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Question-50640




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=(√(h^2 +k^2 ))  A=((hk)/2)+((√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))/4)
$${when}\:{the}\:{angle}\:{between}\:{k}\:{and}\:{h}\:{is} \\ $$$${fixed},\:{then}\:{the}\:{shape}\:{and}\:{therefore} \\ $$$${also}\:{the}\:{area}\:{of}\:{the}\:{quadrilateral}\:{is} \\ $$$${fixed}.\:{the}\:{area}\:{is}\:{a}\:{constant}. \\ $$$${c}=\sqrt{{h}^{\mathrm{2}} +{k}^{\mathrm{2}} } \\ $$$${A}=\frac{{hk}}{\mathrm{2}}+\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{\mathrm{4}} \\ $$
Commented by ajfour last updated on 18/Dec/18
yes mrW Sir, sorry for the  unnecesary, thanks yet.
$${yes}\:{mrW}\:{Sir},\:{sorry}\:{for}\:{the} \\ $$$${unnecesary},\:{thanks}\:{yet}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 18/Dec/18
sinθ=(l/a)    area shaded =(1/2)(k+l)×(h+acosθ)−(1/2)(acosθ)×(asinθ)  =(1/2)(k+asinθ)(h+acosθ)−(a^2 /4)sin2θ  =(1/2)(kh+akcosθ+ahsinθ+(a^2 /2)sin2θ)−(a^2 /4)sin2θ  =((kh)/2)+(a/2)×(√(k^2 +h^2 )) {(k/( (√(k^2 +h^2 ))))cosθ+(h/( (√(k^2 +h^2 ))))sinθ}  =((kh)/2)+((a(√(k^2 +h^2 )))/2)sin(θ+α)  so max area  =((kh)/2)+((a(√(k^2 +h^2 )))/2)  pls check...
$${sin}\theta=\frac{{l}}{{a}}\:\: \\ $$$${area}\:{shaded}\:=\frac{\mathrm{1}}{\mathrm{2}}\left({k}+{l}\right)×\left({h}+{acos}\theta\right)−\frac{\mathrm{1}}{\mathrm{2}}\left({acos}\theta\right)×\left({asin}\theta\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({k}+{asin}\theta\right)\left({h}+{acos}\theta\right)−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}{sin}\mathrm{2}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({kh}+{akcos}\theta+{ahsin}\theta+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{sin}\mathrm{2}\theta\right)−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}{sin}\mathrm{2}\theta \\ $$$$=\frac{{kh}}{\mathrm{2}}+\frac{{a}}{\mathrm{2}}×\sqrt{{k}^{\mathrm{2}} +{h}^{\mathrm{2}} }\:\left\{\frac{{k}}{\:\sqrt{{k}^{\mathrm{2}} +{h}^{\mathrm{2}} }}{cos}\theta+\frac{{h}}{\:\sqrt{{k}^{\mathrm{2}} +{h}^{\mathrm{2}} }}{sin}\theta\right\} \\ $$$$=\frac{{kh}}{\mathrm{2}}+\frac{{a}\sqrt{{k}^{\mathrm{2}} +{h}^{\mathrm{2}} }}{\mathrm{2}}{sin}\left(\theta+\alpha\right) \\ $$$${so}\:{max}\:{area} \\ $$$$=\frac{{kh}}{\mathrm{2}}+\frac{{a}\sqrt{{k}^{\mathrm{2}} +{h}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${pls}\:{check}… \\ $$$$ \\ $$

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