Question Number 199433 by ajfour last updated on 03/Nov/23
Commented by ajfour last updated on 03/Nov/23
$${Outer}\:{figure}\:{is}\:{square}\:{of}\:{side}\:\boldsymbol{{a}}. \\ $$$${Coloured}\:{triangles}\:{are}\:{equilateral}. \\ $$$${Find}\:{radius}\:{of}\:{circle}\:{inscribed}. \\ $$
Answered by mr W last updated on 03/Nov/23
Commented by mr W last updated on 03/Nov/23
$$\frac{{r}}{\mathrm{sin}\:\mathrm{30}°}=\frac{{a}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{75}°} \\ $$$$\Rightarrow{r}=\frac{{a}}{\mathrm{4}\:\mathrm{cos}\:\mathrm{15}°}=\frac{{a}}{\:\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}=\frac{\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right){a}}{\mathrm{4}} \\ $$
Commented by ajfour last updated on 03/Nov/23
$${yes},\:{thanks}\:{sir},\:{amounts}\:{to}\:{same}\checkmark \\ $$