Question-212878

Question Number 212878 by Spillover last updated on 25/Oct/24

Answered by som(math1967) last updated on 26/Oct/24

Commented by som(math1967) last updated on 26/Oct/24

let rad of green circle=R purple circle=r ∴ a^2 =2R×2r=4Rr ∴ a=2(√(Rr)) again b^2 =(R+r)^2 −(R−r)^2 b=(√(4Rr))=2(√(Rr)) ∴ a:b=1:1

$${let}\:{rad}\:{of}\:{green}\:{circle}={R} \\ $$$$\:{purple}\:{circle}={r} \\ $$$$\:\therefore\:{a}^{\mathrm{2}} =\mathrm{2}{R}×\mathrm{2}{r}=\mathrm{4}{Rr} \\ $$$$\therefore\:{a}=\mathrm{2}\sqrt{{Rr}} \\ $$$$\:{again}\:{b}^{\mathrm{2}} =\left({R}+{r}\right)^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} \\ $$$$\:\:{b}=\sqrt{\mathrm{4}{Rr}}=\mathrm{2}\sqrt{{Rr}} \\ $$$$\:\therefore\:{a}:{b}=\mathrm{1}:\mathrm{1} \\ $$

Answered by Spillover last updated on 26/Oct/24