Circle Circumference, Area and Sectors

How much fencing is needed around a circular garden? Calculate its Circumference. To find how much ground can be covered with grass, calculate its Area. For a portion shaped like a slice of pizza, use the formulas for a Sector.

We will study circles, semicircles, rings and sectors, solve simple problems, and explore their relationships with squares and equilateral triangles.

👉 Circle

A circle is the set of all points in a plane at a fixed positive distance from a fixed point. The fixed point is the centre, and the distance is the radius.

A line segment passing through the centre with both endpoints on the circle is a diameter. Its length is twice the radius.

Note: Circumference is the length of the circular boundary. The area of a circle means the area of the plane region enclosed by that boundary.

For every circle, circumferencediameter = a constant.

This constant is π (pi), an irrational number with a non-terminating, non-repeating decimal expansion.

For calculations, use or π ≈ 3.14 as specified in the question. 227 and 3.14 are approximations, not exact values of π.

Circle

Let the diameter = AB = d and radius = OB = r.

🔹 Radius = × Diameter (∴ )

🔹 Circumference = π × Diameter = πd = 2πr (∵ d = 2r)

🔹 Area of circle = π × (Radius)² = πr²

👉 Semicircle

A diameter divides the region inside a circle into 2 equal semicircular regions. Each is bounded by a semicircular arc and a diameter.

Semicircle

🔹 Semicircle area = × Circle area = πr²

🔹 Semicircular arc length, or half the circle's circumference = πr

🔹 Semicircle perimeter = Semicircular arc length + Diameter = πr + 2r = r(π + 2)

👉 Circular Ring or Annulus

A circular ring is the region between 2 concentric circles of different radii. The outer radius is greater than the inner radius: R > r > 0.

Circular ring

Let the outer radius = R and inner radius = r.

🔹 Ring width = R − r

🔹 Ring area = Outer circle area − Inner circle area = πR² − πr² = π(R + r)(R − r)

👉 Semicircular Ring

A line through the centre divides a circular ring into 2 equal semicircular rings. Each boundary consists of 2 semicircular arcs and 2 straight segments.

Semicircular ring

Let the outer radius = R and inner radius = r.

🔹 Width = R − r

🔹 Semicircular ring area = .

🔹 Semicircular ring perimeter = Outer semicircular arc length + Inner semicircular arc length + 2 × Width

🔹 Semicircular ring perimeter = πR + πr + 2(R − r) = π(R + r) + 2(R − r)

👉 Sector of a Circle

Sector of a circle

A sector is a region bounded by 2 radii and an arc joining their endpoints. Depending on the chosen arc, it may be a minor or major sector; a 180° sector is a semicircle.

Let the centre be O, radius r and central angle θ°. Denote the corresponding arc length by l.

Arc and chord are different: An arc is part of the curved circumference; a chord is the straight segment joining its endpoints. Use the arc length in the following formulas.

⭐ When the Central Angle Is in Degrees

A full circle subtends 360° at its centre. A sector of angle θ° is therefore of the full circle.

🔹 Arc length:

🔹 Sector area:

🔹 When the arc length is known:

🔹 Sector perimeter = Arc length + 2 radii:

This perimeter formula applies to sectors with angles greater than 0° and less than 360°. Do not add 2 radii to the circumference of a full circle.

⭐ Extra Insight: Angles in Radians

If the same angle has degree measure θ and radian measure φ:

Then:

Remember: In l = rφ, the angle must be in radians. Do not substitute a degree value directly.

👉 Square Inscribed in a Circle

Square inscribed in a circle

All 4 vertices of an inscribed square lie on the circle.

🔹 Diagonal of the inscribed square = Circle diameter = 2r.

🔹 Square area = = = 2r²

🔹 Square perimeter = 4 × √2 × r = 4 √2 r

👉 Square Circumscribing a Circle

Square circumscribing a circle

For a square whose 4 sides touch a circle lying inside it, Side of square = Circle diameter = 2r.

🔹 Circumscribed square area = (2r)² = 4r²

🔹 Circumscribed square perimeter = 4 × 2r = 8r

👉 Incircle of a Square

The circle inside a square that touches all 4 sides is its incircle.

Let the side of the square = a.

Incircle of a square

🔹 Inradius r =

🔹 Circumference = πa

🔹 Area =

👉 Circumcircle of a Square

The circle passing through all 4 vertices of a square is its circumcircle. The side of the square is again a.

Circumcircle of a square

🔹 Circumcircle diameter = Square diagonal = a × √2

🔹 Circumcircle area = =

🔹 Circumcircle circumference = 2π × = πa × √2 = √2πa

👉 Incircle of an Equilateral Triangle

Let the inradius of equilateral triangle ABC be = r.

Incircle of an equilateral triangle

In an equilateral triangle, the incentre, circumcentre and centroid coincide. The centroid divides a median in the ratio 2 : 1 from the vertex. Each altitude is also a median, so for height h, the inradius is r = h/3.

🔹 Triangle height AD = h = 3r.

🔹 Triangle side = a =

🔹 Triangle area =

🔹 Triangle perimeter

👉 Circumcircle of an Equilateral Triangle

The distance from the circumcentre to a vertex is of the height. Thus, using R = 2h/3 and r = h/3:

🔹 Circumradius of the equilateral triangle R = 2r.

🔹 Circumcircle area = πR² = 4πr²

👉 Simple Solved Examples

Calculation rule: Where 22/7 is used for π, the numerical answers are based on that approximation. Use the value specified in the question; otherwise, you may leave an exact answer in terms of π.

✏️ Example 1: Circumference and Area of a Circle

A circle has radius 7 cm. Find its circumference and area. Use .

Solution:

Circumference = 2π × 7 = 14π cm.

cm.

Area = π × 7² = 49π cm².

cm².

Answer: Circumference ≈ 44 cm and area ≈ 154 cm².

✏️ Example 2: Area from Diameter

A circle has diameter 10 cm. Find its area.

Solution: Radius = 10 ÷ 2 = 5 cm.

Area = π × 5² = 25π cm².

Answer: 25π cm². Using π ≈ 3.14, the area is approximately 78.5 cm².

Note: Substitute the radius 5, not the diameter 10, in the area formula.

✏️ Example 3: Area and Perimeter of a Semicircle

A semicircle has radius 7 cm. Find its area and perimeter. Use .

Solution:

Area = cm².

Semicircular arc length = 7π ≈ 22 cm.

Perimeter = Arc length + Diameter = 7π + 14 ≈ 36 cm.

Answer: Area ≈ 77 cm² and perimeter ≈ 36 cm. The curved part alone is approximately 22 cm long.

✏️ Example 4: Area of a Circular Ring

Two concentric circles have outer radius 7 cm and inner radius 3 cm. Find the ring's width and area.

Solution: Width = 7 − 3 = 4 cm.

Area = π(7² − 3²) = π(49 − 9) = 40π cm².

Answer: Width = 4 cm, area = 40π cm².

Remember: The area is not π(R − r)²; the correct formula is π(R² − r²).

✏️ Example 5: Area and Perimeter of a Semicircular Ring

A semicircular ring has outer radius 5 cm and inner radius 3 cm. Find its area and perimeter.

Solution:

Area = cm².

Perimeter = π(5 + 3) + 2(5 − 3) = 8π + 4 cm.

Answer: Area = 8π cm², perimeter = (8π + 4) cm.

✏️ Example 6: Arc Length, Area and Perimeter of a Sector

A circle has radius 14 cm. For a sector with central angle 90°, find the arc length, area and perimeter. Use .

Solution: The sector is 90/360 = 1/4 of the full circle.

Arc length:

cm.

Area:

cm².

Perimeter = l + 2r = 7π + 28 ≈ 50 cm.

Answer: Arc length ≈ 22 cm, area ≈ 154 cm², perimeter ≈ 50 cm.

✏️ Example 7: Inscribed and Circumscribed Squares

A circle has radius 5 cm. Find the areas of its inscribed square and the square circumscribing it.

Solution:

Inscribed square area = 2r² = 2 × 5² = 50 cm².

Circumscribed square area = 4r² = 4 × 5² = 100 cm².

Answer: The inner square has area 50 cm², and the outer square has area 100 cm².

✏️ Example 8: Incircle and Circumcircle of an Equilateral Triangle

An equilateral triangle has inradius 2 cm. Find its height, side length and circumradius.

Solution:

Height = 3r = 3 × 2 = 6 cm.

Side = cm.

Circumradius = R = 2r = 4 cm.

Answer: Height = 6 cm, side = cm, circumradius = 4 cm.

🧠 Important Points to Remember

⭐ Diameter and radius: d = 2r. Use the radius in the area formula.

⭐ Units: Express circumference, arc length and perimeter in cm or m; express area in cm² or m².

⭐ Same units: Convert all lengths to the same unit before substitution.

⭐ Value of π: 22/7 and 3.14 are approximations. Leave π in your answer when an exact value is required.

⭐ Semicircle perimeter: Add the diameter to the curved length.

⭐ Ring area: Subtract the inner circle's area from the outer circle's area.

⭐ Sector perimeter: Add 2 radii to the arc length. Arc length alone is not the perimeter.

⭐ Degrees and radians: θ is the degree measure and φ the radian measure. Do not interchange their numerical values.

⭐ Equilateral triangle: R = 2r holds for an equilateral triangle. Do not assume this equality for every triangle.

✍️ Try It Yourself

Give exact answers in terms of π; decimal values are not required.

🔹 1. A circle has radius 3 cm. Find its circumference and area.

🔹 2. A semicircle has radius 4 cm. Find its area and perimeter.

🔹 3. A circular ring has outer radius 6 cm and inner radius 4 cm. Find its width and area.

🔹 4. A sector has radius 6 cm and central angle 60°. Find its arc length, area and perimeter.

🔹 5. A square has side 8 cm. Find its incircle's radius and area.

🔹 6. An equilateral triangle has inradius 3 cm. Find its circumcircle's radius and area.

Check Your Answers

🔸 1. Circumference 6π cm; area 9π cm².

🔸 2. Area 8π cm²; perimeter (4π + 8) cm.

🔸 3. Width 2 cm; area 20π cm².

🔸 4. Arc length 2π cm; area 6π cm²; perimeter (2π + 12) cm.

🔸 5. Radius 4 cm; area 16π cm².

🔸 6. Circumradius 6 cm; area 36π cm².

💡 Quick Revision

⭐ Circle circumference = 2πr

⭐ Circle area = πr²

⭐ Semicircle perimeter = πr + 2r

⭐ Ring area = π(R + r)(R − r)

⭐ Sector arc length = × 2πr

⭐ Sector area = × πr²

⭐ Inscribed square area = 2r²

⭐ Circumscribed square area = 4r²

⭐ Sector perimeter = l + 2r (central angle greater than 0° and less than 360°).

⭐ For an equilateral triangle, circumradius R = 2r, where r is the inradius.