A polygon is inscribed in a circle when all its vertices lie on the circle. Using the constructions from Constructing Perpendicular and Parallel Lines, you can inscribe three regular polygons exactly: a hexagon, an equilateral triangle and a square.
Why it works
Keep the compass at the radius $r$ and mark a point $B$ on the circle, one compass width from $A$. Triangle $OAB$ has $OA = OB = r$ (radii) and $AB = r$ (the compass), so it is equilateral and its angle at $O$ is $60^\circ$. Six such steps use $6 \cdot 60^\circ = 360^\circ$, exactly once around, so you land back on $A$ with six equal sides.
- Hexagon: step around the circle with the radius as the compass width; connect the six points. Each side equals $r$.
- Equilateral triangle: make the hexagon's points, then connect every other one. Each side skips one $60^\circ$ step, so each central angle is $120^\circ$.
- Square: draw a diameter, construct the diameter perpendicular to it, and connect the four endpoints. The central angles are $90^\circ$ each.
Key idea
A regular polygon with $n$ sides, inscribed in a circle, has $n$ equal central angles of $\frac{360^\circ}{n}$. Constructions produce those angles: $60^\circ$ by stepping with the radius, $120^\circ$ by skipping a point, $90^\circ$ with perpendicular diameters.
Worked example: The side of the inscribed square
A square is inscribed in a circle of radius $6$. How long is each side?
- Two radii to neighboring corners meet at a right angle, since the diagonals are perpendicular diameters.
- The side is the hypotenuse of a right triangle with legs $6$ and $6$: $\sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2}$.
Answer: $6\sqrt{2} \approx 8.49$
Worked example: The inscribed equilateral triangle
An equilateral triangle is inscribed in a circle of radius $4$. Find its perimeter.
- Draw radii to two vertices and the perpendicular from the center to the side between them. The central angle $120^\circ$ is split into two $60^\circ$ angles.
- In the $30$-$60$-$90$ triangle with hypotenuse $4$, half of the side is opposite $60^\circ$: $2\sqrt{3}$. So each side is $4\sqrt{3}$.
- Perimeter: $3 \cdot 4\sqrt{3} = 12\sqrt{3}$.
Answer: $12\sqrt{3} \approx 20.8$
You will meet the $30$-$60$-$90$ ratios properly in 45-45-90 and 30-60-90 Triangles; here it is enough to see that the side comes from a right triangle.