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

Commented by naka3546 last updated on 13/Feb/20

$$3,4,6areAreaofthreetrianglesrespectively.$$$$Shadedareais...$$

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

$$\frac{\mathrm{x}+\mathrm{y}+4}{3+6}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}\Rightarrow \frac{\mathrm{x}+\mathrm{y}+4}{9}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{\mathrm{x}}{3}\left(\mathrm{i}\right)$$$$\frac{\mathrm{x}+\mathrm{y}+3}{6+4}=\frac{{\mathrm{b}}_3}{{\mathrm{b}}_4}=\frac{\mathrm{y}}{4}\left(\mathrm{ii}\right)$$$$\left(\mathrm{i}\right)3\mathrm{x}+3\mathrm{y}+12=9\mathrm{x}\Rightarrow 6\mathrm{x}-3\mathrm{y}=12$$$$2\mathrm{x}-\mathrm{y}=4$$$$\left(\mathrm{ii}\right)4\mathrm{x}+4\mathrm{y}+12=10\mathrm{y}\Rightarrow 4\mathrm{x}-6\mathrm{y}=-12$$$$2\mathrm{x}-3\mathrm{y}=-6$$$$\mathrm{elimination}\left(\mathrm{i}\right)\wedge \left(\mathrm{ii}\right)$$$$\Rightarrow \mathrm{y}=5\wedge \mathrm{x}=\frac{9}{2}$$$$\mathrm{shaded}\mathrm{area}=\mathrm{x}+\mathrm{y}=\frac{19}{2}$$$$$$

Commented by naka3546 last updated on 13/Feb/20

$$9.5,sir?$$

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

$$9.5iscorrect.$$

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

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

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

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

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

$$\frac{y+4}{x}=\frac{6}{3}=2\Rightarrow 2x=y+4...\left(i\right)$$$$\frac{x+3}{y}=\frac{6}{4}=\frac{3}{2}\Rightarrow 2x+6=3y...\left(ii\right)$$$$from\left(i\right)and\left(ii\right)weget$$$$x=4.5,y=5$$$$$$$$\Rightarrow shadedarea=x+y=9.5$$

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

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

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