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The Olympic symbol comprising five interlocking rings represents the union of the five continents of the world and the meeting of athletes from all over the world at the Olympic games. In order to spread awareness about Olympic games, students of Class-X took part in various activities organised by the school. One such group of students made $5$ circular rings in the school lawn with the help of ropes. Each circular ring required $44$ m of rope.
Also, in the shaded regions as shown in the figure, students made rangoli showcasing various sports and games. It is given that $\triangle OAB$ is an equilateral triangle and all unshaded regions are congruent.
Based on above information, answer the following questions :
(i) Find the radius of each circular ring.
(ii) What is the measure of $\angle AOB$ ?
(iii) (a) Find the area of shaded region $R_1$.
OR
(iii) (b) Find the length of rope around the unshaded regions.
Also, in the shaded regions as shown in the figure, students made rangoli showcasing various sports and games. It is given that $\triangle OAB$ is an equilateral triangle and all unshaded regions are congruent.
Based on above information, answer the following questions :
(i) Find the radius of each circular ring.
(ii) What is the measure of $\angle AOB$ ?
(iii) (a) Find the area of shaded region $R_1$.
OR
(iii) (b) Find the length of rope around the unshaded regions.
Show SolutionHide Solution↓
(i) $2\pi r = 44$
$\Rightarrow r = 7$ m
(ii) $\angle AOB = 60^\circ$
(iii) (a) Area of shaded region $R_1$ = area of circle - area of $2$ segments
$= \pi \times 7 \times 7 - 2 \times (\frac{60}{360} \times \frac{22}{7} \times 7 \times 7 - \frac{\sqrt{3}}{4} \times 7 \times 7)$
$= (308 + 49\sqrt{3}) \text{ m}^2$ or $145.05 \text{ m}^2$ (approx.)
OR
(iii) (b) Length of rope around unshaded regions
$= 8 \times$ length of arc
$= 8\times\frac{60}{360}\times 2\times\frac{22}{7}\times 7$
$= \frac{176}{3}$ m or $58.66$ m (approx.)
$\Rightarrow r = 7$ m
(ii) $\angle AOB = 60^\circ$
(iii) (a) Area of shaded region $R_1$ = area of circle - area of $2$ segments
$= \pi \times 7 \times 7 - 2 \times (\frac{60}{360} \times \frac{22}{7} \times 7 \times 7 - \frac{\sqrt{3}}{4} \times 7 \times 7)$
$= (308 + 49\sqrt{3}) \text{ m}^2$ or $145.05 \text{ m}^2$ (approx.)
OR
(iii) (b) Length of rope around unshaded regions
$= 8 \times$ length of arc
$= 8\times\frac{60}{360}\times 2\times\frac{22}{7}\times 7$
$= \frac{176}{3}$ m or $58.66$ m (approx.)