Question-88820

Question Number 88820 by ajfour last updated on 13/Apr/20

Commented by ajfour last updated on 13/Apr/20

$${If}\:{both}\:{coloured}\:{regions}\:{have} \\ $$$${equal}\:{area},\:{find}\:{B}\left({x}_{{B}} \:,\:{y}_{{B}} \right)\:{in} \\ $$$${terms}\:{of}\:{c}. \\ $$

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

$${total}\:{area}\:=\:\frac{\mathrm{4}}{\mathrm{3}}{c}\sqrt{{c}} \\ $$$${line}\:{BQ}\::\:{y}=\left(\sqrt{{b}}−\sqrt{{c}}\right){x}\:+\sqrt{{bc}} \\ $$$${blue}\:{area}\:=\:\frac{\left({b}+{c}+\mathrm{2}\sqrt{{bc}}\right)^{\mathrm{3}/\mathrm{2}} }{\mathrm{6}}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\left({c}\right)^{\mathrm{3}/\mathrm{2}} \\ $$$$ \\ $$

Commented by jagoll last updated on 13/Apr/20

$$\left(\sqrt{{b}}+\sqrt{{c}}\right)^{\mathrm{3}} \:=\:\mathrm{4}\left(\sqrt{{c}}\right)^{\mathrm{3}} \\ $$$$\sqrt{{b}}\:+\:\sqrt{\:{c}}\:=\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{4}}\:\sqrt{{c}} \\ $$$$\sqrt{{b}}\:=\:\sqrt{{c}}\:\left[\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{4}}\:−\mathrm{1}\:\right] \\ $$$$\sqrt{\frac{{b}}{{c}}}\:=\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{4}}\:−\mathrm{1} \\ $$

Commented by jagoll last updated on 13/Apr/20

$${i}\:{continue}\:{to}\:{work}\:{mr}\:{john} \\ $$

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

$${good}\:{mr}\:{jagoll}.\:{thank}\:{you} \\ $$

Answered by ajfour last updated on 13/Apr/20

$${A}_{{total}} =\mathrm{2}×\left({c}\sqrt{{c}}−\frac{{c}\sqrt{{c}}}{\mathrm{3}}\right)=\frac{\mathrm{4}{c}\sqrt{{c}}}{\mathrm{3}}\: \\ $$$${A}_{{blue}} =\frac{\mathrm{1}}{\mathrm{2}}\left({x}_{{B}} ^{\mathrm{2}} +{c}\right)\left({x}_{{B}} +\sqrt{{c}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−\left(\frac{{x}_{{B}} ^{\mathrm{3}} }{\mathrm{3}}+\frac{{c}\sqrt{{c}}}{\mathrm{3}}\right)\:\:=\:\frac{\mathrm{2}{c}\sqrt{{c}}}{\mathrm{3}} \\ $$$$\:\:\:\:{say}\:\:{x}_{{B}} =\:{t} \\ $$$$\:\:\:\:\:\frac{{t}^{\mathrm{3}} }{\mathrm{6}}+\frac{\sqrt{{c}}}{\mathrm{2}}{t}^{\mathrm{2}} +\frac{{c}}{\mathrm{2}}{t}−\frac{{c}\sqrt{{c}}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow\:\:{t}^{\mathrm{3}} +\mathrm{3}\sqrt{{c}}{t}^{\mathrm{2}} +\mathrm{3}{ct}+{c}\sqrt{{c}}=\mathrm{4}{c}\sqrt{{c}} \\ $$$$\Rightarrow\:\:\:\left({t}+\sqrt{{c}}\right)^{\mathrm{3}} =\mathrm{4}{c}\sqrt{{c}} \\ $$$$\Rightarrow\:\:\:{x}_{{B}} =\:{t}\:=\:\sqrt{{c}}\left(\:\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{1}\right)\:. \\ $$

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

$$\frac{\mathrm{2}}{\mathrm{3}}\left({x}_{{B}} +\sqrt{{c}}\right)\left[\frac{{c}+{x}_{{B}} ^{\mathrm{2}} }{\mathrm{2}}−\left(\frac{{x}_{{B}} −\sqrt{{c}}}{\mathrm{2}}\right)^{\mathrm{2}} \right]=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{2}{c}\sqrt{{c}} \\ $$$${let}\:{C}=\sqrt{{c}} \\ $$$$\left({x}_{{B}} +{C}\right)\left[\frac{{C}^{\mathrm{2}} +{x}_{{B}} ^{\mathrm{2}} +\mathrm{2}{Cx}_{{B}} }{\mathrm{4}}\right]={C}^{\mathrm{3}} \\ $$$$\left(\frac{{x}_{{B}} }{{C}}\right)^{\mathrm{3}} +\mathrm{3}\left(\frac{{x}_{{B}} }{{C}}\right)^{\mathrm{2}} +\mathrm{3}\left(\frac{{x}_{{B}} }{{C}}\right)−\mathrm{3}=\mathrm{0} \\ $$$$\left(\frac{{x}_{{B}} }{{C}}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{4} \\ $$$$\frac{{x}_{{B}} }{{C}}=\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{1}\approx\mathrm{0}.\mathrm{5874} \\ $$$$\Rightarrow{x}_{{B}} \approx\mathrm{0}.\mathrm{5874}\:\sqrt{{c}} \\ $$

Commented by ajfour last updated on 13/Apr/20

$$\frac{{x}_{{B}} }{{C}}+\mathrm{1}\:=\:\:\sqrt[{\mathrm{3}}]{\mathrm{4}} \\ $$$${x}_{{B}} =\sqrt{{c}}\:\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{1}\right)\:.\:\:{Isn}'{t}\:{it}\:{Sir}. \\ $$

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

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

$${yes}.\:{the}\:{exact}\:{value}\:{is}\:{x}_{{B}} =\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{1}\right)\sqrt{{c}}. \\ $$

Commented by ajfour last updated on 13/Apr/20

$${sir}\:{i}\:{will}\:{need}\:{some}\:{time}\: \\ $$$${understanding}\:{your}\:{method} \\ $$$${thoroughly},\:{this}\:{on}\:{and}\:{circle}\:{one}.. \\ $$

Terms of Service
Privacy Policy
Contact: info@tinkutara.com