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Question-206677




Question Number 206677 by mr W last updated on 22/Apr/24
Answered by mr W last updated on 22/Apr/24
Commented by mr W last updated on 23/Apr/24
b=a−c  (√(R^2 −b^2 ))=a+b−(√(R^2 −a^2 ))  ⇒b=(√(R^2 −a^2 ))  ((a+c)/2)=(√(R^2 −a^2 ))  ((a+c)/2)=b=a−c  ⇒a=3c  ((3c+c)/2)=(√(R^2 −(3c)^2 ))  4c^2 =R^2 −9c^2   ⇒c^2 =(R^2 /(13))=13 =area of green square
$${b}={a}−{c} \\ $$$$\sqrt{{R}^{\mathrm{2}} −{b}^{\mathrm{2}} }={a}+{b}−\sqrt{{R}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$$\Rightarrow{b}=\sqrt{{R}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$$\frac{{a}+{c}}{\mathrm{2}}=\sqrt{{R}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$$\frac{{a}+{c}}{\mathrm{2}}={b}={a}−{c} \\ $$$$\Rightarrow{a}=\mathrm{3}{c} \\ $$$$\frac{\mathrm{3}{c}+{c}}{\mathrm{2}}=\sqrt{{R}^{\mathrm{2}} −\left(\mathrm{3}{c}\right)^{\mathrm{2}} } \\ $$$$\mathrm{4}{c}^{\mathrm{2}} ={R}^{\mathrm{2}} −\mathrm{9}{c}^{\mathrm{2}} \\ $$$$\Rightarrow{c}^{\mathrm{2}} =\frac{{R}^{\mathrm{2}} }{\mathrm{13}}=\mathrm{13}\:={area}\:{of}\:{green}\:{square} \\ $$
Answered by A5T last updated on 23/Apr/24
Commented by A5T last updated on 23/Apr/24
(a(√2))^2 +(b(√2))^2 =R^2 +R^2 ⇒a^2 +b^2 =R^2 =169  c=a−b ∧ ((a+c)/2)=(√(R^2 −a^2 ))⇒((2a−b)/2)=b⇒3b=2a  ⇒(2a)^2 +4b^2 =169×4⇒b^2 =13×4⇒a^2 =13×9  ⇒c=3(√(13))−2(√(13))=(√(13))⇒c^2 =13
$$\left({a}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left({b}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} +{R}^{\mathrm{2}} \Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={R}^{\mathrm{2}} =\mathrm{169} \\ $$$${c}={a}−{b}\:\wedge\:\frac{{a}+{c}}{\mathrm{2}}=\sqrt{{R}^{\mathrm{2}} −{a}^{\mathrm{2}} }\Rightarrow\frac{\mathrm{2}{a}−{b}}{\mathrm{2}}={b}\Rightarrow\mathrm{3}{b}=\mathrm{2}{a} \\ $$$$\Rightarrow\left(\mathrm{2}{a}\right)^{\mathrm{2}} +\mathrm{4}{b}^{\mathrm{2}} =\mathrm{169}×\mathrm{4}\Rightarrow{b}^{\mathrm{2}} =\mathrm{13}×\mathrm{4}\Rightarrow{a}^{\mathrm{2}} =\mathrm{13}×\mathrm{9} \\ $$$$\Rightarrow{c}=\mathrm{3}\sqrt{\mathrm{13}}−\mathrm{2}\sqrt{\mathrm{13}}=\sqrt{\mathrm{13}}\Rightarrow{c}^{\mathrm{2}} =\mathrm{13} \\ $$

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