Question Number 50640 by ajfour last updated on 18/Dec/18
Commented by mr W last updated on 18/Dec/18
$${when}\:{the}\:{angle}\:{between}\:{k}\:{and}\:{h}\:{is} \\ $$$${fixed},\:{then}\:{the}\:{shape}\:{and}\:{therefore} \\ $$$${also}\:{the}\:{area}\:{of}\:{the}\:{quadrilateral}\:{is} \\ $$$${fixed}.\:{the}\:{area}\:{is}\:{a}\:{constant}. \\ $$$${c}=\sqrt{{h}^{\mathrm{2}} +{k}^{\mathrm{2}} } \\ $$$${A}=\frac{{hk}}{\mathrm{2}}+\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{\mathrm{4}} \\ $$
Commented by ajfour last updated on 18/Dec/18
$${yes}\:{mrW}\:{Sir},\:{sorry}\:{for}\:{the} \\ $$$${unnecesary},\:{thanks}\:{yet}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 18/Dec/18
$${sin}\theta=\frac{{l}}{{a}}\:\: \\ $$$${area}\:{shaded}\:=\frac{\mathrm{1}}{\mathrm{2}}\left({k}+{l}\right)×\left({h}+{acos}\theta\right)−\frac{\mathrm{1}}{\mathrm{2}}\left({acos}\theta\right)×\left({asin}\theta\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({k}+{asin}\theta\right)\left({h}+{acos}\theta\right)−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}{sin}\mathrm{2}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({kh}+{akcos}\theta+{ahsin}\theta+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{sin}\mathrm{2}\theta\right)−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}{sin}\mathrm{2}\theta \\ $$$$=\frac{{kh}}{\mathrm{2}}+\frac{{a}}{\mathrm{2}}×\sqrt{{k}^{\mathrm{2}} +{h}^{\mathrm{2}} }\:\left\{\frac{{k}}{\:\sqrt{{k}^{\mathrm{2}} +{h}^{\mathrm{2}} }}{cos}\theta+\frac{{h}}{\:\sqrt{{k}^{\mathrm{2}} +{h}^{\mathrm{2}} }}{sin}\theta\right\} \\ $$$$=\frac{{kh}}{\mathrm{2}}+\frac{{a}\sqrt{{k}^{\mathrm{2}} +{h}^{\mathrm{2}} }}{\mathrm{2}}{sin}\left(\theta+\alpha\right) \\ $$$${so}\:{max}\:{area} \\ $$$$=\frac{{kh}}{\mathrm{2}}+\frac{{a}\sqrt{{k}^{\mathrm{2}} +{h}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${pls}\:{check}… \\ $$$$ \\ $$