Question Number 173644 by infinityaction last updated on 15/Jul/22
Answered by mr W last updated on 16/Jul/22
Commented by mr W last updated on 16/Jul/22
$$\frac{\mathrm{8}−{h}_{\mathrm{1}} }{{h}_{\mathrm{1}} }=\frac{{x}}{\mathrm{18}} \\ $$$$\Rightarrow{h}_{\mathrm{1}} =\frac{\mathrm{144}}{\mathrm{18}+{x}} \\ $$$$\frac{\mathrm{8}−{h}_{\mathrm{2}} }{{h}_{\mathrm{2}} }=\frac{{x}}{\mathrm{12}} \\ $$$$\Rightarrow{h}_{\mathrm{2}} =\frac{\mathrm{96}}{\mathrm{12}+{x}} \\ $$$$\frac{\mathrm{8}−{h}_{\mathrm{3}} }{{h}_{\mathrm{3}} }=\frac{{x}}{\mathrm{6}} \\ $$$$\Rightarrow{h}_{\mathrm{3}} =\frac{\mathrm{48}}{\mathrm{6}+{x}} \\ $$$${shaded}\:{area}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{18}{h}_{\mathrm{1}} −\mathrm{12}{h}_{\mathrm{2}} −\mathrm{6}{h}_{\mathrm{3}} \right) \\ $$$$\mathrm{10}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{18}×\mathrm{144}}{\mathrm{18}+{x}}−\frac{\mathrm{12}×\mathrm{96}}{\mathrm{12}+{x}}−\frac{\mathrm{6}×\mathrm{48}}{\mathrm{6}+{x}}\right) \\ $$$$\frac{\mathrm{5}}{\mathrm{72}}=\frac{\mathrm{9}}{\mathrm{18}+{x}}−\frac{\mathrm{4}}{\mathrm{12}+{x}}−\frac{\mathrm{1}}{\mathrm{6}+{x}} \\ $$$$\frac{\mathrm{5}}{\mathrm{72}}=\frac{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{36}{x}}{\left(\mathrm{18}+{x}\right)\left(\mathrm{12}+{x}\right)\left(\mathrm{6}+{x}\right)} \\ $$$$\left({x}−\mathrm{6}\right)\left(\mathrm{5}{x}^{\mathrm{2}} −\mathrm{78}{x}−\mathrm{1080}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{6}\:\checkmark \\ $$$$\Rightarrow{x}=\frac{\mathrm{39}+\mathrm{3}\sqrt{\mathrm{769}}}{\mathrm{5}}\approx\mathrm{24}.\mathrm{439}\:\checkmark \\ $$
Commented by infinityaction last updated on 16/Jul/22
$${thanks}\:{sir} \\ $$
Commented by Tawa11 last updated on 16/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$