Question Number 59000 by Learner last updated on 03/May/19
Answered by MJS last updated on 03/May/19
![the centers of the small circles form a square with side length 2r ⇒ its diagonal =2(√2)r ⇒ ⇒ the radius of the big circle =(1+(√2))r ⇒ ⇒ the shaded area =[(1+(√2))r]^2 π−4r^2 π= =(−1+2(√2))πr^2](Q59001.png)
$$\mathrm{the}\:\mathrm{centers}\:\mathrm{of}\:\mathrm{the}\:\mathrm{small}\:\mathrm{circles}\:\mathrm{form}\:\mathrm{a}\:\mathrm{square} \\ $$$$\mathrm{with}\:\mathrm{side}\:\mathrm{length}\:\mathrm{2}{r}\:\Rightarrow\:\mathrm{its}\:\mathrm{diagonal}\:=\mathrm{2}\sqrt{\mathrm{2}}{r}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{big}\:\mathrm{circle}\:=\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){r}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{shaded}\:\mathrm{area}\:=\left[\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){r}\right]^{\mathrm{2}} \pi−\mathrm{4}{r}^{\mathrm{2}} \pi= \\ $$$$=\left(−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}\right)\pi{r}^{\mathrm{2}} \\ $$
Commented by Learner last updated on 03/May/19
$${thank}\:{you}\:{sir} \\ $$