Question Number 226469 by ajfour last updated on 30/Nov/25
Commented by ajfour last updated on 30/Nov/25
$${Find}\:{the}\:{common}\:{area}\:{of}\:{the} \\ $$$${smaller}\:{circles}.\:{Outer}\:{square} \\ $$$${has}\:{side}\:{length}\:{s}=\mathrm{2}\:{units}. \\ $$
Answered by fantastic2 last updated on 30/Nov/25
Commented by fantastic2 last updated on 30/Nov/25
![method 1:only geometry no calculus r=small C radius (1+r)^2 =(1−r)^2 +(2−r)^2 ⇒4−4r+r^2 =(1+r)^2 −(1−r)^2 =4r r^2 −8r+4=0 ⇒r=((8±4(√3))/2)=4−2(√3)[∵4+2(√3)>2 ,rejected] lets say the opposite of α is x x^2 +(1−r)^2 =r^2 x=(√(2r−1)) α=sin^(−1) (((√(2r−1))/r))=sin^(−1) (((√(2(4−2(√3))−1))/(4−2(√3))))=sin^(−1) 0.5=30^0 Green area=(1/6)πr^2 −(1/2)×2(√(2r−1))×(1−r) total area=2((1/6)πr^2 −(√(2r−1))×(1−r)) ≈0.052030299](https://www.tinkutara.com/question/Q226475.png)
$${method}\:\mathrm{1}:{only}\:{geometry}\:{no}\:{calculus} \\ $$$${r}={small}\:{C}\:{radius} \\ $$$$\left(\mathrm{1}+{r}\right)^{\mathrm{2}} =\left(\mathrm{1}−{r}\right)^{\mathrm{2}} +\left(\mathrm{2}−{r}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}−\mathrm{4}{r}+{r}^{\mathrm{2}} =\left(\mathrm{1}+{r}\right)^{\mathrm{2}} −\left(\mathrm{1}−{r}\right)^{\mathrm{2}} =\mathrm{4}{r} \\ $$$${r}^{\mathrm{2}} −\mathrm{8}{r}+\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{r}=\frac{\mathrm{8}\pm\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}\left[\because\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}>\mathrm{2}\:,{rejected}\right] \\ $$$${lets}\:{say}\:{the}\:{opposite}\:{of}\:\alpha\:{is}\:{x} \\ $$$${x}^{\mathrm{2}} +\left(\mathrm{1}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${x}=\sqrt{\mathrm{2}{r}−\mathrm{1}} \\ $$$$\alpha=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}{r}−\mathrm{1}}}{{r}}\right)=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}\right)−\mathrm{1}}}{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}\right)=\mathrm{sin}^{−\mathrm{1}} \mathrm{0}.\mathrm{5}=\mathrm{30}^{\mathrm{0}} \\ $$$${Green}\:{area}=\frac{\mathrm{1}}{\mathrm{6}}\pi{r}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}\sqrt{\mathrm{2}{r}−\mathrm{1}}×\left(\mathrm{1}−{r}\right) \\ $$$${total}\:{area}=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{6}}\pi{r}^{\mathrm{2}} −\sqrt{\mathrm{2}{r}−\mathrm{1}}×\left(\mathrm{1}−{r}\right)\right) \\ $$$$\approx\mathrm{0}.\mathrm{052030299} \\ $$
Commented by ajfour last updated on 30/Nov/25
$${yes}!\:{great}\:{work}. \\ $$
Answered by mr W last updated on 30/Nov/25
Commented by ajfour last updated on 30/Nov/25
$${Exactly}!\:{Thank}\:{you}\:{sir}. \\ $$
Commented by mr W last updated on 30/Nov/25
![square′s side length =s=2 R=(s/2)=1 (√((R+r)^2 −(R−r)^2 ))=2R−r 4R^2 −8Rr+r^2 =0 r=(4−2(√3))R cos α=((R−r)/r)=(1/(4−2(√3)))−1=((√3)/2) ⇒α=30° shaded area =2(((πr^2 )/6)−(((√3)r^2 )/4))=((π/3)−((√3)/2))r^2 =((π/3)−((√3)/2))(4−2(√3))^2 R^2 =4((π/3)−((√3)/2))(7−4(√3))R^2 =((π/3)−((√3)/2))(7−4(√3))s^2 [≈1.3% of square′s area] ≈0.05203](https://www.tinkutara.com/question/Q226477.png)
$${square}'{s}\:{side}\:{length}\:={s}=\mathrm{2} \\ $$$${R}=\frac{{s}}{\mathrm{2}}=\mathrm{1} \\ $$$$\sqrt{\left({R}+{r}\right)^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} }=\mathrm{2}{R}−{r} \\ $$$$\mathrm{4}{R}^{\mathrm{2}} −\mathrm{8}{Rr}+{r}^{\mathrm{2}} =\mathrm{0} \\ $$$${r}=\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}\right){R} \\ $$$$\mathrm{cos}\:\alpha=\frac{{R}−{r}}{{r}}=\frac{\mathrm{1}}{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}−\mathrm{1}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\: \\ $$$$\Rightarrow\alpha=\mathrm{30}° \\ $$$${shaded}\:{area} \\ $$$$=\mathrm{2}\left(\frac{\pi{r}^{\mathrm{2}} }{\mathrm{6}}−\frac{\sqrt{\mathrm{3}}{r}^{\mathrm{2}} }{\mathrm{4}}\right)=\left(\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right){r}^{\mathrm{2}} \\ $$$$=\left(\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} {R}^{\mathrm{2}} \\ $$$$=\mathrm{4}\left(\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right){R}^{\mathrm{2}} \\ $$$$=\left(\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right){s}^{\mathrm{2}} \:\:\left[\approx\mathrm{1}.\mathrm{3\%}\:{of}\:{square}'{s}\:{area}\right] \\ $$$$\approx\mathrm{0}.\mathrm{05203} \\ $$
Commented by ajfour last updated on 30/Nov/25
