Question-212654

Question Number 212654 by Spillover last updated on 20/Oct/24

Answered by mr W last updated on 20/Oct/24

Commented by mr W last updated on 20/Oct/24

R=a+b (a+r)^2 −(a−r)^2 =(R−r)^2 −(2a−R−r)^2 Rr=a(R−a) ⇒r=((a(R−a))/R)=((ab)/(a+b))=((6×3)/(6+3))=2 ⇒both blue circles are of same size. Σarea =2πr^2 =((2π(ab)^2 )/((a+b)^2 ))=((2π(6×3)^2 )/((6+3)^2 ))=8π

$${R}={a}+{b} \\ $$$$\left({a}+{r}\right)^{\mathrm{2}} −\left({a}−{r}\right)^{\mathrm{2}} =\left({R}−{r}\right)^{\mathrm{2}} −\left(\mathrm{2}{a}−{R}−{r}\right)^{\mathrm{2}} \\ $$$${Rr}={a}\left({R}−{a}\right) \\ $$$$\Rightarrow{r}=\frac{{a}\left({R}−{a}\right)}{{R}}=\frac{{ab}}{{a}+{b}}=\frac{\mathrm{6}×\mathrm{3}}{\mathrm{6}+\mathrm{3}}=\mathrm{2} \\ $$$$\Rightarrow{both}\:{blue}\:{circles}\:{are}\:{of}\:{same}\:{size}. \\ $$$$\Sigma{area}\:=\mathrm{2}\pi{r}^{\mathrm{2}} \\ $$$$\:\:\:=\frac{\mathrm{2}\pi\left({ab}\right)^{\mathrm{2}} }{\left({a}+{b}\right)^{\mathrm{2}} }=\frac{\mathrm{2}\pi\left(\mathrm{6}×\mathrm{3}\right)^{\mathrm{2}} }{\left(\mathrm{6}+\mathrm{3}\right)^{\mathrm{2}} }=\mathrm{8}\pi \\ $$

Commented by Frix last updated on 20/Oct/24