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Question Number 81435 by naka3546 last updated on 13/Feb/20

Commented by naka3546 last updated on 13/Feb/20

3, 4, 6 are Area of three triangles respectively . Shaded area is ...

$$\mathrm{3},\:\mathrm{4},\:\mathrm{6}\:\:{are}\:\:{Area}\:\:{of}\:\:{three}\:\:{triangles}\:\:{respectively}\:. \\ $$$${Shaded}\:\:{area}\:\:{is}\:\:...\: \\ $$

Commented by john santu last updated on 13/Feb/20

((x+y+4)/(3+6)) = (b_1 /b_2 ) ⇒ ((x+y+4)/9) =(b_1 /b_2 ) =(x/3)(i) ((x+y+3)/(6+4)) = (b_3 /b_4 ) = (y/4)(ii) (i) 3x+3y+12 = 9x ⇒6x−3y=12 2x−y=4 (ii) 4x+4y+12 = 10y ⇒4x−6y=−12 2x−3y=−6 elimination (i) ∧(ii) ⇒y = 5 ∧ x = (9/2) shaded area = x+y = ((19)/2)

$$\frac{\mathrm{x}+\mathrm{y}+\mathrm{4}}{\mathrm{3}+\mathrm{6}}\:=\:\frac{\mathrm{b}_{\mathrm{1}} }{\mathrm{b}_{\mathrm{2}} }\:\Rightarrow\:\frac{\mathrm{x}+\mathrm{y}+\mathrm{4}}{\mathrm{9}}\:=\frac{\mathrm{b}_{\mathrm{1}} }{\mathrm{b}_{\mathrm{2}} }\:=\frac{\mathrm{x}}{\mathrm{3}}\left(\mathrm{i}\right) \\ $$$$\frac{\mathrm{x}+\mathrm{y}+\mathrm{3}}{\mathrm{6}+\mathrm{4}}\:=\:\frac{\mathrm{b}_{\mathrm{3}} }{\mathrm{b}_{\mathrm{4}} }\:=\:\frac{\mathrm{y}}{\mathrm{4}}\left(\mathrm{ii}\right) \\ $$$$\left(\mathrm{i}\right)\:\mathrm{3x}+\mathrm{3y}+\mathrm{12}\:=\:\mathrm{9x}\:\Rightarrow\mathrm{6x}−\mathrm{3y}=\mathrm{12} \\ $$$$\mathrm{2x}−\mathrm{y}=\mathrm{4} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{4x}+\mathrm{4y}+\mathrm{12}\:=\:\mathrm{10y}\:\Rightarrow\mathrm{4x}−\mathrm{6y}=−\mathrm{12} \\ $$$$\mathrm{2x}−\mathrm{3y}=−\mathrm{6} \\ $$$$\mathrm{elimination}\:\left(\mathrm{i}\right)\:\wedge\left(\mathrm{ii}\right) \\ $$$$\Rightarrow\mathrm{y}\:=\:\mathrm{5}\:\wedge\:\mathrm{x}\:=\:\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\mathrm{shaded}\:\mathrm{area}\:=\:\mathrm{x}+\mathrm{y}\:=\:\frac{\mathrm{19}}{\mathrm{2}} \\ $$$$ \\ $$

Commented by naka3546 last updated on 13/Feb/20

9.5 , sir ?

$$\mathrm{9}.\mathrm{5}\:\:,\:\:{sir}\:? \\ $$

Commented by mr W last updated on 13/Feb/20

9.5 is correct.

$$\mathrm{9}.\mathrm{5}\:{is}\:{correct}. \\ $$

Commented by john santu last updated on 13/Feb/20

post you work?

$$\mathrm{post}\:\mathrm{you}\:\mathrm{work}? \\ $$

Commented by john santu last updated on 13/Feb/20

yes right, i misscalculated the second equation

$$\mathrm{yes}\:\mathrm{right},\:\mathrm{i}\:\mathrm{misscalculated}\:\mathrm{the} \\ $$$$\mathrm{second}\:\mathrm{equation} \\ $$

Answered by john santu last updated on 13/Feb/20

Answered by mr W last updated on 13/Feb/20

Commented by mr W last updated on 13/Feb/20

((y+4)/x)=(6/3)=2 ⇒2x=y+4 ...(i) ((x+3)/y)=(6/4)=(3/2) ⇒2x+6=3y ...(ii) from (i) and (ii) we get x=4.5, y=5 ⇒shaded area=x+y=9.5

$$\frac{{y}+\mathrm{4}}{{x}}=\frac{\mathrm{6}}{\mathrm{3}}=\mathrm{2}\:\Rightarrow\mathrm{2}{x}={y}+\mathrm{4}\:\:\:\:...\left({i}\right) \\ $$$$\frac{{x}+\mathrm{3}}{{y}}=\frac{\mathrm{6}}{\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow\mathrm{2}{x}+\mathrm{6}=\mathrm{3}{y}\:\:\:...\left({ii}\right) \\ $$$${from}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{we}\:{get} \\ $$$${x}=\mathrm{4}.\mathrm{5},\:{y}=\mathrm{5} \\ $$$$ \\ $$$$\Rightarrow{shaded}\:{area}={x}+{y}=\mathrm{9}.\mathrm{5} \\ $$

Commented by john santu last updated on 13/Feb/20

yes. i misscalculated

$$\mathrm{yes}.\:\mathrm{i}\:\mathrm{misscalculated} \\ $$

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