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Question Number 187961 by normans last updated on 24/Feb/23

Answered by HeferH last updated on 24/Feb/23

Commented by HeferH last updated on 24/Feb/23

i. (R−3)^2 = 9 + (R^2 /4) R^2 +9−6R = 9 + (R^2 /4) 4R^2 −24R = R^3 3R^2 = 24R R = 8 ii. (R−r)^2 −r^2 = ((R/( (√2)))+r)^2 −((R/( (√2)))−r)^2 R^2 −2Rr= ((4Rr)/( 2))(√2) R−2r= 2r(√2) R = 2r + 2r(√2) r = (R/(2 + 2(√2))) r = (8/(2+2(√2))) = (4/(1+(√2))) = ((4((√2)−1))/(2−1)) = 4((√2)−1)

$${i}.\:\left({R}−\mathrm{3}\right)^{\mathrm{2}} \:=\:\mathrm{9}\:+\:\frac{{R}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:{R}^{\mathrm{2}} +\mathrm{9}−\mathrm{6}{R}\:=\:\mathrm{9}\:+\:\frac{{R}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\mathrm{4}{R}^{\mathrm{2}} −\mathrm{24}{R}\:=\:{R}^{\mathrm{3}} \\ $$$$\:\mathrm{3}{R}^{\mathrm{2}} \:=\:\mathrm{24}{R} \\ $$$$\:{R}\:=\:\mathrm{8}\: \\ $$$$\:{ii}.\:\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} =\:\left(\frac{{R}}{\:\sqrt{\mathrm{2}}}+{r}\right)^{\mathrm{2}} −\left(\frac{{R}}{\:\sqrt{\mathrm{2}}}−{r}\right)^{\mathrm{2}} \\ $$$$\:{R}^{\mathrm{2}} −\mathrm{2}{Rr}=\:\frac{\mathrm{4}{Rr}}{\:\mathrm{2}}\sqrt{\mathrm{2}} \\ $$$$\:{R}−\mathrm{2}{r}=\:\mathrm{2}{r}\sqrt{\mathrm{2}} \\ $$$$\:{R}\:=\:\mathrm{2}{r}\:+\:\mathrm{2}{r}\sqrt{\mathrm{2}} \\ $$$$\:{r}\:=\:\frac{{R}}{\mathrm{2}\:+\:\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\:{r}\:=\:\frac{\mathrm{8}}{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}\:=\:\frac{\mathrm{4}}{\mathrm{1}+\sqrt{\mathrm{2}}}\:=\:\frac{\mathrm{4}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{2}−\mathrm{1}}\:=\:\mathrm{4}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$

Commented by normans last updated on 25/Feb/23

nice

$${nice} \\ $$

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