Question Number 47407 by ajfour last updated on 09/Nov/18 | ||
Answered by ajfour last updated on 10/Nov/18 | ||
$${let}\:{BD}\:=\:{x}\:,\:\angle{A}\:=\:\phi,\:\angle{C}\:=\:\theta \\ $$$${S}_{{ABCD}} \:=\:\frac{{ab}\mathrm{sin}\:\theta}{\mathrm{2}}+\frac{{cd}\mathrm{sin}\:\phi}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} =\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\mathrm{cos}\:\theta \\ $$$$\:\:\:\:=\:{c}^{\mathrm{2}} +{d}^{\mathrm{2}} −\mathrm{2}{cd}\mathrm{cos}\:\phi \\ $$$$\Rightarrow\:\left({ab}\mathrm{sin}\:\theta\right)\frac{{d}\theta}{{dx}}\:=\:\left({cd}\mathrm{sin}\:\phi\right)\frac{{d}\phi}{{dx}}\:=\:\mathrm{2}{x} \\ $$$$\frac{{dS}}{{dx}}\:=\:\mathrm{0}\:\Rightarrow \\ $$$$\:\:\:\left({ab}\mathrm{cos}\:\theta\right)\frac{{d}\theta}{{dx}}+\left({cd}\mathrm{cos}\:\phi\right)\frac{{d}\phi}{{dx}}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{cos}\:\theta+\left(\mathrm{cos}\:\phi\right)\left(\frac{\mathrm{sin}\:\theta}{\mathrm{sin}\:\phi}\right)=\:\mathrm{0} \\ $$$$\Rightarrow\:\:\theta+\phi\:=\:\pi \\ $$$$\Rightarrow\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\mathrm{cos}\:\theta\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +\mathrm{2}{cd}\mathrm{cos}\:\theta \\ $$$${or}\:\:\:\:\mathrm{cos}\:\theta\:=\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} −{d}^{\mathrm{2}} }{\mathrm{2}\left({ab}+{cd}\right)} \\ $$$$\left({S}_{{ABCD}} \right)_{{max}} \:=\:\frac{\mathrm{sin}\:\theta}{\mathrm{2}}\left({ab}+{cd}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\:\frac{\left({ab}+{cd}\right)}{\mathrm{2}}\sqrt{\mathrm{1}−\left[\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} −{d}^{\mathrm{2}} }{\mathrm{2}\left({ab}+{cd}\right)}\right]^{\mathrm{2}} }\: \\ $$$$\:\:{S}_{{max}} =\:\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\mathrm{4}\left({ab}+{cd}\right)^{\mathrm{2}} −\left[\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\left({c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)\right]^{\mathrm{2}} }\:. \\ $$ | ||
Commented by mr W last updated on 10/Nov/18 | ||
$${thank}\:{you}\:{sir}! \\ $$$${it}\:{shows}\:{the}\:{area}\:{reaches}\:{maximum} \\ $$$${when}\:{the}\:{quadrilateral}\:{is}\:{inscribed}\:{in} \\ $$$${a}\:{circle}. \\ $$ | ||