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Question Number 189743 by normans last updated on 21/Mar/23

Answered by mr W last updated on 21/Mar/23

Commented by mr W last updated on 21/Mar/23

(2h)^2 =4a×a ⇒h=a h_1 =((5a)/2) (h_2 /(2h))=((5a−h_2 )/(4a)) ⇒h_2 =((5a)/3) A_3 =((5ah_2 )/2)=((5a)/2)×((5a)/3)=((25a^2 )/6) A_1 +A_3 =((5ah_1 )/2)=((5a×5a)/(2×2))=((25a^2 )/4) ⇒A_1 =((25a^2 )/4)−((25a^2 )/6)=((25a^2 )/(12)) A_2 +A_3 =((5a(2h))/2)=5a^2 ⇒A_2 =5a^2 −((25a^2 )/6)=((5a^2 )/6) ⇒(A_1 /A_2 )=((25)/(12))×(6/5)=(5/2) ✓

$$\left(\mathrm{2}{h}\right)^{\mathrm{2}} =\mathrm{4}{a}×{a} \\ $$$$\Rightarrow{h}={a} \\ $$$${h}_{\mathrm{1}} =\frac{\mathrm{5}{a}}{\mathrm{2}} \\ $$$$\frac{{h}_{\mathrm{2}} }{\mathrm{2}{h}}=\frac{\mathrm{5}{a}−{h}_{\mathrm{2}} }{\mathrm{4}{a}} \\ $$$$\Rightarrow{h}_{\mathrm{2}} =\frac{\mathrm{5}{a}}{\mathrm{3}} \\ $$$${A}_{\mathrm{3}} =\frac{\mathrm{5}{ah}_{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{5}{a}}{\mathrm{2}}×\frac{\mathrm{5}{a}}{\mathrm{3}}=\frac{\mathrm{25}{a}^{\mathrm{2}} }{\mathrm{6}} \\ $$$${A}_{\mathrm{1}} +{A}_{\mathrm{3}} =\frac{\mathrm{5}{ah}_{\mathrm{1}} }{\mathrm{2}}=\frac{\mathrm{5}{a}×\mathrm{5}{a}}{\mathrm{2}×\mathrm{2}}=\frac{\mathrm{25}{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow{A}_{\mathrm{1}} =\frac{\mathrm{25}{a}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{25}{a}^{\mathrm{2}} }{\mathrm{6}}=\frac{\mathrm{25}{a}^{\mathrm{2}} }{\mathrm{12}} \\ $$$${A}_{\mathrm{2}} +{A}_{\mathrm{3}} =\frac{\mathrm{5}{a}\left(\mathrm{2}{h}\right)}{\mathrm{2}}=\mathrm{5}{a}^{\mathrm{2}} \\ $$$$\Rightarrow{A}_{\mathrm{2}} =\mathrm{5}{a}^{\mathrm{2}} −\frac{\mathrm{25}{a}^{\mathrm{2}} }{\mathrm{6}}=\frac{\mathrm{5}{a}^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\Rightarrow\frac{{A}_{\mathrm{1}} }{{A}_{\mathrm{2}} }=\frac{\mathrm{25}}{\mathrm{12}}×\frac{\mathrm{6}}{\mathrm{5}}=\frac{\mathrm{5}}{\mathrm{2}}\:\checkmark \\ $$

Answered by HeferH last updated on 21/Mar/23

(2x)^2 = (2a)^2 a=x ((Green)/(Blue)) = (((((5a)/2)(√2))/(a(√5))))^2 = ((5/2)(√2)∙(1/( (√5))))^2 = ((25∙2)/(4∙5)) =(5/2)

$$\left(\mathrm{2x}\right)^{\mathrm{2}} =\:\left(\mathrm{2a}\right)^{\mathrm{2}} \\ $$$$\:\mathrm{a}=\mathrm{x} \\ $$$$\:\frac{\mathrm{Green}}{\mathrm{Blue}}\:=\:\left(\frac{\frac{\mathrm{5a}}{\mathrm{2}}\sqrt{\mathrm{2}}}{\mathrm{a}\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} \:=\:\left(\frac{\mathrm{5}}{\mathrm{2}}\sqrt{\mathrm{2}}\centerdot\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{25}\centerdot\mathrm{2}}{\mathrm{4}\centerdot\mathrm{5}}\:=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\: \\ $$

Commented by HeferH last updated on 21/Mar/23

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