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Question-194463




Question Number 194463 by Mingma last updated on 08/Jul/23
Answered by mr W last updated on 08/Jul/23
Commented by Mingma last updated on 08/Jul/23
How did you calculate 1%?
Commented by mr W last updated on 08/Jul/23
d=2c  R=((a+b+2c+2d)/2)=((a+b)/2)+3c  a^2 =2c(a+b+2d)=2c(a+b)+8c^2   b^2 =2d(a+b+2c)=4c(a+b)+8c^2   b^2 −a^2 =2c(a+b)  ⇒c=((b−a)/2)  a^2 =(b−a)(a+b)+2(b−a)^2   ⇒3b=4a  ⇒b=8c, a=6c  (c/R)=(c/(((a+b)/2)+3c))=(1/(7+3))=(1/(10))  ((area small circle)/(area big circle))=((πc^2 )/(πR^2 ))=((c/R))^2 =(1/(100))=1%
$${d}=\mathrm{2}{c} \\ $$$${R}=\frac{{a}+{b}+\mathrm{2}{c}+\mathrm{2}{d}}{\mathrm{2}}=\frac{{a}+{b}}{\mathrm{2}}+\mathrm{3}{c} \\ $$$${a}^{\mathrm{2}} =\mathrm{2}{c}\left({a}+{b}+\mathrm{2}{d}\right)=\mathrm{2}{c}\left({a}+{b}\right)+\mathrm{8}{c}^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} =\mathrm{2}{d}\left({a}+{b}+\mathrm{2}{c}\right)=\mathrm{4}{c}\left({a}+{b}\right)+\mathrm{8}{c}^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} −{a}^{\mathrm{2}} =\mathrm{2}{c}\left({a}+{b}\right) \\ $$$$\Rightarrow{c}=\frac{{b}−{a}}{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} =\left({b}−{a}\right)\left({a}+{b}\right)+\mathrm{2}\left({b}−{a}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{3}{b}=\mathrm{4}{a} \\ $$$$\Rightarrow{b}=\mathrm{8}{c},\:{a}=\mathrm{6}{c} \\ $$$$\frac{{c}}{{R}}=\frac{{c}}{\frac{{a}+{b}}{\mathrm{2}}+\mathrm{3}{c}}=\frac{\mathrm{1}}{\mathrm{7}+\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{10}} \\ $$$$\frac{{area}\:{small}\:{circle}}{{area}\:{big}\:{circle}}=\frac{\pi{c}^{\mathrm{2}} }{\pi{R}^{\mathrm{2}} }=\left(\frac{{c}}{{R}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{100}}=\mathrm{1\%} \\ $$
Commented by Mingma last updated on 09/Jul/23
Very detailed solution!

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