Question 21

We can construct this figure by drawing two overlapping equilateral triangles inside a circle.

Step 1 — Draw a circle and mark points

Let us draw a circle with a suitable radius. Let the radius be 5 cm. Let us mark a point on the circle. Let us call it point A. Now, let us use a compass with the same radius (5 cm). Place the compass needle on point A. Draw an arc to cut the circle at a new point. Let us call it point B. Repeat this process, placing the needle on point B to find point C, and so on. We will get 6 equally spaced points around the circle. Let us label them A, B, C, D, E, F in order.

Step 2 — Draw the first equilateral triangle

Let us join point A to point C using a ruler. Then, let us join point C to point E. Finally, let us join point E to point A. This forms the first equilateral triangle, ACE.

Step 3 — Draw the second equilateral triangle

Now, let us join point B to point D using a ruler. Then, let us join point D to point F. Finally, let us join point F to point B. This forms the second equilateral triangle, BDF.

Step 4 — Find the angles

Let us find the angle at each point of the star. Consider the point A of the star. It is formed by lines AE and AC. The angle $\angle CAE$ is subtended by the arc CE. The arc CE covers two segments, CD and DE. Each segment is $60^\circ$ of the circle. So arc CE is $60^\circ + 60^\circ = 120^\circ$. The angle at the circumference is half the arc. So, the angle at point A is $120^\circ / 2$.

$= \frac{120^\circ}{2}$

$\boxed{60^\circ}$

So, the angle at each point of the star is 60 degrees.

Now, let us find the interior angles of the central hexagon. Let P be one vertex of the central hexagon. P is formed by the intersection of lines AC and BD. Consider the triangle formed by points B, P, C. The angle $\angle PCB$ is subtended by arc AB. Arc AB is $60^\circ$. So, $\angle PCB = 60^\circ / 2 = 30^\circ$. The angle $\angle PBC$ is subtended by arc CD. Arc CD is $60^\circ$. So, $\angle PBC = 60^\circ / 2 = 30^\circ$. The sum of angles in triangle BPC is $180^\circ$. So, the angle $\angle BPC = 180^\circ - (\angle PCB + \angle PBC)$.

$= 180^\circ - (30^\circ + 30^\circ)$

$= 180^\circ - 60^\circ$

$\boxed{120^\circ}$

This angle $\angle BPC$ is an interior angle of the central hexagon. So, each interior angle of the central hexagon is 120 degrees.

Answer

(i) The figure is constructed by drawing two overlapping equilateral triangles. (ii) The angle at each point of the star is 60 degrees. (iii) The interior angle of the central hexagon is 120 degrees.