Red-Area-

Question Number 198210 by a.lgnaoui last updated on 14/Oct/23

$$\:\boldsymbol{\mathrm{Red}}\:\boldsymbol{\mathrm{Area}}? \\ $$

Commented by a.lgnaoui last updated on 14/Oct/23

Answered by mr W last updated on 14/Oct/23

Commented by mr W last updated on 14/Oct/23

$${c}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${A}=\frac{{ab}}{\mathrm{2}} \\ $$$$\frac{{A}_{\mathrm{1}} }{{A}}=\left(\frac{{c}}{{a}}\right)^{\mathrm{2}} \:\Rightarrow{A}_{\mathrm{1}} =\frac{{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} }{A} \\ $$$$\frac{{x}}{{c}}=\frac{{b}}{{a}}\:\Rightarrow{x}=\frac{{bc}}{{a}} \\ $$$${y}={c}−{x}={c}−\frac{{bc}}{{a}}=\left(\mathrm{1}−\frac{{b}}{{a}}\right){c} \\ $$$$\frac{{A}_{\mathrm{2}} }{{A}}=\left(\frac{{y}}{{a}}\right)^{\mathrm{2}} \:\Rightarrow{A}_{\mathrm{2}} =\left(\mathrm{1}−\frac{{b}}{{a}}\right)^{\mathrm{2}} \frac{{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} }{A} \\ $$$${A}_{\mathrm{1}} +{A}_{\mathrm{2}} =\left[\mathrm{1}+\left(\mathrm{1}−\frac{{b}}{{a}}\right)^{\mathrm{2}} \right]\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }×\frac{{ab}}{\mathrm{2}} \\ $$$${A}_{\mathrm{1}} +{A}_{\mathrm{2}} =\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left(\mathrm{2}{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\right){b}}{\mathrm{2}{a}^{\mathrm{3}} } \\ $$$${A}_{{shaded}} ={c}^{\mathrm{2}} −{A}_{\mathrm{1}} −{A}_{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left(\mathrm{2}{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\right){b}}{\mathrm{2}{a}^{\mathrm{3}} } \\ $$$${A}_{{shaded}} =\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left(\mathrm{2}{a}^{\mathrm{3}} −\mathrm{2}{a}^{\mathrm{2}} {b}+\mathrm{2}{ab}^{\mathrm{2}} −{b}^{\mathrm{3}} \right)}{\mathrm{2}{a}^{\mathrm{3}} }\:\checkmark \\ $$

Commented by a.lgnaoui last updated on 14/Oct/23

$$\mathrm{thanks} \\ $$

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