Question Number 224168 by mr W last updated on 23/Aug/25
Commented by mr W last updated on 23/Aug/25
$${find}\:{the}\:{shaded}\:{area}. \\ $$
Answered by fantastic last updated on 23/Aug/25
Commented by fantastic last updated on 23/Aug/25
$$\Box{ABCD}={x}+\mathrm{6}+\mathrm{9}+\mathrm{63}=\mathrm{78}+{x} \\ $$$$\bigtriangleup{AFC}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{78}+{x}\right)−\left(\mathrm{6}+\mathrm{9}+{x}\right) \\ $$$$=\mathrm{39}+\frac{{x}}{\mathrm{2}}−\mathrm{15}−{x} \\ $$$$=\mathrm{24}−\frac{{x}}{\mathrm{2}} \\ $$$${AF}/{FE}={x}/\mathrm{6} \\ $$$${AF}/{FE}=\left(\mathrm{24}−\frac{{x}}{\mathrm{2}}\right)/\mathrm{9} \\ $$$$\therefore\:\frac{{x}}{\mathrm{6}}=\frac{\mathrm{24}−\frac{{x}}{\mathrm{2}}}{\mathrm{9}} \\ $$$$\Rightarrow{x}=\mathrm{12}\checkmark \\ $$
Answered by A5T last updated on 23/Aug/25
![Let [ACF]=y and [ABF]=x (y/9)=(x/6)⇒y=((3x)/2) x+6+9+y=63−y⇒x+2y=48⇒4x=48 ⇒x=12](https://www.tinkutara.com/question/Q224174.png)
$$\mathrm{Let}\:\left[\mathrm{ACF}\right]=\mathrm{y}\:\mathrm{and}\:\left[\mathrm{ABF}\right]=\mathrm{x} \\ $$$$\frac{\mathrm{y}}{\mathrm{9}}=\frac{\mathrm{x}}{\mathrm{6}}\Rightarrow\mathrm{y}=\frac{\mathrm{3x}}{\mathrm{2}} \\ $$$$\mathrm{x}+\mathrm{6}+\mathrm{9}+\mathrm{y}=\mathrm{63}−\mathrm{y}\Rightarrow\mathrm{x}+\mathrm{2y}=\mathrm{48}\Rightarrow\mathrm{4x}=\mathrm{48} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{12} \\ $$