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Question Number 174664 by dragan91 last updated on 07/Aug/22

Answered by mr W last updated on 07/Aug/22

((b−5)/5)=(5/(a−5))  ⇒ab=5(a+b)  let u=a+b, v=ab  ⇒v=5u    ((a(1−((√(24))/( (√(a^2 +b^2 ))))))/( (√(24))))=((√(24))/((b(√(24)))/( (√(a^2 +b^2 )))))  ab(((√((a+b)^2 −2ab))/( (√(24))))−1)=(a+b)^2 −2ab  u^2 −10u−600=0  (u+20)(u−30)=0  ⇒u=30, v=150  A_(yellow) =ab−24−25=150−24−25=101 ✓

$$\frac{{b}−\mathrm{5}}{\mathrm{5}}=\frac{\mathrm{5}}{{a}−\mathrm{5}} \\ $$$$\Rightarrow{ab}=\mathrm{5}\left({a}+{b}\right) \\ $$$${let}\:{u}={a}+{b},\:{v}={ab} \\ $$$$\Rightarrow{v}=\mathrm{5}{u} \\ $$$$ \\ $$$$\frac{{a}\left(\mathrm{1}−\frac{\sqrt{\mathrm{24}}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\right)}{\:\sqrt{\mathrm{24}}}=\frac{\sqrt{\mathrm{24}}}{\frac{{b}\sqrt{\mathrm{24}}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}} \\ $$$${ab}\left(\frac{\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}{ab}}}{\:\sqrt{\mathrm{24}}}−\mathrm{1}\right)=\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}{ab} \\ $$$${u}^{\mathrm{2}} −\mathrm{10}{u}−\mathrm{600}=\mathrm{0} \\ $$$$\left({u}+\mathrm{20}\right)\left({u}−\mathrm{30}\right)=\mathrm{0} \\ $$$$\Rightarrow{u}=\mathrm{30},\:{v}=\mathrm{150} \\ $$$${A}_{{yellow}} ={ab}−\mathrm{24}−\mathrm{25}=\mathrm{150}−\mathrm{24}−\mathrm{25}=\mathrm{101}\:\checkmark \\ $$

Commented by Tawa11 last updated on 08/Aug/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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