Question Number 217495 by Hanuda354 last updated on 14/Mar/25
Answered by mr W last updated on 17/Mar/25
Commented by Hanuda354 last updated on 18/Mar/25
$$\mathrm{Thanks}\:\:\mathrm{a}\:\:\mathrm{lot},\:\mathrm{Prof}. \\ $$
Commented by mr W last updated on 17/Mar/25
$${eqn}.\:{of}\:{PR}: \\ $$$${y}=\frac{\mathrm{7}{x}}{\mathrm{2}\left({a}−\mathrm{3}\right)} \\ $$$${eqn}.\:{of}\:{SW}: \\ $$$${y}={a}−\frac{{x}}{\mathrm{2}} \\ $$$${y}_{{Q}} ={a}−\frac{{x}_{{Q}} }{\mathrm{2}}=\frac{\mathrm{7}{x}_{{Q}} }{\mathrm{2}\left({a}−\mathrm{3}\right)} \\ $$$$\Rightarrow{x}_{{Q}} =\frac{\mathrm{2}{a}\left({a}−\mathrm{3}\right)}{\left({a}+\mathrm{4}\right)} \\ $$$${A}=\frac{{a}}{\mathrm{2}}×\frac{\mathrm{2}{a}\left({a}−\mathrm{3}\right)}{\left({a}+\mathrm{4}\right)}=\frac{{a}^{\mathrm{2}} \left({a}−\mathrm{3}\right)}{{a}+\mathrm{4}} \\ $$$${A}+{B}=\frac{\left(\mathrm{2}{a}−\mathrm{6}\right)\left({a}+{a}−\mathrm{7}\right)}{\mathrm{2}}=\left({a}−\mathrm{3}\right)\left(\mathrm{2}{a}−\mathrm{7}\right) \\ $$$$\frac{{A}+{B}}{{A}}=\mathrm{1}+\frac{{B}}{{A}}=\mathrm{1}+\frac{\mathrm{41}}{\mathrm{50}}=\frac{\mathrm{91}}{\mathrm{50}} \\ $$$$\frac{\left({a}−\mathrm{3}\right)\left(\mathrm{2}{a}−\mathrm{7}\right)}{\frac{{a}^{\mathrm{2}} \left({a}−\mathrm{3}\right)}{{a}+\mathrm{4}}}=\frac{\mathrm{91}}{\mathrm{50}} \\ $$$$\mathrm{9}{a}^{\mathrm{2}} +\mathrm{50}{a}−\mathrm{1400}=\mathrm{0} \\ $$$${a}=\frac{−\mathrm{50}+\sqrt{\mathrm{50}^{\mathrm{2}} +\mathrm{4}×\mathrm{9}×\mathrm{1400}}}{\mathrm{18}}=\mathrm{10} \\ $$$${SW}=\sqrt{{a}^{\mathrm{2}} +\left(\mathrm{2}{a}\right)^{\mathrm{2}} }=\mathrm{10}\sqrt{\mathrm{5}} \\ $$$$\frac{{SQ}}{{SW}}=\frac{{x}_{{Q}} }{\mathrm{2}{a}}=\frac{{a}−\mathrm{3}}{{a}+\mathrm{4}}=\frac{\mathrm{10}−\mathrm{3}}{\mathrm{10}+\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${QW}={SQ}=\frac{\mathrm{10}\sqrt{\mathrm{5}}}{\mathrm{2}}=\mathrm{5}\sqrt{\mathrm{5}} \\ $$$$\frac{{PQ}}{{PR}}=\frac{{x}_{{Q}} }{\mathrm{2}{a}−\mathrm{6}}=\frac{{a}}{{a}+\mathrm{4}}=\frac{\mathrm{10}}{\mathrm{14}}=\frac{\mathrm{5}}{\mathrm{7}} \\ $$$$\Rightarrow\frac{{QR}}{{PR}}=\frac{\mathrm{2}}{\mathrm{7}} \\ $$$$\frac{{a}}{{a}}×\frac{{x}}{{y}}×\frac{{QR}}{{RP}}=\mathrm{1} \\ $$$$\frac{{x}}{{y}}×\frac{\mathrm{2}}{\mathrm{7}}=\mathrm{1} \\ $$$$\Rightarrow\frac{{x}}{{y}}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{x}}{{x}+{y}}=\frac{\mathrm{7}}{\mathrm{9}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{7}×{QW}}{\mathrm{9}}=\frac{\mathrm{7}×\mathrm{5}\sqrt{\mathrm{5}}}{\mathrm{9}}=\frac{\mathrm{35}\sqrt{\mathrm{5}}}{\mathrm{9}}\:\checkmark \\ $$