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Question Number 139967 by I want to learn more last updated on 02/May/21

Answered by mr W last updated on 02/May/21

Commented by mr W last updated on 02/May/21

((AF)/(AD))=((AB)/(AH))=((3k)/(3k+(√2)k+(√2)k))=(3/(3+2(√2))) ((ED)/(AD))=((BC)/(AC))=(((√2)k)/(3k+(√2)k))=((√2)/(3+(√2))) (x/y)=((AF)/(ED))=(3/(3+2(√2)))×((3+(√2))/( (√2)))=((15)/( (√2)))−9≈1.6066

$$\frac{{AF}}{{AD}}=\frac{{AB}}{{AH}}=\frac{\mathrm{3}{k}}{\mathrm{3}{k}+\sqrt{\mathrm{2}}{k}+\sqrt{\mathrm{2}}{k}}=\frac{\mathrm{3}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\frac{{ED}}{{AD}}=\frac{{BC}}{{AC}}=\frac{\sqrt{\mathrm{2}}{k}}{\mathrm{3}{k}+\sqrt{\mathrm{2}}{k}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{3}+\sqrt{\mathrm{2}}} \\ $$$$\frac{{x}}{{y}}=\frac{{AF}}{{ED}}=\frac{\mathrm{3}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}×\frac{\mathrm{3}+\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}}=\frac{\mathrm{15}}{\:\sqrt{\mathrm{2}}}−\mathrm{9}\approx\mathrm{1}.\mathrm{6066} \\ $$

Commented by I want to learn more last updated on 02/May/21

I appreciate sir. Thanks.

$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}.\:\mathrm{Thanks}. \\ $$

Commented by Tawa11 last updated on 14/Sep/21

nice

$$\mathrm{nice} \\ $$

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