Question-212721

Question Number 212721 by Spillover last updated on 22/Oct/24

Answered by A5T last updated on 22/Oct/24

Commented by A5T last updated on 22/Oct/24

In[ABC]; tanθ=(R/(R+2r));AC=(√(2R^2 +4r^2 +4Rr)) 2r(R+2r)=MC×(√(2R^2 +4r^2 +4Rr)) In[MNC];tanθ=((MN)/(MC))⇒MN=((2Rr)/( (√(2R^2 +4r^2 +4Rr)))) 2r(2r+2R)=LC×(√(2R^2 +4r^2 +4Rr)) ⇒LM=LC−MC=((2r(2r+2R−R−2r))/( (√(2R^2 +4r^2 +4Rr))))=MN ⇒[LMN]=((LM×MN)/2)=((MN^2 )/2)=((R^2 r^2 )/(R^2 +2r^2 +2Rr))

$${In}\left[{ABC}\right];\:{tan}\theta=\frac{{R}}{{R}+\mathrm{2}{r}};{AC}=\sqrt{\mathrm{2}{R}^{\mathrm{2}} +\mathrm{4}{r}^{\mathrm{2}} +\mathrm{4}{Rr}} \\ $$$$\mathrm{2}{r}\left({R}+\mathrm{2}{r}\right)={MC}×\sqrt{\mathrm{2}{R}^{\mathrm{2}} +\mathrm{4}{r}^{\mathrm{2}} +\mathrm{4}{Rr}} \\ $$$${In}\left[{MNC}\right];{tan}\theta=\frac{{MN}}{{MC}}\Rightarrow{MN}=\frac{\mathrm{2}{Rr}}{\:\sqrt{\mathrm{2}{R}^{\mathrm{2}} +\mathrm{4}{r}^{\mathrm{2}} +\mathrm{4}{Rr}}} \\ $$$$\mathrm{2}{r}\left(\mathrm{2}{r}+\mathrm{2}{R}\right)={LC}×\sqrt{\mathrm{2}{R}^{\mathrm{2}} +\mathrm{4}{r}^{\mathrm{2}} +\mathrm{4}{Rr}} \\ $$$$\Rightarrow{LM}={LC}−{MC}=\frac{\mathrm{2}{r}\left(\mathrm{2}{r}+\mathrm{2}{R}−{R}−\mathrm{2}{r}\right)}{\:\sqrt{\mathrm{2}{R}^{\mathrm{2}} +\mathrm{4}{r}^{\mathrm{2}} +\mathrm{4}{Rr}}}={MN} \\ $$$$\Rightarrow\left[{LMN}\right]=\frac{{LM}×{MN}}{\mathrm{2}}=\frac{{MN}^{\mathrm{2}} }{\mathrm{2}}=\frac{{R}^{\mathrm{2}} {r}^{\mathrm{2}} }{{R}^{\mathrm{2}} +\mathrm{2}{r}^{\mathrm{2}} +\mathrm{2}{Rr}} \\ $$

Commented by Spillover last updated on 23/Oct/24

$${great}\:{work}.{thanks} \\ $$

Answered by mr W last updated on 22/Oct/24

Commented by mr W last updated on 22/Oct/24

AB=(√(R^2 +(R+2r)^2 )) ((AL)/(2R))=(R/(AB)) ⇒AL=((2R^2 )/(AB)) ((MB)/(2r))=((R+2r)/(AB)) ⇒MB=((2r(R+2r))/(AB)) ((NM)/(2r))=(R/(AB)) ⇒NM=((2Rr)/(AB)) ΔLMN=(((AB−AL−MB)×NM)/2) =(1/2)(AB−((2R^2 +2r(R+2r))/(AB)))×((2Rr)/(AB)) =((Rr(R^2 +(R+2r)^2 −2R^2 −2r(R+2r)))/(R^2 +(R+2r)^2 )) =((2R^2 r^2 )/(R^2 +(R+2r)^2 ))=((R^2 r^2 )/(R^2 +2Rr+2r^2 ))

$${AB}=\sqrt{{R}^{\mathrm{2}} +\left({R}+\mathrm{2}{r}\right)^{\mathrm{2}} } \\ $$$$\frac{{AL}}{\mathrm{2}{R}}=\frac{{R}}{{AB}}\:\Rightarrow{AL}=\frac{\mathrm{2}{R}^{\mathrm{2}} }{{AB}} \\ $$$$\frac{{MB}}{\mathrm{2}{r}}=\frac{{R}+\mathrm{2}{r}}{{AB}}\:\Rightarrow{MB}=\frac{\mathrm{2}{r}\left({R}+\mathrm{2}{r}\right)}{{AB}} \\ $$$$\frac{{NM}}{\mathrm{2}{r}}=\frac{{R}}{{AB}}\:\Rightarrow{NM}=\frac{\mathrm{2}{Rr}}{{AB}} \\ $$$$\Delta{LMN}=\frac{\left({AB}−{AL}−{MB}\right)×{NM}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({AB}−\frac{\mathrm{2}{R}^{\mathrm{2}} +\mathrm{2}{r}\left({R}+\mathrm{2}{r}\right)}{{AB}}\right)×\frac{\mathrm{2}{Rr}}{{AB}} \\ $$$$=\frac{{Rr}\left({R}^{\mathrm{2}} +\left({R}+\mathrm{2}{r}\right)^{\mathrm{2}} −\mathrm{2}{R}^{\mathrm{2}} −\mathrm{2}{r}\left({R}+\mathrm{2}{r}\right)\right)}{{R}^{\mathrm{2}} +\left({R}+\mathrm{2}{r}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}{R}^{\mathrm{2}} {r}^{\mathrm{2}} }{{R}^{\mathrm{2}} +\left({R}+\mathrm{2}{r}\right)^{\mathrm{2}} }=\frac{{R}^{\mathrm{2}} {r}^{\mathrm{2}} }{{R}^{\mathrm{2}} +\mathrm{2}{Rr}+\mathrm{2}{r}^{\mathrm{2}} } \\ $$

Commented by Spillover last updated on 23/Oct/24

$${great}\:{work}.{thanks} \\ $$

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