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Question Number 73919 by necxxx last updated on 16/Nov/19
Commented by necxxx last updated on 16/Nov/19
Good day sirs. This question was formed and solved by some of us here. I really do not remember the question or approaches applied. Please help. Thanks in advance.
$${Good}\:{day}\:{sirs}.\:{This}\:{question}\:{was}\:{formed} \\ $$$${and}\:{solved}\:{by}\:{some}\:{of}\:{us}\:{here}.\:{I}\:{really}\: \\ $$$${do}\:{not}\:{remember}\:{the}\:{question}\:{or} \\ $$$${approaches}\:{applied}.\:{Please}\:{help}. \\ $$$${Thanks}\:{in}\:{advance}. \\ $$
Commented by mr W last updated on 16/Nov/19
creature of ajfour sir
$${creature}\:{of}\:{ajfour}\:{sir} \\ $$
Commented by necxxx last updated on 16/Nov/19
yes sir
$${yes}\:{sir} \\ $$
Answered by mind is power last updated on 16/Nov/19
((ab)/2)=(a/2)+(b/2)+((√(a^2 +b^2 ))/2) ⇒ab=a+b+(√(a^2 +b^2 ))...E ⇒(ab−a−b)^2 =a^2 +b^2 ⇒a^2 b^2 +2ab−2a^2 b−2ab^2 =0 ⇒ab(ab+2−2a−2b)=0 ⇒ab+2−2a−2b=0 ⇒b=((2a−2)/(a−2))...G DT=DP=b−1 by thales⇒((b−1)/b)=((b−1)/( (√(b^2 +a^2 ))−b+1)) ⇒b=(√(b^2 +a^2 ))−b+1⇒2b−1=(√(b^2 +a^2 )) E⇔ab=a+b+2b−1⇒b(a−3)=a−1⇒b=((a−1)/(a−3)) witheG⇒((a−1)/(a−3))=((2a−2)/(a−2))⇒a−2=2a−6⇒a=4 b=((4−1)/(4−3))=3
$$\frac{\mathrm{ab}}{\mathrm{2}}=\frac{\mathrm{a}}{\mathrm{2}}+\frac{\mathrm{b}}{\mathrm{2}}+\frac{\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{ab}=\mathrm{a}+\mathrm{b}+\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }…\mathrm{E} \\ $$$$\Rightarrow\left(\mathrm{ab}−\mathrm{a}−\mathrm{b}\right)^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} +\mathrm{2ab}−\mathrm{2a}^{\mathrm{2}} \mathrm{b}−\mathrm{2ab}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\mathrm{ab}\left(\mathrm{ab}+\mathrm{2}−\mathrm{2a}−\mathrm{2b}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{ab}+\mathrm{2}−\mathrm{2a}−\mathrm{2b}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{b}=\frac{\mathrm{2a}−\mathrm{2}}{\mathrm{a}−\mathrm{2}}…\mathrm{G} \\ $$$$ \\ $$$$\mathrm{DT}=\mathrm{DP}=\mathrm{b}−\mathrm{1} \\ $$$$\mathrm{by}\:\mathrm{thales}\Rightarrow\frac{\mathrm{b}−\mathrm{1}}{\mathrm{b}}=\frac{\mathrm{b}−\mathrm{1}}{\:\sqrt{\mathrm{b}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} }−\mathrm{b}+\mathrm{1}} \\ $$$$\Rightarrow\mathrm{b}=\sqrt{\mathrm{b}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} }−\mathrm{b}+\mathrm{1}\Rightarrow\mathrm{2b}−\mathrm{1}=\sqrt{\mathrm{b}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} } \\ $$$$\mathrm{E}\Leftrightarrow\mathrm{ab}=\mathrm{a}+\mathrm{b}+\mathrm{2b}−\mathrm{1}\Rightarrow\mathrm{b}\left(\mathrm{a}−\mathrm{3}\right)=\mathrm{a}−\mathrm{1}\Rightarrow\mathrm{b}=\frac{\mathrm{a}−\mathrm{1}}{\mathrm{a}−\mathrm{3}} \\ $$$$\mathrm{witheG}\Rightarrow\frac{\mathrm{a}−\mathrm{1}}{\mathrm{a}−\mathrm{3}}=\frac{\mathrm{2a}−\mathrm{2}}{\mathrm{a}−\mathrm{2}}\Rightarrow\mathrm{a}−\mathrm{2}=\mathrm{2a}−\mathrm{6}\Rightarrow\mathrm{a}=\mathrm{4} \\ $$$$\mathrm{b}=\frac{\mathrm{4}−\mathrm{1}}{\mathrm{4}−\mathrm{3}}=\mathrm{3} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by necxxx last updated on 17/Nov/19
Thank you so much but what do G and E represent?
$${Thank}\:{you}\:{so}\:{much}\:{but}\:{what}\:{do}\:{G}\:{and}\:{E} \\ $$$${represent}? \\ $$
Commented by mr W last updated on 17/Nov/19
he means equation G and equation E. other people may use eqn. (i) and (ii) in such cases.
$${he}\:{means}\:{equation}\:{G}\:{and}\:{equation}\:{E}. \\ $$$${other}\:{people}\:{may}\:{use}\:{eqn}.\:\left({i}\right)\:{and}\:\left({ii}\right) \\ $$$${in}\:{such}\:{cases}. \\ $$
Commented by necxxx last updated on 17/Nov/19
ok Thanks
$${ok}\:{Thanks} \\ $$$$ \\ $$

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