Question-225666

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Question Number 225666 by mr W last updated on 05/Nov/25

Commented by mr W last updated on 05/Nov/25

$${an}\:{alternative}\:{solution} \\ $$

Commented by ajfour last updated on 06/Nov/25

$${Thank}\:{you}\:{sir}.\:{Good}\:{way}. \\ $$

Answered by mr W last updated on 05/Nov/25

Commented by mr W last updated on 05/Nov/25

((AB)/(AD))=((CE)/(CB))=(3/4) ((AB)/p)=(p/(CB))=(3/4) ⇒AB=((3p)/4), CB=((4p)/3) BD=(√(p^2 +(((3p)/4))^2 ))=((5p)/4) AB×AD=(AB+AD+BD)×3 ((3p)/4)×p=(((3p)/4)+p+((5p)/4))×3 ⇒p=12 q=((3p)/4)+((4p)/3)=((25p)/(12))=25

$$\frac{{AB}}{{AD}}=\frac{{CE}}{{CB}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\frac{{AB}}{{p}}=\frac{{p}}{{CB}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow{AB}=\frac{\mathrm{3}{p}}{\mathrm{4}},\:{CB}=\frac{\mathrm{4}{p}}{\mathrm{3}} \\ $$$${BD}=\sqrt{{p}^{\mathrm{2}} +\left(\frac{\mathrm{3}{p}}{\mathrm{4}}\right)^{\mathrm{2}} }=\frac{\mathrm{5}{p}}{\mathrm{4}} \\ $$$${AB}×{AD}=\left({AB}+{AD}+{BD}\right)×\mathrm{3} \\ $$$$\frac{\mathrm{3}{p}}{\mathrm{4}}×{p}=\left(\frac{\mathrm{3}{p}}{\mathrm{4}}+{p}+\frac{\mathrm{5}{p}}{\mathrm{4}}\right)×\mathrm{3} \\ $$$$\Rightarrow{p}=\mathrm{12} \\ $$$${q}=\frac{\mathrm{3}{p}}{\mathrm{4}}+\frac{\mathrm{4}{p}}{\mathrm{3}}=\frac{\mathrm{25}{p}}{\mathrm{12}}=\mathrm{25} \\ $$

Commented by fantastic last updated on 05/Nov/25