathrm{Thanks}. \\ $$

Commented by tawa last updated on 07/May/17

Please sir. What is the name of the application you are using to draw the image

$$\mathrm{Please}\:\mathrm{sir}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{name}\:\mathrm{of}\:\mathrm{the}\:\mathrm{application}\:\mathrm{you}\:\mathrm{are}\:\mathrm{using}\:\mathrm{to}\:\mathrm{draw}\:\mathrm{the}\: \\ $$$$\mathrm{image} \\ $$

Commented by tawa tawa last updated on 11/May/17

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

Answered by Joel577 last updated on 08/May/17

Commented by Joel577 last updated on 08/May/17

$$\Delta{BCE}\:\cong\:\Delta{DEF} \\ $$$${AB}\:=\:\mathrm{24},\:{BF}\:=\:\mathrm{26},\:{AF}\:=\:\mathrm{10} \\ $$

Answered by sandy_suhendra last updated on 09/May/17

Commented by sandy_suhendra last updated on 09/May/17

EF=(√(13^2 −12^2 ))=5 EF=((DA×FB+CB×AF)/(AF+FB)) 5=((x.12+y.12)/(12+12)) 12(x+y)=120 x+y=10=DA+CB A_(ABCD) =((DA+CB)/2)×AB =((10)/2)×24=120

$$\mathrm{EF}=\sqrt{\mathrm{13}^{\mathrm{2}} −\mathrm{12}^{\mathrm{2}} }=\mathrm{5} \\ $$$$\mathrm{EF}=\frac{\mathrm{DA}×\mathrm{FB}+\mathrm{CB}×\mathrm{AF}}{\mathrm{AF}+\mathrm{FB}} \\ $$$$\mathrm{5}=\frac{\mathrm{x}.\mathrm{12}+\mathrm{y}.\mathrm{12}}{\mathrm{12}+\mathrm{12}} \\ $$$$\mathrm{12}\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{120} \\ $$$$\mathrm{x}+\mathrm{y}=\mathrm{10}=\mathrm{DA}+\mathrm{CB} \\ $$$$\mathrm{A}_{\mathrm{ABCD}} =\frac{\mathrm{DA}+\mathrm{CB}}{\mathrm{2}}×\mathrm{AB} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{10}}{\mathrm{2}}×\mathrm{24}=\mathrm{120} \\ $$

Commented by chux last updated on 10/May/17

$$\mathrm{wow}…..\:\mathrm{thanks}\:\mathrm{alot}. \\ $$

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