Construct this figure. [Hint: Find the angles in this figure.]

Question

Homi · Accepted Answer

Understand the Question

The given figure is a six-pointed star (hexagram), formed by two overlapping equilateral triangles inscribed in a circle.
A circle's radius divides the circumference into exactly $6$ equal arcs of $60^\circ$ each.
Connecting alternate points creates two equilateral triangles ( $\Delta \text{ACE}$ and $\Delta \text{BDF}$ ), which intersect to form the star and a regular hexagon at the center.
Each star point has an angle of $60^\circ$ , and each interior angle of the central hexagon is $120^\circ$ .

Step 1 · Divide Circle into Six Equal Arcs

Draw a circle of a convenient radius (e.g., $5\text{ cm}$ ).

Keeping the compass set to the same radius, start at point $\text{A}$ on the circumference and mark off successive points $\text{B, C, D, E, F}$ around the circle. Diagram 1

Step 2 · Draw the Overlapping Equilateral Triangles

Join alternate points to create two equilateral triangles:

Join $\text{A}$ to $\text{C}$ , $\text{C}$ to $\text{E}$ , and $\text{E}$ to $\text{A}$ to form the first equilateral triangle $\Delta \text{ACE}$ .
Join $\text{B}$ to $\text{D}$ , $\text{D}$ to $\text{F}$ , and $\text{F}$ to $\text{B}$ to form the second equilateral triangle $\Delta \text{BDF}$ .

Step 3 · Calculate the Angles of the Figure

1. Angle at each point of the star:

For vertex $\text{A}$ , angle $\angle \text{CAE}$ subtends arc $\text{CDE} = 60^\circ + 60^\circ = 120^\circ$ .

\begin{aligned} \text{Angle at point A} &= \dfrac{120^\circ}{2} \\[0.6em] &= 60^\circ \end{aligned}

Thus, each point of the star has an angle of $60^\circ$ .

2. Interior angles of the central hexagon:

Let $\text{P}$ be the intersection point of $\text{AC}$ and $\text{BD}$ . In $\Delta \text{BPC}$ : $\angle \text{PCB} = \dfrac{60^\circ}{2} = 30^\circ$ $\angle \text{PBC} = \dfrac{60^\circ}{2} = 30^\circ$

Using the angle sum property of $\Delta \text{BPC}$ :

\begin{aligned} \angle \text{BPC} &= 180^\circ - (\angle \text{PCB} + \angle \text{PBC}) \\[0.6em] &= 180^\circ - (30^\circ + 30^\circ) \\[0.6em] &= 180^\circ - 60^\circ \\[0.6em] &= 120^\circ \end{aligned}

Thus, each interior angle of the central hexagon is $120^\circ$ .

Answer

The figure is constructed by drawing two overlapping equilateral triangles ( $\Delta \text{ACE}$ and $\Delta \text{BDF}$ ) inside a circle. The angle at each point of the star is $60^\circ$ , and each interior angle of the central hexagon is $120^\circ$ .

Common Mistakes

Changing Compass Radius: Changing the compass width while marking the $6$ points will distort the equal spacing. The radius must remain strictly unchanged throughout.
Joining Adjacent Points: Connecting consecutive points ( $\text{A-B-C-D-E-F}$ ) produces a simple regular hexagon instead of two overlapping equilateral triangles.

Question 21

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