Question-28779

Question Number 28779 by ajfour last updated on 30/Jan/18

Commented by naka3546 last updated on 30/Jan/18

substitute x and y in a and b.

Answered by ajfour last updated on 30/Jan/18

Area = (1/2)(a+b)^2 cos θsin θ −(1/2)(a+b)a =(1/2)(a+b)^2 ×((ab)/(a^2 +b^2 ))−((a(a+b))/2) =((a(a+b))/2)[((b(a+b)−(a^2 +b^2 ))/(a^2 +b^2 ))] Area = ((a^2 (b^2 −a^2 ))/(2(a^2 +b^2 ))) .

$${Area}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{b}\right)^{\mathrm{2}} \mathrm{cos}\:\theta\mathrm{sin}\:\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{b}\right){a} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{b}\right)^{\mathrm{2}} ×\frac{{ab}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }−\frac{{a}\left({a}+{b}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:=\frac{{a}\left({a}+{b}\right)}{\mathrm{2}}\left[\frac{{b}\left({a}+{b}\right)−\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right] \\ $$$$\boldsymbol{{Area}}\:=\:\frac{\boldsymbol{{a}}^{\mathrm{2}} \left(\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} \right)}{\mathrm{2}\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} \right)}\:. \\ $$

Commented by naka3546 last updated on 02/Feb/18

$${thank}\:{you},\:{sir}. \\ $$

Answered by naka3546 last updated on 30/Jan/18

Answered by mrW2 last updated on 30/Jan/18

Commented by mrW2 last updated on 30/Jan/18

EC=(√(a^2 +b^2 )) ((AB)/a)=((a+b)/b)=1+(a/b) ⇒AB=a(1+(a/b)) ((BD)/(AB))=(a/( (√(a^2 +b^2 )))) ⇒BD=(a^2 /( (√(a^2 +b^2 ))))(1+(a/b)) ((AD)/(AB))=(b/( (√(a^2 +b^2 )))) ⇒AD=((ab)/( (√(a^2 +b^2 ))))(1+(a/b)) ((BC)/(a+b))=((√(a^2 +b^2 ))/b) ⇒BC=(1+(a/b))(√(a^2 +b^2 )) DE=(1+(a/b))(√(a^2 +b^2 ))−(√(a^2 +b^2 ))−(a^2 /( (√(a^2 +b^2 ))))(1+(a/b)) =((a(b−a))/( (√(a^2 +b^2 )))) A_(Blue) =(1/2)×((a(b−a))/( (√(a^2 +b^2 ))))×((ab)/( (√(a^2 +b^2 ))))(1+(a/b)) =((a^2 (b^2 −a^2 ))/(2(a^2 +b^2 )))

$${EC}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\frac{{AB}}{{a}}=\frac{{a}+{b}}{{b}}=\mathrm{1}+\frac{{a}}{{b}} \\ $$$$\Rightarrow{AB}={a}\left(\mathrm{1}+\frac{{a}}{{b}}\right) \\ $$$$\frac{{BD}}{{AB}}=\frac{{a}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$$\Rightarrow{BD}=\frac{{a}^{\mathrm{2}} }{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\left(\mathrm{1}+\frac{{a}}{{b}}\right) \\ $$$$\frac{{AD}}{{AB}}=\frac{{b}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$$\Rightarrow{AD}=\frac{{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\left(\mathrm{1}+\frac{{a}}{{b}}\right) \\ $$$$\frac{{BC}}{{a}+{b}}=\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{b}} \\ $$$$\Rightarrow{BC}=\left(\mathrm{1}+\frac{{a}}{{b}}\right)\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${DE}=\left(\mathrm{1}+\frac{{a}}{{b}}\right)\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }−\frac{{a}^{\mathrm{2}} }{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\left(\mathrm{1}+\frac{{a}}{{b}}\right) \\ $$$$=\frac{{a}\left({b}−{a}\right)}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$$ \\ $$$${A}_{{Blue}} =\frac{\mathrm{1}}{\mathrm{2}}×\frac{{a}\left({b}−{a}\right)}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}×\frac{{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\left(\mathrm{1}+\frac{{a}}{{b}}\right) \\ $$$$=\frac{{a}^{\mathrm{2}} \left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)}{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)} \\ $$

Commented by ajfour last updated on 30/Jan/18

$${Thank}\:{you},\:{Sir}. \\ $$

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