Question and Answers Forum

All Questions      Topic List

Mensuration Questions

Previous in All Question      Next in All Question      

Previous in Mensuration      Next in Mensuration      

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!

Terms of Service

Privacy Policy

Contact: info@tinkutara.com