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Question Number 195704 by sonukgindia last updated on 08/Aug/23

Answered by mr W last updated on 08/Aug/23

A=(a+b)^2 A_1 =(((a+b)b)/2) A_2 =((b/( (√((a+b)^2 +b^2 )))))^2 A_1 =((b^2 A_1 )/((a+b)^2 +b^2 )) A_3 =A_1 −2A_2 A_(shade) =A−4A_2 −4A_3 A_(shade) =A−4A_2 −4(A_1 −2A_2 ) A_(shade) =A−4A_1 +4A_2 A_(shade) =A−4A_1 (1−(b^2 /((a+b)^2 +b^2 ))) A_(shade) =A−A_1 ((4(a+b)^2 )/((a+b)^2 +b^2 )) A_(shade) =(a+b)^2 −((2(a+b)^3 b)/((a+b)^2 +b^2 )) A_(shade) =((a^2 (a+b)^2 )/((a+b)^2 +b^2 )) (A_(shaded) /A)=(a^2 /((a+b)^2 +b^2 ))=(a^2 /(a^2 +2b^2 +2ab)) ✓

$${A}=\left({a}+{b}\right)^{\mathrm{2}} \\ $$$${A}_{\mathrm{1}} =\frac{\left({a}+{b}\right){b}}{\mathrm{2}} \\ $$$${A}_{\mathrm{2}} =\left(\frac{{b}}{\:\sqrt{\left({a}+{b}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} }}\right)^{\mathrm{2}} {A}_{\mathrm{1}} =\frac{{b}^{\mathrm{2}} {A}_{\mathrm{1}} }{\left({a}+{b}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${A}_{\mathrm{3}} ={A}_{\mathrm{1}} −\mathrm{2}{A}_{\mathrm{2}} \\ $$$${A}_{{shade}} ={A}−\mathrm{4}{A}_{\mathrm{2}} −\mathrm{4}{A}_{\mathrm{3}} \\ $$$${A}_{{shade}} ={A}−\mathrm{4}{A}_{\mathrm{2}} −\mathrm{4}\left({A}_{\mathrm{1}} −\mathrm{2}{A}_{\mathrm{2}} \right) \\ $$$${A}_{{shade}} ={A}−\mathrm{4}{A}_{\mathrm{1}} +\mathrm{4}{A}_{\mathrm{2}} \\ $$$${A}_{{shade}} ={A}−\mathrm{4}{A}_{\mathrm{1}} \left(\mathrm{1}−\frac{{b}^{\mathrm{2}} }{\left({a}+{b}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} }\right) \\ $$$${A}_{{shade}} ={A}−{A}_{\mathrm{1}} \frac{\mathrm{4}\left({a}+{b}\right)^{\mathrm{2}} }{\left({a}+{b}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${A}_{{shade}} =\left({a}+{b}\right)^{\mathrm{2}} −\frac{\mathrm{2}\left({a}+{b}\right)^{\mathrm{3}} {b}}{\left({a}+{b}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${A}_{{shade}} =\frac{{a}^{\mathrm{2}} \left({a}+{b}\right)^{\mathrm{2}} }{\left({a}+{b}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\frac{{A}_{{shaded}} }{{A}}=\frac{{a}^{\mathrm{2}} }{\left({a}+{b}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} }=\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{ab}}\:\checkmark \\ $$

Commented by mr W last updated on 08/Aug/23

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