Question-54334

Question Number 54334 by ajfour last updated on 02/Feb/19

Commented by ajfour last updated on 02/Feb/19

Sir d^2 ≠a^2 +b^2 (coz d isn′t diagonal)

$${Sir}\:{d}^{\mathrm{2}} \neq{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:\:\:\left({coz}\:{d}\:{isn}'{t}\:{diagonal}\right) \\ $$

Answered by mr W last updated on 02/Feb/19

(a−2r)^2 +(b−2r)^2 =d^2 ⇒a^2 +b^2 −4r(a+b)=d^2 −8r^2 (√(a^2 +b^2 ))=(a+b)−2r ⇒a^2 +b^2 =(a+b)^2 −4r(a+b)+4r^2 let X=a+b, Y=a^2 +b^2 ⇒Y−4rX=d^2 −8r^2 ⇒Y=X^2 −4rX+4r^2 ⇒d^2 −8r^2 +4rX=X^2 −4rX+4r^2 ⇒X^2 −8rX+12r^2 −d^2 =0 ⇒X=a+b=4r+(√(4r^2 +d^2 )) ⇒Y=a^2 +b^2 =d^2 +8r^2 +4r(√(4r^2 +d^2 )) a−b=(√(2(a^2 +b^2 )−(a+b)^2 )) a−b=(√(2(d^2 +8r^2 +4r(√(4r^2 +d^2 )))−(4r+(√(4r^2 +d^2 )))^2 )) ⇒a−b=(√(d^2 −4r^2 )) ⇒a=2r+(√(r^2 +((d/2))^2 ))+(√(((d/2))^2 −r^2 )) ⇒b=2r+(√(r^2 +((d/2))^2 ))−(√(((d/2))^2 −r^2 ))

$$\left({a}−\mathrm{2}{r}\right)^{\mathrm{2}} +\left({b}−\mathrm{2}{r}\right)^{\mathrm{2}} ={d}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{4}{r}\left({a}+{b}\right)={d}^{\mathrm{2}} −\mathrm{8}{r}^{\mathrm{2}} \\ $$$$ \\ $$$$\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\left({a}+{b}\right)−\mathrm{2}{r} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{4}{r}\left({a}+{b}\right)+\mathrm{4}{r}^{\mathrm{2}} \\ $$$${let}\:{X}={a}+{b},\:{Y}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$$\Rightarrow{Y}−\mathrm{4}{rX}={d}^{\mathrm{2}} −\mathrm{8}{r}^{\mathrm{2}} \\ $$$$\Rightarrow{Y}={X}^{\mathrm{2}} −\mathrm{4}{rX}+\mathrm{4}{r}^{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow{d}^{\mathrm{2}} −\mathrm{8}{r}^{\mathrm{2}} +\mathrm{4}{rX}={X}^{\mathrm{2}} −\mathrm{4}{rX}+\mathrm{4}{r}^{\mathrm{2}} \\ $$$$\Rightarrow{X}^{\mathrm{2}} −\mathrm{8}{rX}+\mathrm{12}{r}^{\mathrm{2}} −{d}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{X}={a}+{b}=\mathrm{4}{r}+\sqrt{\mathrm{4}{r}^{\mathrm{2}} +{d}^{\mathrm{2}} } \\ $$$$\Rightarrow{Y}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={d}^{\mathrm{2}} +\mathrm{8}{r}^{\mathrm{2}} +\mathrm{4}{r}\sqrt{\mathrm{4}{r}^{\mathrm{2}} +{d}^{\mathrm{2}} } \\ $$$${a}−{b}=\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\left({a}+{b}\right)^{\mathrm{2}} } \\ $$$${a}−{b}=\sqrt{\mathrm{2}\left({d}^{\mathrm{2}} +\mathrm{8}{r}^{\mathrm{2}} +\mathrm{4}{r}\sqrt{\mathrm{4}{r}^{\mathrm{2}} +{d}^{\mathrm{2}} }\right)−\left(\mathrm{4}{r}+\sqrt{\mathrm{4}{r}^{\mathrm{2}} +{d}^{\mathrm{2}} }\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{a}−{b}=\sqrt{{d}^{\mathrm{2}} −\mathrm{4}{r}^{\mathrm{2}} } \\ $$$$\Rightarrow{a}=\mathrm{2}{r}+\sqrt{{r}^{\mathrm{2}} +\left(\frac{{d}}{\mathrm{2}}\right)^{\mathrm{2}} }+\sqrt{\left(\frac{{d}}{\mathrm{2}}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$$\Rightarrow{b}=\mathrm{2}{r}+\sqrt{{r}^{\mathrm{2}} +\left(\frac{{d}}{\mathrm{2}}\right)^{\mathrm{2}} }−\sqrt{\left(\frac{{d}}{\mathrm{2}}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$

Commented by ajfour last updated on 02/Feb/19

$${Thank}\:{you}\:{Sir},\:{i}\:{hope}\:{you}\:{liked}\:{it}. \\ $$$${I}\:{liked}\:{it}\:{myself}. \\ $$

Commented by mr W last updated on 02/Feb/19

$${like}\:{all}\:{others}\:{of}\:{yours},\:{this}\:{is}\:{also}\:{a} \\ $$$${nice}\:{and}\:{interesting}\:{question}\:{sir}! \\ $$

Commented by ajfour last updated on 02/Feb/19

$${Thanks}\:{for}\:{the}\:{appreciation},\:{Sir}, \\ $$$${thats}\:{how}\:{i}\:{learn}\:{from}\:{you}. \\ $$

Commented by Rasheed.Sindhi last updated on 03/Feb/19

Sir Mr W, I didn′t understand the 3rd line: (√(a^2 +b^2 ))=(a+b)−2r Pl explain.

$${Sir}\:{Mr}\:{W},\:{I}\:{didn}'{t}\:{understand}\:{the} \\ $$$$\mathrm{3}{rd}\:{line}:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\left({a}+{b}\right)−\mathrm{2}{r} \\ $$$${Pl}\:{explain}. \\ $$

Commented by mr W last updated on 03/Feb/19