Tenth Grade Mathematics: Circles Inscribed in and Circumscribed About Polygons
The interaction between rectilinear polygons and smooth circular curves represents one of the oldest optimization themes in geometry. A polygon is inscribed in a circle (or the circle is circumscribed about the polygon) when all vertices of the polygon lie on the circle. Conversely, a circle is inscribed in a polygon (an incircle) when it touches every side of the polygon tangentially. In this chapter, we analyze the area and perimeter relationships for regular n-gons, equilateral triangles, and squares interacting with inscribed and circumscribed circles.
Regular Polygons and Circular Enclosures
A regular polygon with n sides possesses full rotational and reflective symmetry. It has an apothem (the perpendicular distance from the center to any side, which equals the inradius r) and a circumradius R (the distance from the center to any vertex).
Geometry of a Regular n-gon:
Vertex
/ R / \ R
/ / r +---+----+ Side s
Apothem r
The central angle subtended by one side at the center is theta = 360° / n. Partitioning the polygon into n congruent isosceles triangles gives:
Fundamental Relations for Regular n-gon:
Side Length: s = 2 * R * sin(180° / n) = 2 * r * tan(180° / n)
Apothem: r = R * cos(180° / n)
Perimeter: P = n * s
Area of n-gon: A = (1/2) * Perimeter * Apothem = (1/2) * n * s * r
As the number of sides n approaches infinity, the perimeter P approaches 2piR, the apothem r approaches R, and the area approaches (1/2)(2piR)R = pi*R^2, seamlessly recovering the circle.
Inscribed and Circumscribed Equilateral Triangles
For an equilateral triangle of side length a:
Equilateral Triangle Summary:
Height of Triangle: h = a * sqrt(3) / 2
Area of Triangle: Delta = a^2 * sqrt(3) / 4
1. Circumscribed Circle (Circumradius R):
The circumcenter is the centroid, dividing the altitude in ratio 2 : 1:
R = (2/3) * h = (2/3) * [a * sqrt(3) / 2] = a / sqrt(3)
Area(Circumcircle) = pi * R^2 = pi * a^2 / 3
2. Inscribed Circle (Inradius r):
The inradius is the lower third of the altitude:
r = (1/3) * h = (1/3) * [a * sqrt(3) / 2] = a / (2 * sqrt(3))
Area(Incircle) = pi * r^2 = pi * a^2 / 12
Ratio of Radii: R : r = 2 : 1
Ratio of Areas: Area(Circumcircle) : Area(Incircle) = R^2 : r^2 = 4 : 1
Area between Circumcircle and Incircle (The Concentric Ring):
Area(Ring) = Area(Circumcircle) - Area(Incircle)
= pi * a^2 / 3 - pi * a^2 / 12 = pi * a^2 / 4
Inscribed and Circumscribed Squares
For a square of side length s:
Square Summary:
1. Inscribed Circle:
The diameter of the inscribed circle equals the side length of the square:
2*r = s ===> r = s / 2
Area(Inscribed Circle) = pi * (s / 2)^2 = pi * s^2 / 4
2. Circumscribed Circle:
The diameter of the circumcircle equals the diagonal of the square:
Diagonal d = s * sqrt(2)
2*R = s * sqrt(2) ===> R = s / sqrt(2)
Area(Circumscribed Circle) = pi * (s / sqrt(2))^2 = pi * s^2 / 2
Ratio of Areas:
Area(Circumcircle) : Area(Inscribed Circle) = (pi * s^2 / 2) : (pi * s^2 / 4) = 2 : 1
The area of the circle circumscribed about a square is exactly DOUBLE the area of the circle inscribed in the square!
Regular Hexagon Enclosures
A regular hexagon of side a can be divided into six congruent equilateral triangles of side a: Area of Hexagon = 6 * (sqrt(3)/4)a^2 = (3sqrt(3)/2)a^2. Circumradius R = a (Circumcircle area = pi * a^2). Inradius r = a * sqrt(3)/2 (Incircle area = 3pi*a^2 / 4).
Chapter Exercises
Exercise 1: An equilateral triangle is inscribed in a circle of radius 8 cm. Find the length of each side of the triangle and calculate the area of the region of the circle outside the triangle. (Use pi = 3.14 and sqrt(3) = 1.732).
Exercise 2: Prove that the area of the circle inscribed in an equilateral triangle is equal to one-fourth the area of the circle circumscribed about the same triangle.
Exercise 3: A square is inscribed in a circle of radius 7 cm. Find the area of the square, and find the area of the four circular segments outside the square.
Solutions and Explanations
Solution 1: We are given circumradius R = 8 cm. For an equilateral triangle inscribed in a circle, the circumradius formula is R = a / sqrt(3), which means side length a = R * sqrt(3) = 8 * sqrt(3) cm. Using sqrt(3) = 1.732: a = 8 * 1.732 = 13.856 cm. The total area of the circle is Area(Circle) = pi * R^2 = 3.14 * 8^2 = 3.14 * 64 = 200.96 cm^2. The area of the equilateral triangle is Area(triangle) = (sqrt(3)/4) * a^2 = (sqrt(3)/4) * (8*sqrt(3))^2 = (sqrt(3)/4) * (64 * 3) = 48 * sqrt(3) = 48 * 1.732 = 83.136 cm^2. The area of the circular region outside the triangle is: Area(Remaining) = Area(Circle) - Area(triangle) = 200.96 - 83.136 = 117.824 cm^2.
Solution 2: Let the side length of the equilateral triangle be a. The altitude is h = asqrt(3)/2. The centroid coincides with the incenter and circumcenter. The circumradius is R = (2/3)h = a / sqrt(3). The area of the circumscribed circle is Area(Circumcircle) = pi * R^2 = pi * (a / sqrt(3))^2 = pi * a^2 / 3. The inradius is r = (1/3)h = a / (2sqrt(3)). The area of the inscribed circle is Area(Incircle) = pi * r^2 = pi * [a / (2*sqrt(3))]^2 = pi * a^2 / 12. Taking the ratio: Area(Incircle) / Area(Circumcircle) = (pi * a^2 / 12) / (pi * a^2 / 3) = (1 / 12) * (3 / 1) = 3 / 12 = 1 / 4. Therefore, the area of the inscribed circle is exactly one-fourth of the circumscribed circle area.
Solution 3: The circle has radius R = 7 cm. The diameter of the circle is d = 2 * R = 14 cm. When a square is inscribed in a circle, the diagonal of the square is identically equal to the diameter of the circle: Diagonal d = 14 cm. The area of any square in terms of its diagonal is Area(Square) = (1/2) * d^2 = (1/2) * 14^2 = (1/2) * 196 = 98 cm^2. The total area of the circle is Area(Circle) = pi * R^2 = (22/7) * 7^2 = 22 * 7 = 154 cm^2. The total area of the four circular segments outside the square is the circle area minus the square area: Area(4 Segments) = Area(Circle) - Area(Square) = 154 - 98 = 56 cm^2. (The area of each individual segment is 56 / 4 = 14 cm^2).