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Question-186030




Question Number 186030 by Rupesh123 last updated on 31/Jan/23
Answered by ajfour last updated on 31/Jan/23
x^2 +y^2 =d^2   (a−x)y=d^2   (b−y)x=d^2   d((x/y)+1+(y/x))=c  ⇒ ay=bx  ⇒  (a/b)+1+(b/a)=(c/d)
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={d}^{\mathrm{2}} \\ $$$$\left({a}−{x}\right){y}={d}^{\mathrm{2}} \\ $$$$\left({b}−{y}\right){x}={d}^{\mathrm{2}} \\ $$$${d}\left(\frac{{x}}{{y}}+\mathrm{1}+\frac{{y}}{{x}}\right)={c} \\ $$$$\Rightarrow\:{ay}={bx} \\ $$$$\Rightarrow\:\:\frac{{a}}{{b}}+\mathrm{1}+\frac{{b}}{{a}}=\frac{{c}}{{d}} \\ $$
Answered by mr W last updated on 31/Jan/23
generally for any triangle:  ((ch)/2)=Δ=((√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))/4)  h=((√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))/(2c))  ((h−d)/h)=(d/c)  ⇒(1/d)=(1/c)+(1/h)  ⇒(1/d)=(1/c)+((2c)/( (√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))))    in case of right angled triangle:  h=((ab)/c)  ⇒(1/d)=(1/c)+(c/(ab))
$${generally}\:{for}\:{any}\:{triangle}: \\ $$$$\frac{{ch}}{\mathrm{2}}=\Delta=\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{\mathrm{4}} \\ $$$${h}=\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{\mathrm{2}{c}} \\ $$$$\frac{{h}−{d}}{{h}}=\frac{{d}}{{c}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{d}}=\frac{\mathrm{1}}{{c}}+\frac{\mathrm{1}}{{h}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{d}}=\frac{\mathrm{1}}{{c}}+\frac{\mathrm{2}{c}}{\:\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}} \\ $$$$ \\ $$$${in}\:{case}\:{of}\:{right}\:{angled}\:{triangle}: \\ $$$${h}=\frac{{ab}}{{c}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{d}}=\frac{\mathrm{1}}{{c}}+\frac{{c}}{{ab}} \\ $$

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