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Question-61825




Question Number 61825 by Tawa1 last updated on 09/Jun/19
Answered by mr W last updated on 10/Jun/19
Commented by mr W last updated on 10/Jun/19
(q/4)=(p/7)  ⇒q=((4p)/7)  ((4+p)/a)=((7+q)/9)  ⇒9(4+p)=a(7+q)  ⇒36+9p=7a+((4ap)/7)  ⇒p=((49a−252)/(63−4a))  ((q×a)/4)+(√(7^2 −p^2 ))=9  ⇒((ap)/7)+(√(7^2 −p^2 ))=9  ⇒((a(49a−252))/(7(63−4a)))+(√(49−(((49a−252)/(63−4a)))^2 ))=9  ⇒a≈6.9282  A_(green) =((7×a)/2)≈24.25
$$\frac{{q}}{\mathrm{4}}=\frac{{p}}{\mathrm{7}} \\ $$$$\Rightarrow{q}=\frac{\mathrm{4}{p}}{\mathrm{7}} \\ $$$$\frac{\mathrm{4}+{p}}{{a}}=\frac{\mathrm{7}+{q}}{\mathrm{9}} \\ $$$$\Rightarrow\mathrm{9}\left(\mathrm{4}+{p}\right)={a}\left(\mathrm{7}+{q}\right) \\ $$$$\Rightarrow\mathrm{36}+\mathrm{9}{p}=\mathrm{7}{a}+\frac{\mathrm{4}{ap}}{\mathrm{7}} \\ $$$$\Rightarrow{p}=\frac{\mathrm{49}{a}−\mathrm{252}}{\mathrm{63}−\mathrm{4}{a}} \\ $$$$\frac{{q}×{a}}{\mathrm{4}}+\sqrt{\mathrm{7}^{\mathrm{2}} −{p}^{\mathrm{2}} }=\mathrm{9} \\ $$$$\Rightarrow\frac{{ap}}{\mathrm{7}}+\sqrt{\mathrm{7}^{\mathrm{2}} −{p}^{\mathrm{2}} }=\mathrm{9} \\ $$$$\Rightarrow\frac{{a}\left(\mathrm{49}{a}−\mathrm{252}\right)}{\mathrm{7}\left(\mathrm{63}−\mathrm{4}{a}\right)}+\sqrt{\mathrm{49}−\left(\frac{\mathrm{49}{a}−\mathrm{252}}{\mathrm{63}−\mathrm{4}{a}}\right)^{\mathrm{2}} }=\mathrm{9} \\ $$$$\Rightarrow{a}\approx\mathrm{6}.\mathrm{9282} \\ $$$${A}_{{green}} =\frac{\mathrm{7}×{a}}{\mathrm{2}}\approx\mathrm{24}.\mathrm{25} \\ $$
Commented by Tawa1 last updated on 10/Jun/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by MJS last updated on 10/Jun/19
the red eq can be solved exactly  a=±4(√3)∨a=±9  4(√3)≈6.9282
$$\mathrm{the}\:\mathrm{red}\:\mathrm{eq}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{exactly} \\ $$$${a}=\pm\mathrm{4}\sqrt{\mathrm{3}}\vee{a}=\pm\mathrm{9} \\ $$$$\mathrm{4}\sqrt{\mathrm{3}}\approx\mathrm{6}.\mathrm{9282} \\ $$
Commented by mr W last updated on 10/Jun/19
thank you sir!  i didn′t go further, since it led to a  quadratic equation.
$${thank}\:{you}\:{sir}! \\ $$$${i}\:{didn}'{t}\:{go}\:{further},\:{since}\:{it}\:{led}\:{to}\:{a} \\ $$$${quadratic}\:{equation}. \\ $$

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