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Question-187901

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Question Number 187901 by Rupesh123 last updated on 23/Feb/23
Commented by Rupesh123 last updated on 23/Feb/23
Find R and r
Commented by BaliramKumar last updated on 23/Feb/23
R = ((3(√6))/4)
$${R}\:=\:\frac{\mathrm{3}\sqrt{\mathrm{6}}}{\mathrm{4}} \\ $$
Answered by mr W last updated on 23/Feb/23
s=((7+10+15)/2)=16 Δ=(√(16×9×6×1))=12(√6) R=(Δ/s)=((12(√6))/(16))=((3(√6))/4) cos C=((15^2 +10^2 −7^2 )/(2×15×10))=((23)/(25)) ((R−r)/(R+r))=sin (C/2)=(√((1−cos C)/2))=(1/5) ⇒r=((2R)/3)=((√6)/2) ✓
$${s}=\frac{\mathrm{7}+\mathrm{10}+\mathrm{15}}{\mathrm{2}}=\mathrm{16} \\ $$$$\Delta=\sqrt{\mathrm{16}×\mathrm{9}×\mathrm{6}×\mathrm{1}}=\mathrm{12}\sqrt{\mathrm{6}} \\ $$$${R}=\frac{\Delta}{{s}}=\frac{\mathrm{12}\sqrt{\mathrm{6}}}{\mathrm{16}}=\frac{\mathrm{3}\sqrt{\mathrm{6}}}{\mathrm{4}} \\ $$$$\mathrm{cos}\:{C}=\frac{\mathrm{15}^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }{\mathrm{2}×\mathrm{15}×\mathrm{10}}=\frac{\mathrm{23}}{\mathrm{25}} \\ $$$$\frac{{R}−{r}}{{R}+{r}}=\mathrm{sin}\:\frac{{C}}{\mathrm{2}}=\sqrt{\frac{\mathrm{1}−\mathrm{cos}\:{C}}{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{2}{R}}{\mathrm{3}}=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\:\checkmark \\ $$
Commented by Rupesh123 last updated on 23/Feb/23
Very nice solution!
Answered by HeferH last updated on 23/Feb/23
A = s∙I; I = R s = ((7+10+15)/2) = 16 R = (A/s) = ((√(16(16−10)(16−15)(16−7)))/(16)) = ((√(16∙6∙9))/(16)) = ((3(√6))/4) ((12r)/( (√6))) + 2(√((3r(√6))/4)) = 9 2r(√6) + (√(3r(√6))) = 9 3r(√6) = (9−2r(√6))^2 3r(√6) = 81 + 24r^2 − 36r(√6) 24r^2 −39r(√6) + 81 = 0 8r^2 −13r(√6) + 27 = 0 r = ((13(√6)±(√(169∙6−4(8)(27))))/(16)) = ((13(√6)±(√(6∙169−6∙144)))/(16)) = ((13(√6)±5(√6))/(16)) R > r ⇒ r = ((√6)/2);
$$\:{A}\:=\:{s}\centerdot{I};\:{I}\:=\:{R} \\ $$$$\:{s}\:=\:\frac{\mathrm{7}+\mathrm{10}+\mathrm{15}}{\mathrm{2}}\:=\:\mathrm{16} \\ $$$$\:{R}\:=\:\frac{{A}}{{s}}\:=\:\frac{\sqrt{\mathrm{16}\left(\mathrm{16}−\mathrm{10}\right)\left(\mathrm{16}−\mathrm{15}\right)\left(\mathrm{16}−\mathrm{7}\right)}}{\mathrm{16}}\:=\:\frac{\sqrt{\mathrm{16}\centerdot\mathrm{6}\centerdot\mathrm{9}}}{\mathrm{16}}\:=\:\frac{\mathrm{3}\sqrt{\mathrm{6}}}{\mathrm{4}} \\ $$$$\:\frac{\mathrm{12}{r}}{\:\sqrt{\mathrm{6}}}\:+\:\mathrm{2}\sqrt{\frac{\mathrm{3}{r}\sqrt{\mathrm{6}}}{\mathrm{4}}}\:=\:\mathrm{9} \\ $$$$\:\mathrm{2}{r}\sqrt{\mathrm{6}}\:+\:\sqrt{\mathrm{3}{r}\sqrt{\mathrm{6}}}\:=\:\mathrm{9} \\ $$$$\:\mathrm{3}{r}\sqrt{\mathrm{6}}\:=\:\left(\mathrm{9}−\mathrm{2}{r}\sqrt{\mathrm{6}}\right)^{\mathrm{2}} \\ $$$$\:\mathrm{3}{r}\sqrt{\mathrm{6}}\:=\:\mathrm{81}\:+\:\mathrm{24}{r}^{\mathrm{2}} \:−\:\mathrm{36}{r}\sqrt{\mathrm{6}} \\ $$$$\:\mathrm{24}{r}^{\mathrm{2}} −\mathrm{39}{r}\sqrt{\mathrm{6}}\:+\:\mathrm{81}\:=\:\mathrm{0} \\ $$$$\:\mathrm{8}{r}^{\mathrm{2}} −\mathrm{13}{r}\sqrt{\mathrm{6}}\:+\:\mathrm{27}\:=\:\mathrm{0}\: \\ $$$$\:{r}\:=\:\frac{\mathrm{13}\sqrt{\mathrm{6}}\pm\sqrt{\mathrm{169}\centerdot\mathrm{6}−\mathrm{4}\left(\mathrm{8}\right)\left(\mathrm{27}\right)}}{\mathrm{16}}\:=\:\frac{\mathrm{13}\sqrt{\mathrm{6}}\pm\sqrt{\mathrm{6}\centerdot\mathrm{169}−\mathrm{6}\centerdot\mathrm{144}}}{\mathrm{16}} \\ $$$$\:=\:\frac{\mathrm{13}\sqrt{\mathrm{6}}\pm\mathrm{5}\sqrt{\mathrm{6}}}{\mathrm{16}}\:\: \\ $$$$\:{R}\:>\:{r}\:\Rightarrow\:{r}\:\:=\:\frac{\sqrt{\mathrm{6}}}{\mathrm{2}};\: \\ $$
Commented by Rupesh123 last updated on 23/Feb/23
Very nice solution!
Commented by HeferH last updated on 23/Feb/23
Commented by Rupesh123 last updated on 23/Feb/23
Very detailed, sir

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