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Areas Related to Circles

NCERT Class 10 · Mathematics Based on NCERT Class 10 Mathematics textbook · Free CBSE study kit

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Chapter Notes

Overview
This chapter extends your knowledge of circles to calculate areas of sectors and segments, which are parts of a circle bounded by radii and arcs or chords and arcs. It exists in the syllabus to apply the unitary method to real-life problems involving circular regions, such as grazing fields, clock hands, and umbrella ribs. The one big idea is that the area of a sector is a fraction of the whole circle's area, determined by the angle at the centre, and the area of a segment is found by subtracting the area of the corresponding triangle from the sector area.
Key Concepts (10)
Sector of a circle
The portion (or part) of the circular region enclosed by two radii and the corresponding arc is called a sector of the circle.
e.g. A slice of pizza cut from a round pizza is a sector of the circle.
Angle of the sector
The angle subtended at the centre by the two radii of a sector, denoted as ∠AOB, is called the angle of the sector.
e.g. In a clock, the angle swept by the minute hand in 5 minutes is 30°, which is the angle of the sector formed.
Minor sector
The sector corresponding to the smaller angle (less than 180°) is called the minor sector.
e.g. In a circle divided by two radii forming a 60° angle, the smaller region is the minor sector.
Major sector
The sector corresponding to the larger angle (greater than 180°) is called the major sector. Its angle is 360° – angle of the minor sector.
e.g. If the minor sector has a 60° angle, the major sector has a 300° angle.
Segment of a circle
The portion (or part) of the circular region enclosed between a chord and the corresponding arc is called a segment of the circle.
e.g. The region between a chord of a circle and its arc, like the area under a bowstring, is a segment.
Minor segment
The segment formed by the smaller arc (corresponding to the minor arc) is called the minor segment.
e.g. In a circle with a chord subtending a 60° angle at the centre, the smaller region between the chord and the arc is the minor segment.
Major segment
The segment formed by the larger arc (corresponding to the major arc) is called the major segment.
e.g. The larger region between the same chord and the major arc is the major segment.
Unitary method
A method used to derive the area of a sector by first finding the area for 1° angle (πr²/360) and then multiplying by the given angle θ.
e.g. To find the area of a 30° sector of radius 4 cm, first find area for 1° = (3.14 × 16)/360, then multiply by 30.
Arc of a sector
The curved part of the boundary of a sector, which is a part of the circumference of the circle.
e.g. The outer edge of a pizza slice is the arc of the sector.
Chord
A line segment joining two points on the circumference of a circle, which forms the straight boundary of a segment.
e.g. The straight line connecting the two ends of a bowstring is a chord.
Formulas & Equations (5)
Area of the sector of angle θ = (θ/360) × πr²
Reads as: Area of the sector of angle theta equals theta divided by 360 multiplied by pi r squared
Variables: θ = angle of the sector in degrees, r = radius of the circle (in cm or m), π = constant (22/7 or 3.14)
Use when: Use this formula when you need to find the area of a sector given its central angle and radius.
⚠ Students often forget to convert the angle to degrees if given in radians, or they use the wrong fraction (e.g., θ/360 instead of θ/360).
Length of an arc of a sector of angle θ = (θ/360) × 2πr
Reads as: Length of an arc of a sector of angle theta equals theta divided by 360 multiplied by 2 pi r
Variables: θ = angle of the sector in degrees, r = radius of the circle (in cm or m), π = constant (22/7 or 3.14)
Use when: Use this formula when you need to find the length of the arc of a sector given its central angle and radius.
⚠ Students often confuse this with the area formula and use πr² instead of 2πr, or they forget to multiply by 2.
Area of the segment APB = Area of sector OAPB – Area of ΔOAB
Reads as: Area of the segment APB equals area of sector OAPB minus area of triangle OAB
Variables: O = centre of circle, A and B = endpoints of chord, P = point on arc, r = radius, θ = angle of sector
Use when: Use this formula when you need to find the area of a minor segment formed by a chord and its corresponding arc.
⚠ Students forget to subtract the area of the triangle or incorrectly calculate the triangle's area, especially when the triangle is not right-angled.
Area of major sector OAQB = πr² – Area of minor sector OAPB
Reads as: Area of major sector OAQB equals pi r squared minus area of minor sector OAPB
Variables: r = radius of the circle, π = constant (22/7 or 3.14)
Use when: Use this formula when you need to find the area of the major sector given the minor sector area.
⚠ Students sometimes subtract the wrong way or forget that the major sector area is the whole circle minus the minor sector.
Area of major segment AQB = πr² – Area of minor segment APB
Reads as: Area of major segment AQB equals pi r squared minus area of minor segment APB
Variables: r = radius of the circle, π = constant (22/7 or 3.14)
Use when: Use this formula when you need to find the area of the major segment given the minor segment area.
⚠ Students often confuse this with the major sector formula and subtract the sector area instead of the segment area.
How It Works
Deriving the area of a sector using the unitary method
  1. Step 1: Recall that the area of the whole circle (angle 360°) is πr².
  2. Step 2: For an angle of 1°, the area of the sector is πr²/360.
  3. Step 3: For an angle of θ°, the area of the sector is (θ/360) × πr².
💡 The unitary method works because the area of a sector is directly proportional to its central angle; always divide by 360 first.
Deriving the length of an arc of a sector using the unitary method
  1. Step 1: Recall that the circumference of the whole circle (angle 360°) is 2πr.
  2. Step 2: For an angle of 1°, the length of the arc is 2πr/360.
  3. Step 3: For an angle of θ°, the length of the arc is (θ/360) × 2πr.
💡 The arc length is also proportional to the angle; use 2πr, not πr², because it is a length, not an area.
Finding the area of a segment of a circle
  1. Step 1: Identify the sector OAPB formed by the chord AB and the radii OA and OB.
  2. Step 2: Calculate the area of the sector OAPB using the formula (θ/360) × πr².
  3. Step 3: Calculate the area of triangle OAB. For this, draw a perpendicular from O to AB, find its length using trigonometric ratios (e.g., cos and sin of half the angle), then use area = (1/2) × base × height.
  4. Step 4: Subtract the area of triangle OAB from the area of sector OAPB to get the area of the segment APB.
💡 The triangle OAB is isosceles (OA = OB = r), so the perpendicular from O bisects AB and the angle at O. Use cos and sin of half the angle to find the height and half the base.
Finding the area of a major sector or major segment
  1. Step 1: Calculate the area of the whole circle: πr².
  2. Step 2: Subtract the area of the minor sector to get the area of the major sector.
  3. Step 3: Subtract the area of the minor segment to get the area of the major segment.
💡 The major sector and major segment are always the larger parts; they are found by subtracting the minor parts from the total area of the circle.
Exam Traps — Don't Make These Mistakes
⚠ Using the wrong formula for arc length vs. sector area
✗ Students write 'Area of sector = (θ/360) × 2πr' or 'Length of arc = (θ/360) × πr²'.
→ The area formula uses πr² (area of whole circle), while the arc length formula uses 2πr (circumference). Swapping them gives incorrect units (length vs. area).
✓ For area of sector: (θ/360) × πr². For length of arc: (θ/360) × 2πr.
⚠ Forgetting to subtract the triangle area for a segment
✗ Students directly write 'Area of segment = (θ/360) × πr²' without subtracting the triangle area.
→ The segment is the region between the chord and the arc; the sector includes the triangle, so the triangle area must be subtracted to get the segment area.
✓ Area of segment = Area of sector – Area of triangle OAB.
⚠ Incorrectly calculating the triangle area in a segment problem
✗ Students write 'Area of ΔOAB = (1/2) × r × r × sin θ' but use θ instead of half the angle for the perpendicular.
→ In an isosceles triangle with vertex angle θ, the perpendicular from the centre bisects the angle, so the height is r × cos(θ/2) and half the base is r × sin(θ/2). Using sin θ directly gives the wrong height.
✓ Draw OM perpendicular to AB. Then OM = r cos(θ/2) and AM = r sin(θ/2). Area = (1/2) × AB × OM = (1/2) × (2r sin(θ/2)) × (r cos(θ/2)) = r² sin(θ/2) cos(θ/2).
⚠ Confusing minor and major sector/segment
✗ Students write 'Area of major sector = (θ/360) × πr²' where θ is the minor angle.
→ The major sector corresponds to the larger angle (360° – θ), not the minor angle θ. Using θ gives the minor sector area.
✓ Area of major sector = πr² – Area of minor sector, or use (360° – θ)/360 × πr².
⚠ Using π = 22/7 when the problem specifies π = 3.14
✗ Students use π = 22/7 even when the question says 'Use π = 3.14'.
→ The question explicitly states which value of π to use; using the wrong one gives a different numerical answer and loses marks.
✓ Always check the problem statement: if it says 'Use π = 22/7', use that; if it says 'Use π = 3.14', use that. If not specified, you may use either, but 22/7 is often preferred for exactness.
Compare & Contrast
Definition and boundaries
Sector of a circle
Segment of a circle
A sector is enclosed by two radii and the corresponding arc, while a segment is enclosed by a chord and the corresponding arc.
Angle at the centre
Minor sector
Major sector
The minor sector has an angle less than 180°, while the major sector has an angle greater than 180° (360° – angle of minor sector).
What is measured
Length of an arc
Area of a sector
Length of an arc is a linear measure (in cm or m) of the curved boundary, while area of a sector is a measure of the region (in cm² or m²) enclosed by the radii and the arc.
Exam Focus
Marks pattern: This chapter typically appears in board exams as: 1 MCQ (1 mark) on sector/segment area, 1 short answer (2-3 marks) on arc length or sector area, and 1 long answer (5 marks) on segment area or combined figure problems.
Most asked: Area of a sector given radius and angle · Area of a segment (especially with angles 60°, 90°, 120°) · Length of an arc · Problems combining sectors and triangles (e.g., horse grazing, clock hands, umbrella ribs)
Don't skip: Area of a segment (requires both sector and triangle area calculation); Word problems involving real-life applications (e.g., horse grazing, brooch, umbrella, lighthouse)
One-Line Revision (read aloud before the exam)
  1. A sector is the region enclosed by two radii and the corresponding arc.
  2. A segment is the region enclosed by a chord and the corresponding arc.
  3. The minor sector has an angle less than 180°, and the major sector has an angle greater than 180°.
  4. Area of a sector = (θ/360) × πr², where θ is the angle in degrees.
  5. Length of an arc = (θ/360) × 2πr.
  6. Area of a segment = Area of the corresponding sector – Area of the corresponding triangle.
  7. Area of the major sector = πr² – Area of the minor sector.
  8. Area of the major segment = πr² – Area of the minor segment.
  9. In an isosceles triangle OAB, the perpendicular from O bisects AB and the angle at O.
  10. Use trigonometric ratios (cos and sin of half the angle) to find the height and base of triangle OAB.
  11. Always check whether to use π = 22/7 or π = 3.14 as specified in the question.
  12. The unitary method is used to derive both sector area and arc length formulas.
  13. A quadrant is a sector with angle 90° (one-fourth of a circle).
  14. The angle of the major sector is 360° – angle of the minor sector.

MCQs — 10 Questions with Answers

Q1. A student claims that if a chord subtends an angle of 90° at the centre of a circle, then the area of the minor segment is always equal to the area of the minor sector minus the area of the right triangle formed. Which concept does this claim rely on?

  • A. The definition of a segment as the region between a chord and its arc ✓
  • B. The property that all sectors with 90° angle are congruent
  • C. The fact that the triangle formed is always isosceles
  • D. The relationship between arc length and radius

Answer: A — The claim correctly applies the definition: segment area = sector area − triangle area; this is the fundamental relationship from the chapter, not about congruence or triangle properties.

Q2. A rope of length 7 m is tied to a corner of a square field. The horse grazes a quarter-circle of radius 7 m. A teacher says this area represents a 'sector' while another says it is a 'segment'. Who is using the correct term and why?

  • A. The first teacher is correct because a sector is bounded by two radii and an arc ✓
  • B. The second teacher is correct because the region is bounded by a rope and the field edges
  • C. Both are correct depending on which boundary you emphasise
  • D. Neither is correct; this is neither a sector nor a segment

Answer: A — The grazing area is bounded by two radii (the two sides of the corner) and an arc, which defines a sector; a segment requires a chord, which is not present here.

Q3. Assertion (A): The area of a major sector is always greater than the area of its corresponding minor sector. Reason (R): A major sector has a central angle greater than 180°. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A ✓
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false but R is true

Answer: A — Both statements are true: a major sector has angle > 180° (so < 180° for minor), and since area = (θ/360°)πr², larger angle directly implies larger area; R explains why A holds.

Q4. A construction engineer designs a circular tank and marks an arc of 60° from the centre. She needs to find the area between the chord and the arc. Which formula should she use?

  • A. Area of sector − area of triangle ✓
  • B. Area of circle − area of sector
  • C. Arc length × radius
  • D. Radius squared × angle in radians

Answer: A — The region between a chord and its arc is a segment, defined as sector area minus the triangular area formed by the two radii and chord; option B gives the complementary (major) segment.

Q5. Assertion (A): For a sector with central angle θ, if θ is doubled, the area is also doubled. Reason (R): Sector area is directly proportional to the central angle. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A ✓
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false but R is true

Answer: A — Both are true: sector area = (θ/360°)πr², so doubling θ doubles the area; R directly explains A because area is proportional (linear relationship) to angle.

Q6. A student observes that for a given radius, the area of a 120° sector equals the area of a circle of radius r/√2. Is this observation mathematically correct?

  • A. Yes, because (120/360) × πr² = π(r/√2)² ✓
  • B. No, because (120/360) × πr² = (1/3)πr², not (1/2)πr²
  • C. Yes, only if the radius of the smaller circle is adjusted
  • D. No, because sector area cannot equal any full circle area

Answer: A — Sector area = (120/360)πr² = (1/3)πr²; circle area = π(r/√2)² = (1/2)πr²; these are NOT equal, so the observation is false; option A appears to affirm it but the math shows they differ.

Q7. A farmer ties a 10 m rope to one corner of a rectangular field. The grazing region forms a quarter-circle. If the farmer later moves the tying point to the middle of a side, what happens to the grazing area?

  • A. It increases because the rope can now sweep in a semicircle ✓
  • B. It remains the same because the rope length is unchanged
  • C. It decreases because the corner restricts less than a side's geometry
  • D. It depends on the field's dimensions

Answer: A — From a corner, only a quarter-circle (90°) can be swept; from a side midpoint (not bounded by two edges meeting at 90°), a semicircle (180°) can be swept; with the same radius, sector area increases with larger central angle.

Q8. Assertion (A): A chord that subtends a 60° angle at the centre creates a minor segment whose area equals (π/6 − √3/4)r². Reason (R): The segment area formula subtracts the equilateral triangle area from the 60° sector area. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false ✓
  • D. A is false but R is true

Answer: C — The segment area formula (sector − triangle) is correct, but a 60° sector with two equal radii forms an isosceles triangle, NOT equilateral; the triangle area is (1/2)r²sin(60°) = (√3/4)r², making A true but R's reasoning about the triangle type is false.

Q9. Two circles have radii 7 cm and 14 cm. A sector of 90° is marked in each. A student claims the larger circle's sector is exactly twice the smaller one. Evaluate this claim using the sector area formula.

  • A. The claim is correct because doubling the radius doubles the sector area
  • B. The claim is incorrect because doubling the radius quadruples the sector area ✓
  • C. The claim is correct only if the angle remains 90° in both
  • D. The claim cannot be evaluated without knowing the angle in the larger circle

Answer: B — Sector area ∝ r²: for radius 7, area = (90/360)π(7)² = (49π/4); for radius 14, area = (90/360)π(14)² = (196π/4) = 4 × (49π/4); doubling radius quadruples area, not doubles it.

Q10. Assertion (A): The area of a major segment is always larger than the area of the minor segment for the same chord. Reason (R): Major and minor segments are complementary regions that together form the complete circle. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A ✓
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false but R is true

Answer: A — Both statements are true: major and minor segments partition the circle (R is true), and since the major segment's arc spans > 180°, its area exceeds the minor segment's (A is true); R correctly explains why A holds.

Flashcards

What is the formula for the area of a sector with angle θ (in degrees) and radius r?

Area of sector = (θ/360) × πr², where θ is the angle in degrees and r is the radius.

How do you find the length of an arc in a sector of angle θ?

Arc length = (θ/360) × 2πr, using the same unitary method as sector area.

Define a sector of a circle.

A sector is the region enclosed between two radii and the arc connecting their endpoints.

Define a segment of a circle.

A segment is the region enclosed between a chord and the arc it cuts off.

What is the formula for the area of a segment?

Area of segment = Area of sector − Area of triangle formed by the two radii and chord.

What is the relationship between minor and major sectors?

Minor sector angle + Major sector angle = 360°, and Minor sector area + Major sector area = πr².

If a chord subtends a right angle (90°) at the centre, what is the area of the minor sector?

Area of minor sector = (90/360) × πr² = (1/4)πr², which is called a quadrant.

How do you find the area of a triangle OAB when the two radii OA and OB and angle AOB are known?

Area of triangle OAB = (1/2) × OA × OB × sin(∠AOB) = (1/2)r² sin(θ).

What happens to sector area when the angle is 360°?

When angle = 360°, the sector becomes the entire circle, so sector area = πr².

Why is the unitary method used to derive the sector area formula?

Because a full circle (360°) has known area πr², so any angle θ has proportional area (θ/360) × πr².

Important Board Questions

Define sector and segment of a circle. State the formula for the area of a sector with angle θ (in degrees) and radius r. [2 marks]

A sector is bounded by two radii and an arc; a segment is bounded by a chord and an arc. Formula: Area = (θ/360) × πr².

A chord of a circle with radius 21 cm subtends an angle of 60° at the centre. Find the area of the minor segment. (Use π = 22/7 and √3 = 1.73) [3 marks]

Calculate sector area using (θ/360) × πr², then find triangle area = (1/2)r² sin(60°), and subtract to get segment area = Sector − Triangle.

A chord of a circle with radius 15 cm subtends an angle of 60° at the centre. Find the areas of both the minor and major segments. Also, verify that both segments together equal the circle's area. (Use π = 3.14 and √3 = 1.73) [5 marks]

For minor segment: compute sector area (θ/360)πr², triangle area = (1/2)r² sin(θ), then subtract. For major segment: use total area πr² minus minor segment. Verify: Minor + Major segment = πr².

Assertion-Reason Questions for Areas Related to Circles

Q1. Assertion (A):

The area of a sector with angle 180° and radius r equals half the area of the circle.

Reason (R):

A sector of angle 180° is a semicircle, and its area is (180°/360°) × πr² = (1/2)πr², which is exactly half the circle's area πr².

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true. The reason correctly explains the assertion by showing that 180° is exactly half of 360°, making the sector area exactly half the circle area. The mathematical calculation confirms this.

Q2. Assertion (A):

The length of an arc is directly proportional to the central angle of the sector.

Reason (R):

The arc length formula is l = (θ/360°) × 2πr, showing that as θ increases, the arc length increases proportionally.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason perfectly explains the assertion. The formula clearly shows the direct proportionality relationship between arc length and central angle θ.

Q3. Assertion (A):

The area of a minor segment equals the area of its corresponding sector minus the area of the triangle formed by the two radii and the chord.

Reason (R):

A segment is the region between a chord and an arc, while a sector includes the triangle portion, so removing the triangle from the sector gives the segment.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true. The reason correctly explains why we subtract the triangle area from the sector area. The geometric relationship is accurately described.

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Case-Based Questions (CBQ) for Areas Related to Circles

Case Study:

A city garden is designed with a circular flower bed of radius 14 m. The garden designer wants to create different sections. One section is a sector with central angle 90° where sunflowers are planted. Another section is a segment created by a chord, with the chord subtending a central angle of 120° where roses are planted. The remaining area has grass. The designer needs to calculate the area for each section to determine the required quantity of seeds and fertilizers.

Q1. What is the area of the sunflower sector (90°)? (Use π = 22/7)

  • A. 154 cm²
  • B. 154 m² ✓
  • C. 308 m²
  • D. 77 m²

Answer: B

Q2. Calculate the area of the sector where roses are planted (120° angle). (Use π = 22/7)

  • A. 205.33 m² ✓
  • B. 102.67 m²
  • C. 410.67 m²
  • D. 51.33 m²

Answer: A

Q3. For the rose sector (120°), if we need to find just the segment area (between chord and arc), we need to subtract the triangle area. What is the area of the triangle OAB formed by the two radii and chord? (Use √3 = 1.73)

  • A. 84.7 m² ✓
  • B. 169.4 m²
  • C. 42.35 m²
  • D. 254.1 m²

Answer: A

Q4. What is the area of the grass section? Total garden area = πr². (Use π = 22/7)

  • A. 308 m²
  • B. 384.67 m² ✓
  • C. 462 m²
  • D. 76 m²

Answer: B

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Previous Year Questions (PYQ) for Areas Related to Circles

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 31/4/1 · 1 mark

Chord AB subtends an angle of 40° at the centre of the circle of radius 9 cm. The length of arc AB is:

  • 22/7 cm
  • 198/7 cm
  • 44/7 cm
  • 54/7 cm

CBSE 2026 · Set 31/4/1 · 3 marks

A chord of a circle of radius 14 cm subtends a right angle at the centre. Find the area of the corresponding (i) minor segment (ii) major segment.

CBSE 2026 · Set 31/4/3 · 1 mark

An arc of length 11 cm subtends an angle of 105° at the centre of the circle. The radius of the circle is:

  • 8 cm
  • 4√3 cm
  • 6 cm
  • 7 cm

CBSE 2026 · Set 31/4/3 · 3 marks

A circle of radius 12 cm is divided into 11 identical sectors. Find the total area of 7 such sectors.

CBSE 2026 · Set 31/5/2 · 4 marks

In a circular museum hall of radius 14 m, some statues are displayed. Statues are kept inside the inner concentric circle of radius 7 m. One such statue lying in sector OAB, is fenced along line segments OA, AP, PB and BO where P is a point on outer circle.

(i) Find m∠AOP.

CBSE 2026 · Set 31/5/3 · 3 marks

Chord AB of a circle subtends an angle of 120° at the centre O of the circle. Find the length of arc AB, if radius of the circle is 21 cm.

CBSE 2026 · Set 31/5/3 · 1 mark

In a circular museum hall of radius 14 m, some statues are displayed. Statues are kept inside the inner concentric circle of radius 7 m. One such statue lying in sector OAB, is fenced along line segments OA, AP, PB and BO where P is a point on outer circle.

Find m∠AOP.

CBSE 2026 · Set 31/1164/30 · 1 mark

The perimeter of sector of a circle of radius 21 cm and central angle 60, is

  • 22 cm
  • 44 cm
  • 64 cm
  • 273 cm

CBSE 2026 · Set 31/1164/30 · 1 mark

Assertion (A) : In a circle of radius 21 cm, an arc of length 22 cm subtends an angle of 60° at the centre. Reason (R) : The length of arc of a sector of a circle of radius r and central angle θ is 2πrθ/360.

  • Both, Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
  • Both, Assertion (A) and Reason (R) are true, but Reason (R) is not correct explanation for Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Assertion (A) is false, but Reason (R) is true.

CBSE 2026 · Set 31/1164/30 · 2 marks

A chord AB of a circle of radius 14 cm makes a right angle at the centre of the circle. Find the area of the minor segment.

CBSE 2026 · Set 31/1167/430 · 1 mark

The area (in cm²) of a sector of a circle of radius 14 cm cut off by an arc of length 22 cm is:

  • 77
  • 308
  • 154
  • 462

CBSE 2026 · Set 31/1167/430 · 1 mark

The minute hand of a clock is 21 cm long. The distance covered by the tip of minute hand from 2:10 pm to 2:25 pm is:

  • 346.5 cm
  • 33 cm
  • 66 cm
  • 16.5 cm

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