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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

Thanks a lot, Prof.

$$\mathrm{Thanks}\:\:\mathrm{a}\:\:\mathrm{lot},\:\mathrm{Prof}. \\ $$

Commented by mr W last updated on 17/Mar/25

eqn. of PR: y=((7x)/(2(a−3))) eqn. of SW: y=a−(x/2) y_Q =a−(x_Q /2)=((7x_Q )/(2(a−3))) ⇒x_Q =((2a(a−3))/((a+4))) A=(a/2)×((2a(a−3))/((a+4)))=((a^2 (a−3))/(a+4)) A+B=(((2a−6)(a+a−7))/2)=(a−3)(2a−7) ((A+B)/A)=1+(B/A)=1+((41)/(50))=((91)/(50)) (((a−3)(2a−7))/((a^2 (a−3))/(a+4)))=((91)/(50)) 9a^2 +50a−1400=0 a=((−50+(√(50^2 +4×9×1400)))/(18))=10 SW=(√(a^2 +(2a)^2 ))=10(√5) ((SQ)/(SW))=(x_Q /(2a))=((a−3)/(a+4))=((10−3)/(10+4))=(1/2) QW=SQ=((10(√5))/2)=5(√5) ((PQ)/(PR))=(x_Q /(2a−6))=(a/(a+4))=((10)/(14))=(5/7) ⇒((QR)/(PR))=(2/7) (a/a)×(x/y)×((QR)/(RP))=1 (x/y)×(2/7)=1 ⇒(x/y)=(7/2) ⇒(x/(x+y))=(7/9) ⇒x=((7×QW)/9)=((7×5(√5))/9)=((35(√5))/9) ✓

$${eqn}.\:{of}\:{PR}: \\ $$$${y}=\frac{\mathrm{7}{x}}{\mathrm{2}\left({a}−\mathrm{3}\right)} \\ $$$${eqn}.\:{of}\:{SW}: \\ $$$${y}={a}−\frac{{x}}{\mathrm{2}} \\ $$$${y}_{{Q}} ={a}−\frac{{x}_{{Q}} }{\mathrm{2}}=\frac{\mathrm{7}{x}_{{Q}} }{\mathrm{2}\left({a}−\mathrm{3}\right)} \\ $$$$\Rightarrow{x}_{{Q}} =\frac{\mathrm{2}{a}\left({a}−\mathrm{3}\right)}{\left({a}+\mathrm{4}\right)} \\ $$$${A}=\frac{{a}}{\mathrm{2}}×\frac{\mathrm{2}{a}\left({a}−\mathrm{3}\right)}{\left({a}+\mathrm{4}\right)}=\frac{{a}^{\mathrm{2}} \left({a}−\mathrm{3}\right)}{{a}+\mathrm{4}} \\ $$$${A}+{B}=\frac{\left(\mathrm{2}{a}−\mathrm{6}\right)\left({a}+{a}−\mathrm{7}\right)}{\mathrm{2}}=\left({a}−\mathrm{3}\right)\left(\mathrm{2}{a}−\mathrm{7}\right) \\ $$$$\frac{{A}+{B}}{{A}}=\mathrm{1}+\frac{{B}}{{A}}=\mathrm{1}+\frac{\mathrm{41}}{\mathrm{50}}=\frac{\mathrm{91}}{\mathrm{50}} \\ $$$$\frac{\left({a}−\mathrm{3}\right)\left(\mathrm{2}{a}−\mathrm{7}\right)}{\frac{{a}^{\mathrm{2}} \left({a}−\mathrm{3}\right)}{{a}+\mathrm{4}}}=\frac{\mathrm{91}}{\mathrm{50}} \\ $$$$\mathrm{9}{a}^{\mathrm{2}} +\mathrm{50}{a}−\mathrm{1400}=\mathrm{0} \\ $$$${a}=\frac{−\mathrm{50}+\sqrt{\mathrm{50}^{\mathrm{2}} +\mathrm{4}×\mathrm{9}×\mathrm{1400}}}{\mathrm{18}}=\mathrm{10} \\ $$$${SW}=\sqrt{{a}^{\mathrm{2}} +\left(\mathrm{2}{a}\right)^{\mathrm{2}} }=\mathrm{10}\sqrt{\mathrm{5}} \\ $$$$\frac{{SQ}}{{SW}}=\frac{{x}_{{Q}} }{\mathrm{2}{a}}=\frac{{a}−\mathrm{3}}{{a}+\mathrm{4}}=\frac{\mathrm{10}−\mathrm{3}}{\mathrm{10}+\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${QW}={SQ}=\frac{\mathrm{10}\sqrt{\mathrm{5}}}{\mathrm{2}}=\mathrm{5}\sqrt{\mathrm{5}} \\ $$$$\frac{{PQ}}{{PR}}=\frac{{x}_{{Q}} }{\mathrm{2}{a}−\mathrm{6}}=\frac{{a}}{{a}+\mathrm{4}}=\frac{\mathrm{10}}{\mathrm{14}}=\frac{\mathrm{5}}{\mathrm{7}} \\ $$$$\Rightarrow\frac{{QR}}{{PR}}=\frac{\mathrm{2}}{\mathrm{7}} \\ $$$$\frac{{a}}{{a}}×\frac{{x}}{{y}}×\frac{{QR}}{{RP}}=\mathrm{1} \\ $$$$\frac{{x}}{{y}}×\frac{\mathrm{2}}{\mathrm{7}}=\mathrm{1} \\ $$$$\Rightarrow\frac{{x}}{{y}}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{x}}{{x}+{y}}=\frac{\mathrm{7}}{\mathrm{9}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{7}×{QW}}{\mathrm{9}}=\frac{\mathrm{7}×\mathrm{5}\sqrt{\mathrm{5}}}{\mathrm{9}}=\frac{\mathrm{35}\sqrt{\mathrm{5}}}{\mathrm{9}}\:\checkmark \\ $$

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