Question Number 211167 by cherokeesay last updated on 30/Aug/24
Commented by Frix last updated on 30/Aug/24
$$\mathrm{12} \\ $$
Answered by som(math1967) last updated on 30/Aug/24
Commented by cherokeesay last updated on 30/Aug/24
$${nice}\:! \\ $$$${thank}\:{you}\:{sir}\:! \\ $$
Commented by som(math1967) last updated on 30/Aug/24
![let AB=x,AD=y ⇒AE=((18)/y) ∴BE=x−((18)/y) CF=((10)/x) ⇒BF=y−((10)/x) ∴ (x−((18)/y))(y−((10)/x))=8 ⇒xy−10−18+((180)/(xy))=8 ⇒(xy)^2 −36xy+180=0 ⇒(xy−30)(xy−6)=0 ⇒xy=30 [ xy≠6 ∵xy>9] Orange area=30−(9+5+4) =12sq unit](https://www.tinkutara.com/question/Q211178.png)
$${let}\:{AB}={x},{AD}={y} \\ $$$$\Rightarrow{AE}=\frac{\mathrm{18}}{{y}} \\ $$$$\therefore{BE}={x}−\frac{\mathrm{18}}{{y}} \\ $$$$\:{CF}=\frac{\mathrm{10}}{{x}}\:\Rightarrow{BF}={y}−\frac{\mathrm{10}}{{x}} \\ $$$$\:\therefore\:\left({x}−\frac{\mathrm{18}}{{y}}\right)\left({y}−\frac{\mathrm{10}}{{x}}\right)=\mathrm{8} \\ $$$$\Rightarrow{xy}−\mathrm{10}−\mathrm{18}+\frac{\mathrm{180}}{{xy}}=\mathrm{8} \\ $$$$\Rightarrow\left({xy}\right)^{\mathrm{2}} −\mathrm{36}{xy}+\mathrm{180}=\mathrm{0} \\ $$$$\Rightarrow\left({xy}−\mathrm{30}\right)\left({xy}−\mathrm{6}\right)=\mathrm{0} \\ $$$$\Rightarrow{xy}=\mathrm{30}\:\left[\:{xy}\neq\mathrm{6}\:\because{xy}>\mathrm{9}\right] \\ $$$${Orange}\:{area}=\mathrm{30}−\left(\mathrm{9}+\mathrm{5}+\mathrm{4}\right) \\ $$$$\:\:=\mathrm{12}{sq}\:{unit} \\ $$$$ \\ $$$$ \\ $$
Answered by mr W last updated on 30/Aug/24
$${A}_{{orange}} =\sqrt{\left({A}+{B}+{C}\right)^{\mathrm{2}} −\mathrm{4}{AC}} \\ $$$$\:\:\:\:=\sqrt{\left(\mathrm{9}+\mathrm{4}+\mathrm{5}\right)^{\mathrm{2}} −\mathrm{4}×\mathrm{9}×\mathrm{5}}=\mathrm{12} \\ $$