Question-223101

Question Number 223101 by Jubr last updated on 14/Jul/25

Commented by Tawa11 last updated on 15/Jul/25

? = ((16)/(25))(? + 76.5) ∴ ? = 136 cm^2

$$?\:\:=\:\:\frac{\mathrm{16}}{\mathrm{25}}\left(?\:\:+\:\:\mathrm{76}.\mathrm{5}\right) \\ $$$$\therefore\:\:\:\:?\:\:\:=\:\:\:\mathrm{136}\:\mathrm{cm}^{\mathrm{2}} \\ $$

Commented by mr W last updated on 15/Jul/25

how did you get ((16)/(25))? how did you get ?=((16)/(25))(?+76.5) ?

$${how}\:{did}\:{you}\:{get}\:\frac{\mathrm{16}}{\mathrm{25}}? \\ $$$${how}\:{did}\:{you}\:{get}\:?=\frac{\mathrm{16}}{\mathrm{25}}\left(?+\mathrm{76}.\mathrm{5}\right)\:\:? \\ $$

Commented by A5T last updated on 15/Jul/25

(([CDFN])/([CBMN]))=((DF×NF=BM×NF)/(BM×NM))=((NF)/(NM))=((CD)/(CB)) (([CED])/([CAB]))=(((1/2)×CD×ED)/((1/2)×CB×AB)) but ((ED)/(AB))=((CD)/(CB)) ⇒(([CED])/([CAB]))=((CD^2 )/(CB^2 ))=((NF^2 )/(NM^2 ))=((476^2 )/((476+119)^2 ))=(4^2 /5^2 )=((16)/(25)) ⇒(?/(?+76.5))=((16)/(25))⇒?=((16)/(25))(?+76.5)

$$\frac{\left[\mathrm{CDFN}\right]}{\left[\mathrm{CBMN}\right]}=\frac{\mathrm{DF}×\mathrm{NF}=\mathrm{BM}×\mathrm{NF}}{\mathrm{BM}×\mathrm{NM}}=\frac{\mathrm{NF}}{\mathrm{NM}}=\frac{\mathrm{CD}}{\mathrm{CB}} \\ $$$$\frac{\left[\mathrm{CED}\right]}{\left[\mathrm{CAB}\right]}=\frac{\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{CD}×\mathrm{ED}}{\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{CB}×\mathrm{AB}}\:\mathrm{but}\:\frac{\mathrm{ED}}{\mathrm{AB}}=\frac{\mathrm{CD}}{\mathrm{CB}} \\ $$$$\Rightarrow\frac{\left[\mathrm{CED}\right]}{\left[\mathrm{CAB}\right]}=\frac{\mathrm{CD}^{\mathrm{2}} }{\mathrm{CB}^{\mathrm{2}} }=\frac{\mathrm{NF}^{\mathrm{2}} }{\mathrm{NM}^{\mathrm{2}} }=\frac{\mathrm{476}^{\mathrm{2}} }{\left(\mathrm{476}+\mathrm{119}\right)^{\mathrm{2}} }=\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{5}^{\mathrm{2}} }=\frac{\mathrm{16}}{\mathrm{25}} \\ $$$$\Rightarrow\frac{?}{?+\mathrm{76}.\mathrm{5}}=\frac{\mathrm{16}}{\mathrm{25}}\Rightarrow?=\frac{\mathrm{16}}{\mathrm{25}}\left(?+\mathrm{76}.\mathrm{5}\right) \\ $$

Commented by Tawa11 last updated on 15/Jul/25

The height of the rectangles are in the ratio ((476)/(119)) = 4 : 1 That is: CD : DB = 4 : 1 ((Area of green triangle (small) ΔCDE)/(Area of big triangle ΔCBA )) = (((CD)/(CB)))^2 = ((4/5))^2 = ((16)/(25)) Area of big triangle (ΔCBA) = Area of ΔCDE + Area of trapezium) Area of ΔCDE = ((16)/(25))(Area of ΔCDE + Area of trapezium) ? = ((16)/(25))(? + 76.5)

$$\mathrm{The}\:\mathrm{height}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rectangles}\:\mathrm{are}\:\mathrm{in}\:\mathrm{the}\:\mathrm{ratio}\:\:\frac{\mathrm{476}}{\mathrm{119}}\:\:=\:\:\mathrm{4}\::\:\mathrm{1} \\ $$$$\mathrm{That}\:\mathrm{is}:\:\mathrm{CD}\::\:\mathrm{DB}\:=\:\mathrm{4}\::\:\mathrm{1} \\ $$$$\frac{\mathrm{Area}\:\mathrm{of}\:\mathrm{green}\:\mathrm{triangle}\:\left(\mathrm{small}\right)\:\Delta\mathrm{CDE}}{\mathrm{Area}\:\mathrm{of}\:\mathrm{big}\:\mathrm{triangle}\:\Delta\mathrm{CBA}\:}\:\:=\:\:\left(\frac{\mathrm{CD}}{\mathrm{CB}}\right)^{\mathrm{2}} \:=\:\:\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{\mathrm{2}} \:\:=\:\:\frac{\mathrm{16}}{\mathrm{25}} \\ $$$$\left.\mathrm{Area}\:\mathrm{of}\:\mathrm{big}\:\mathrm{triangle}\:\left(\Delta\mathrm{CBA}\right)\:\:=\:\:\mathrm{Area}\:\mathrm{of}\:\Delta\mathrm{CDE}\:\:+\:\:\mathrm{Area}\:\mathrm{of}\:\mathrm{trapezium}\right) \\ $$$$\mathrm{Area}\:\mathrm{of}\:\Delta\mathrm{CDE}\:\:=\:\:\frac{\mathrm{16}}{\mathrm{25}}\left(\mathrm{Area}\:\mathrm{of}\:\Delta\mathrm{CDE}\:+\:\mathrm{Area}\:\mathrm{of}\:\mathrm{trapezium}\right) \\ $$$$\:\:\:\:\:\:\:\:?\:\:\:=\:\:\:\frac{\mathrm{16}}{\mathrm{25}}\left(?\:\:+\:\:\mathrm{76}.\mathrm{5}\right) \\ $$

Commented by mr W last updated on 15/Jul/25

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Commented by Tawa11 last updated on 15/Jul/25

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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