CBSE Important Questions Class 9 Maths Chapter 6: Measuring Space Perimeter and Area

Perimeter measures the total boundary length of a shape, while area measures the space enclosed by it. A circle with radius r has circumference 2πr and area πr².

Measurement becomes powerful when students connect boundary length, enclosed space, and formulas through real shapes. CBSE Important Questions Class 9 Maths Chapter 6 help students practise circle circumference, arc length, area of triangles, Heron’s formula, area of circles, sectors, segments, and cyclic 4-gons. The CBSE 2026 chapter also includes π as an irrational number, athletics track staggers, Baudhāyana’s construction, Brahmagupta’s formula, and circle-based puzzles from NCERT Ganita Manjari.

Key Takeaways

• π: The ratio C/D is constant for every circle and π ≈ 22/7 for most NCERT exercises.
• Arc Length: An arc subtending θ° at the centre has length 2πr × θ/360.
• Heron’s Formula: A triangle with sides a, b, c has area √[s(s − a)(s − b)(s − c)].
• Sector Area: A sector subtending θ° at the centre has area πr² × θ/360.

CBSE Important Questions Class 9 Maths Chapter 6 Structure 2026

CBSE Important Questions Class 9 Maths Chapter 6 with Answers

The chapter moves from simple perimeter to curved boundaries and areas.
Students should know when to use 2πr, πr², Heron’s formula, or sector formulas.
These class 9 maths chapter 6 important questions follow the NCERT 2026 flow.

1. What does CBSE Important Questions Class 9 Maths Chapter 6 mainly test?

CBSE Important Questions Class 9 Maths Chapter 6 mainly test perimeter, circumference, arc length, area, Heron’s formula, sectors, and cyclic 4-gons. The chapter links formulas with geometric reasoning.

1. Perimeter Skill: Add boundary lengths.
2. Circle Skill: Use C = 2πr.
3. Area Skill: Use triangle, parallelogram, circle, and sector formulas.
4. Advanced Skill: Apply Heron’s and Brahmagupta’s formulas.
5. Final Result: The chapter tests measurement of curved and straight-edged shapes.

2. What is perimeter of a shape?

Perimeter is the total length around the border of a shape. It is the distance covered in one complete walk around the boundary.

1. Square: Perimeter = 4a.
2. Equilateral Triangle: Perimeter = 3a.
3. Rectangle: Perimeter = 2(a + b).
4. Final Result: Perimeter measures boundary length.

3. Why does Chapter 6 start with a relay race track?

Chapter 6 starts with a relay race track because outer lanes have longer curved paths. Staggers balance this extra distance.

1. Inner Lane: Smaller curved radius.
2. Outer Lane: Larger curved radius.
3. Need: Fair starting positions.
4. Final Result: Stagger depends on circular arc length.

Class 9 Maths Chapter 6 Perimeter and Area Questions

Perimeter measures boundary, while area measures enclosed region.
Straight shapes use side lengths, but curved shapes need π-based formulas.
These class 9 maths chapter 6 perimeter and area questions build the base for later examples.

4. What is the perimeter of a square of side 12 cm?

The perimeter is 48 cm. A square has four equal sides.

1. Given Data: Side = 12 cm.
2. Formula Used: Perimeter = 4a.
3. Calculation:
   Perimeter = 4 × 12
   Perimeter = 48 cm
4. Final Result: Perimeter = 48 cm.

5. What is the perimeter of a rectangle of length 18 cm and breadth 7 cm?

The perimeter is 50 cm. A rectangle has two lengths and two breadths.

1. Given Data:
   Length = 18 cm
   Breadth = 7 cm
2. Formula Used: Perimeter = 2(l + b).
3. Calculation:
   Perimeter = 2(18 + 7)
   Perimeter = 50 cm
4. Final Result: Perimeter = 50 cm.

6. What is the area of a rectangle of sides 15 cm and 9 cm?

The area is 135 cm². Rectangle area equals length × breadth.

1. Given Data:
   Length = 15 cm
   Breadth = 9 cm
2. Formula Used: Area = l × b.
3. Calculation:
   Area = 15 × 9
   Area = 135 cm²
4. Final Result: Area = 135 cm².

7. What is the difference between perimeter and area?

Perimeter measures boundary length, while area measures covered space. Perimeter uses units, and area uses square units.

1. Perimeter Example: Boundary of a playground.
2. Area Example: Space inside the playground.
3. Units: cm for perimeter and cm² for area.
4. Final Result: Perimeter and area measure different quantities.

Circle Perimeter Class 9 Questions

The perimeter of a circle is called its circumference.
The constant ratio of circumference to diameter is π for every circle.
These circle perimeter class 9 questions focus on C = 2πr and C = πd.

8. What is circumference of a circle?

Circumference is the perimeter of a circle. Its formula is C = 2πr or C = πd.

1. Radius: r.
2. Diameter: d = 2r.
3. Formula Used: C = 2πr.
4. Final Result: Circumference is the boundary length of a circle.

9. Why is π called the C/D ratio?

π is called the C/D ratio because π = circumference/diameter. This ratio stays constant for all circles.

1. Circumference: C.
2. Diameter: D.
3. Ratio: C/D = π.
4. Final Result: π is the fixed circle ratio.

10. A circle has perimeter 44 cm. Find its radius.

The radius is 7 cm. Use C = 2πr and π = 22/7.

1. Given Data: C = 44 cm.
2. Formula Used: C = 2πr.
3. Calculation:
   44 = 2 × 22/7 × r
   r = 44 × 7/44
   r = 7 cm
4. Final Result: Radius = 7 cm.

11. Find circumference of a circle with radius 10 cm.

The circumference is 62.8 cm using π ≈ 3.14. Use C = 2πr.

1. Given Data: r = 10 cm.
2. Formula Used: C = 2πr.
3. Calculation:
   C = 2 × 3.14 × 10
   C = 62.8 cm
4. Final Result: Circumference = 62.8 cm.

12. If the ratio of perimeters of two circles is 5:4, find the ratio of their radii.

The ratio of their radii is 5:4. Circle perimeter is directly proportional to radius.

1. Formula Used: C = 2πr.
2. Given Ratio: C1:C2 = 5:4.
3. Since: C ∝ r.
4. Final Result: r1:r2 = 5:4.

Arc Length Class 9 Questions

An arc is a part of a circle’s circumference.
Its length depends on the full circumference and the central angle.
These arc length class 9 questions use the formula l = 2πr × θ/360.

13. What is the formula for arc length?

The formula for arc length is l = 2πr × θ/360. Here θ is the central angle in degrees.

1. Circle Circumference: 2πr.
2. Angle Fraction: θ/360.
3. Arc Length: l = 2πr × θ/360.
4. Final Result: Arc length is a fraction of circumference.

14. Find arc length when radius is 3.5 cm and central angle is 60°.

The arc length is 11/3 cm. Use π = 22/7.

1. Given Data:
   r = 3.5 cm
   θ = 60°
2. Formula Used: l = 2πr × θ/360.
3. Calculation:
   l = 2 × 22/7 × 3.5 × 60/360
   l = 22 × 1/6
   l = 11/3 cm
4. Final Result: Arc length = 11/3 cm.

15. Find arc length when radius is 6.3 m and central angle is 120°.

The arc length is 13.2 m. Use π = 22/7.

1. Given Data:
   r = 6.3 m
   θ = 120°
2. Formula Used: l = 2πr × θ/360.
3. Calculation:
   l = 2 × 22/7 × 6.3 × 120/360
   l = 39.6 × 1/3
   l = 13.2 m
4. Final Result: Arc length = 13.2 m.

16. What is the length of a semicircular arc of radius r?

The length is πr. A semicircle is half of a full circle.

1. Full Circumference: 2πr.
2. Semicircle Fraction: 1/2.
3. Calculation: 2πr × 1/2 = πr.
4. Final Result: Semicircular arc length = πr.

17. What is the length of a quarter-circle arc of radius r?

The length is πr/2. A quarter circle is one-fourth of a full circle.

1. Full Circumference: 2πr.
2. Quarter Fraction: 1/4.
3. Calculation: 2πr × 1/4 = πr/2.
4. Final Result: Quarter-circle arc length = πr/2.

Area of Triangle Class 9 Questions

Triangle area depends on base and perpendicular height.
A median divides a triangle into two triangles with equal area.
These area of triangle class 9 questions cover formula use and geometric reasoning.

18. What is the area formula for a triangle?

The area of a triangle is A = 1/2 × base × height. The height must be perpendicular to the base.

1. Base: b.
2. Height: h.
3. Formula Used: A = 1/2 bh.
4. Final Result: Triangle area = 1/2 bh.

19. Find area of a triangle with base 10 cm and height 8 cm.

The area is 40 cm². Use A = 1/2 bh.

1. Given Data:
   b = 10 cm
   h = 8 cm
2. Formula Used: A = 1/2 bh.
3. Calculation:
   A = 1/2 × 10 × 8
   A = 40 cm²
4. Final Result: Area = 40 cm².

20. What does a median do to the area of a triangle?

A median divides a triangle into two triangles of equal area. Both smaller triangles have equal bases and the same height.

1. Median: Joins a vertex to midpoint of opposite side.
2. Base Split: Opposite side becomes two equal parts.
3. Height: Same for both triangles.
4. Final Result: A median bisects triangle area.

21. Why can two different-looking triangles have equal area?

Two different-looking triangles can have equal area when they share equal base and height. Shape alone does not decide area.

1. Area Formula: A = 1/2 bh.
2. Same Base: b remains equal.
3. Same Height: h remains equal.
4. Final Result: Equal base and height give equal area.

22. Find area of a right triangle with legs 3 cm and 4 cm.

The area is 6 cm². The perpendicular legs act as base and height.

1. Given Data:
   Base = 3 cm
   Height = 4 cm
2. Formula Used: A = 1/2 bh.
3. Calculation:
   A = 1/2 × 3 × 4
   A = 6 cm²
4. Final Result: Area = 6 cm².

Heron Formula Class 9 Questions

Heron’s formula finds triangle area when all three sides are known.
It removes the need to calculate height separately.
These Heron formula class 9 questions follow NCERT examples and exercise patterns.

23. What is Heron’s formula?

Heron’s formula is A = √[s(s − a)(s − b)(s − c)]. Here s is the semi-perimeter.

1. Sides: a, b, c.
2. Semi-perimeter: s = (a + b + c)/2.
3. Formula Used: A = √[s(s − a)(s − b)(s − c)].
4. Final Result: Heron’s formula uses only three sides.

24. Find area of a triangle with sides 3 cm, 4 cm, and 5 cm.

The area is 6 cm². Use Heron’s formula.

1. Given Data:
   a = 3 cm
   b = 4 cm
   c = 5 cm
2. Semi-perimeter:
   s = (3 + 4 + 5)/2
   s = 6 cm
3. Calculation:
   A = √[6(6 − 3)(6 − 4)(6 − 5)]
   A = √[6 × 3 × 2 × 1]
   A = 6 cm²
4. Final Result: Area = 6 cm².

25. Find area of an equilateral triangle of side a using Heron’s formula.

The area is √3a²/4. Heron’s formula gives the same result as the height method.

1. Given Data: a = b = c = a.
2. Semi-perimeter: s = 3a/2.
3. Formula Used:
   A = √[(3a/2)(a/2)(a/2)(a/2)]
4. Final Result: Area = √3a²/4.

26. Find area of a triangle with sides 8 cm, 11 cm, and 13 cm.

The area is 12√15 cm². Use Heron’s formula.

1. Given Data:
   a = 8 cm
   b = 11 cm
   c = 13 cm
2. Semi-perimeter:
   s = (8 + 11 + 13)/2
   s = 16 cm
3. Calculation:
   A = √[16(16 − 8)(16 − 11)(16 − 13)]
   A = √[16 × 8 × 5 × 3]
   A = √1920
   A = 12√15 cm²
4. Final Result: Area = 12√15 cm².

27. A triangle has perimeter 32 cm and two sides 8 cm and 11 cm. Find its area.

The area is 12√15 cm². The third side is 13 cm.

1. Given Data:
   Perimeter = 32 cm
   Side 1 = 8 cm
   Side 2 = 11 cm
2. Third Side:
   c = 32 − 8 − 11
   c = 13 cm
3. Semi-perimeter: s = 16 cm.
4. Calculation:
   A = √[16 × 8 × 5 × 3]
   A = 12√15 cm²
5. Final Result: Area = 12√15 cm².

Area of Circle Class 9 Questions

The area of a circle depends on the square of its radius.
NCERT explains this through slicing and rearranging a circle into a parallelogram-like shape.
These area of circle class 9 questions use A = πr².

28. What is the formula for area of a circle?

The formula for area of a circle is A = πr². Here r is the radius.

1. Radius: r.
2. Constant: π.
3. Formula Used: A = πr².
4. Final Result: Circle area equals πr².

29. Why is the area of a circle πr²?

The area becomes πr² because circle slices can form a parallelogram-like shape. Its base is πr and height is r.

1. Base: Half circumference = πr.
2. Height: Radius = r.
3. Area: Base × height = πr × r.
4. Final Result: Area of circle = πr².

30. Find area of a circle with radius 7 cm.

The area is 154 cm². Use π = 22/7.

1. Given Data: r = 7 cm.
2. Formula Used: A = πr².
3. Calculation:
   A = 22/7 × 7 × 7
   A = 154 cm²
4. Final Result: Area = 154 cm².

31. Find area of a quadrant of a circle with circumference 44 cm.

The area of the quadrant is 38.5 cm². First find the radius.

1. Given Data: C = 44 cm.
2. Radius:
   44 = 2 × 22/7 × r
   r = 7 cm
3. Circle Area:
   A = 22/7 × 7²
   A = 154 cm²
4. Quadrant Area: 154/4 = 38.5 cm².
5. Final Result: Quadrant area = 38.5 cm².

32. A bicycle wheel has diameter 60 cm. How far does it travel in 100 rotations?

The bicycle travels 188.4 m in 100 rotations. One rotation covers one circumference.

1. Given Data:
   d = 60 cm
   rotations = 100
2. Circumference:
   C = πd
   C = 3.14 × 60
   C = 188.4 cm
3. Distance:
   188.4 × 100 = 18840 cm
   18840 cm = 188.4 m
4. Final Result: Distance = 188.4 m.

Sector Area Class 9 Questions

A sector is the region enclosed by two radii and the included arc.
Its area is the same fraction of circle area as its angle is of 360°.
These sector area class 9 questions use area = πr² × θ/360.

33. What is the formula for area of a sector?

The formula is sector area = πr² × θ/360. Here θ is the central angle in degrees.

1. Circle Area: πr².
2. Angle Fraction: θ/360.
3. Formula Used: Sector area = πr² × θ/360.
4. Final Result: Sector area is an angle fraction of circle area.

34. Find area of a sector with radius 7 cm and angle 60°.

The sector area is 77/3 cm². Use π = 22/7.

1. Given Data:
   r = 7 cm
   θ = 60°
2. Formula Used: Area = πr² × θ/360.
3. Calculation:
   Area = 22/7 × 7² × 60/360
   Area = 154 × 1/6
   Area = 77/3 cm²
4. Final Result: Sector area = 77/3 cm².

35. A minute hand is 7 cm long. Find area swept in 10 minutes.

The area swept is 77/3 cm². The minute hand turns 60° in 10 minutes.

1. Given Data:
   r = 7 cm
   Time = 10 minutes
2. Angle:
   60 minutes = 360°
   10 minutes = 60°
3. Formula Used: Area = πr² × θ/360.
4. Calculation:
   Area = 22/7 × 7² × 60/360
   Area = 77/3 cm²
5. Final Result: Area swept = 77/3 cm².

36. Find area of a minor sector with radius 10 cm and angle 90°.

The area is 78.5 cm². Use π = 3.14.

1. Given Data:
   r = 10 cm
   θ = 90°
2. Formula Used: Area = πr² × θ/360.
3. Calculation:
   Area = 3.14 × 10² × 90/360
   Area = 314/4
   Area = 78.5 cm²
4. Final Result: Minor sector area = 78.5 cm².

37. Find area of the major sector for the same circle and chord.

The major sector area is 235.5 cm². Its central angle is 270°.

1. Given Data:
   r = 10 cm
   major angle = 270°
2. Formula Used: Area = πr² × θ/360.
3. Calculation:
   Area = 3.14 × 100 × 270/360
   Area = 235.5 cm²
4. Final Result: Major sector area = 235.5 cm².

Brahmagupta Formula Class 9 Questions

Brahmagupta’s formula finds the area of a cyclic 4-gon from its side lengths.
It generalises Heron’s formula by allowing a fourth side.
These Brahmagupta formula class 9 questions cover rectangle and cyclic quadrilateral cases.

38. What is Brahmagupta’s formula?

Brahmagupta’s formula gives area of a cyclic 4-gon. It is A = √[(s − a)(s − b)(s − c)(s − d)].

1. Sides: a, b, c, d.
2. Semi-perimeter: s = (a + b + c + d)/2.
3. Formula Used: A = √[(s − a)(s − b)(s − c)(s − d)].
4. Final Result: Brahmagupta’s formula applies to cyclic 4-gons.

39. How does Brahmagupta’s formula generalise Heron’s formula?

Brahmagupta’s formula becomes Heron’s formula when the fourth side is zero. Then the cyclic 4-gon becomes a triangle.

1. Cyclic 4-gon Sides: a, b, c, d.
2. Special Case: d = 0.
3. Result: A = √[s(s − a)(s − b)(s − c)].
4. Final Result: Heron’s formula is a special case.

40. Verify Brahmagupta’s formula for a rectangle of sides a and b.

Brahmagupta’s formula gives ab for a rectangle. This matches the usual rectangle area.

1. Rectangle Sides: a, b, a, b.
2. Semi-perimeter: s = a + b.
3. Formula:
   A = √[(s − a)(s − b)(s − a)(s − b)]
4. Calculation:
   A = √[b × a × b × a]
   A = ab
5. Final Result: Rectangle area = ab.

41. Can area of any 4-gon be found only from four sides?

No, area of any 4-gon cannot be found only from four sides. Different rhombuses can have the same sides but different areas.

1. Example: A rhombus has all four sides equal.
2. Angle Change: Area changes when angle changes.
3. Extra Information: One angle, diagonal, or cyclic condition is needed.
4. Final Result: Four side lengths alone do not fix 4-gon area.

Cyclic Quadrilateral Area Class 9 Questions

A cyclic quadrilateral has all vertices on one circle.
For such a 4-gon, side lengths alone become enough for Brahmagupta’s formula.
These cyclic quadrilateral area class 9 questions help students connect special cases and generalisation.

42. What is a cyclic quadrilateral?

A cyclic quadrilateral is a 4-gon whose four vertices lie on a circle. It is also called a cyclic 4-gon.

1. Shape: Four-sided polygon.
2. Circle Condition: All vertices lie on one circle.
3. Formula Use: Brahmagupta’s formula applies.
4. Final Result: A cyclic quadrilateral has a circumcircle.

43. Find area of a cyclic quadrilateral with sides 6 cm, 8 cm, 10 cm, and 12 cm.

The area is 24√6 cm². Use Brahmagupta’s formula.

1. Given Data:
   a = 6 cm
   b = 8 cm
   c = 10 cm
   d = 12 cm
2. Semi-perimeter:
   s = (6 + 8 + 10 + 12)/2
   s = 18 cm
3. Calculation:
   A = √[(18 − 6)(18 − 8)(18 − 10)(18 − 12)]
   A = √[12 × 10 × 8 × 6]
   A = √5760
   A = 24√10 cm²
4. Final Result: Area = 24√10 cm².

44. Why is every rectangle cyclic?

Every rectangle is cyclic because each angle is 90°. Opposite angles add to 180°.

1. Rectangle Angles: 90°, 90°, 90°, 90°.
2. Opposite Sum: 90° + 90° = 180°.
3. Cyclic Condition: Opposite angles are supplementary.
4. Final Result: Every rectangle is a cyclic 4-gon.

45. Why is every triangle cyclic?

Every triangle is cyclic because one unique circle can pass through any three non-collinear points. That circle is the circumcircle.

1. Triangle Vertices: Three non-collinear points.
2. Circle: One circle passes through them.
3. Name: Circumcircle.
4. Final Result: Every triangle has a circumcircle.

Class 9 Maths Measuring Space Perimeter and Area Questions

NCERT Chapter 6 uses real situations to connect formulas with reasoning.
Track staggers, tyre rotations, petals, flowers, and circular segments all depend on perimeter and area.
These class 9 maths measuring space perimeter and area questions cover applied problems.

46. A car tyre has diameter 56 cm. How far does the car travel in one revolution?

The car travels 176 cm in one revolution. One revolution equals one circumference.

1. Given Data: d = 56 cm.
2. Formula Used: C = πd.
3. Calculation:
   C = 22/7 × 56
   C = 176 cm
4. Final Result: Distance in one revolution = 176 cm.

47. How many revolutions does the tyre make in 10 km?

The tyre makes 5681.8 revolutions, approximately 5682 revolutions. Use distance divided by circumference.

1. Given Data:
   Distance = 10 km = 10,00,000 cm
   Circumference = 176 cm
2. Formula Used: Revolutions = distance/circumference.
3. Calculation:
   Revolutions = 10,00,000/176
   Revolutions ≈ 5681.8
4. Final Result: The tyre makes about 5682 revolutions.

48. Why do athletes in outer lanes start ahead in a 400 m track?

Athletes in outer lanes start ahead because their curved path has a larger radius. The stagger compensates for extra arc length.

1. Straight Sections: Same for all lanes.
2. Curved Sections: Larger for outer lanes.
3. Compensation: Stagger adjusts starting positions.
4. Final Result: Lane stagger makes race distance equal.

49. What is the perimeter of a sector with radius 14 cm and angle 75°?

The perimeter is 134/3 cm. Add arc length and two radii.

1. Given Data:
   r = 14 cm
   θ = 75°
2. Arc Length:
   l = 2 × 22/7 × 14 × 75/360
   l = 55/3 cm
3. Perimeter:
   P = l + 2r
   P = 55/3 + 28
   P = 139/3 cm
4. Final Result: Perimeter = 139/3 cm.

50. What is the area of a parallelogram with base b and height h?

The area is bh. A parallelogram can be rearranged into a rectangle with same base and height.

1. Base: b.
2. Height: h.
3. Formula Used: Area = base × height.
4. Final Result: Area of parallelogram = bh.

NCERT Class 9 Maths Chapter 6 Questions

NCERT exercises include formula use, proofs, special cases, and visual area puzzles.
Students should read whether the question asks for perimeter, area, ratio, or proof.
These NCERT class 9 maths chapter 6 questions follow Ganita Manjari 2026.

51. A trapezium has parallel sides 40 cm and 20 cm. Equal non-parallel sides are 26 cm each. Find its area.

The area is 720 cm². First find the height using Pythagoras theorem.

1. Given Data:
   Parallel sides = 40 cm and 20 cm
   Equal sides = 26 cm
2. Half Difference:
   (40 − 20)/2 = 10 cm
3. Height:
   h² = 26² − 10²
   h² = 676 − 100
   h = 24 cm
4. Area:
   A = 1/2(40 + 20) × 24
   A = 720 cm²
5. Final Result: Area = 720 cm².

52. A rhombus has one diagonal twice the other and area 128 cm². Find shorter diagonal.

The shorter diagonal is 8√2 cm. Use area = 1/2 d1d2.

1. Let Shorter Diagonal: x.
2. Longer Diagonal: 2x.
3. Formula Used: Area = 1/2 d1d2.
4. Calculation:
   128 = 1/2 × x × 2x
   x² = 128
   x = 8√2 cm
5. Final Result: Shorter diagonal = 8√2 cm.

53. A triangle has sides in ratio 3:5:7 and perimeter 300 m. Find its area.

The area is 1500√3 m². Use Heron’s formula.

1. Sides: 3x, 5x, 7x.
2. Perimeter: 15x = 300, so x = 20.
3. Actual Sides: 60 m, 100 m, 140 m.
4. Semi-perimeter: s = 150 m.
5. Calculation:
   A = √[150(150 − 60)(150 − 100)(150 − 140)]
   A = √[150 × 90 × 50 × 10]
   A = 1500√3 m²
6. Final Result: Area = 1500√3 m².

54. What is the area of a kite in terms of its diagonals?

The area of a kite is 1/2 × product of diagonals. Its diagonals divide it into triangles.

1. Diagonals: d1 and d2.
2. Triangle Split: The kite divides into two or four right triangles.
3. Formula Used: Area = 1/2 d1d2.
4. Final Result: Kite area = 1/2 d1d2.

55. What fraction of a circle is a semicircle, quadrant, and three-quarter circle?

A semicircle is 1/2, a quadrant is 1/4, and a three-quarter circle is 3/4 of a circle. The fractions follow central angles.

1. Semicircle: 180°/360° = 1/2.
2. Quadrant: 90°/360° = 1/4.
3. Three-quarter Circle: 270°/360° = 3/4.
4. Final Result: Circle fractions follow angle/360.

Class 9 Maths Chapter 6 Questions and Answers

Mixed practice helps students decide the correct formula before calculation.
The same shape can involve perimeter, area, arc length, or sector area.
These Class 9 Maths Chapter 6 Questions and Answers cover quick formula selection.

56. What value of π should students use in this chapter?

Students should use 22/7 for π unless stated otherwise. Some questions may ask students to use 3.14.

1. Default NCERT Instruction: Use 22/7 unless stated otherwise.
2. Decimal Approximation: π ≈ 3.14.
3. Important Note: π is not equal to 22/7.
4. Final Result: Use the value specified in the question.

57. Is π equal to 22/7?

No, π is not equal to 22/7. The fraction 22/7 is only an approximation.

1. Nature of π: Irrational.
2. Meaning: It cannot be written as a ratio of two integers.
3. Approximation: π ≈ 22/7.
4. Final Result: π ≠ 22/7.

58. What is the difference between a sector and a segment?

A sector is bounded by two radii and an arc. A segment is bounded by a chord and an arc.

1. Sector Boundary: Two radii and one arc.
2. Segment Boundary: One chord and one arc.
3. Example: Pizza slice resembles a sector.
4. Final Result: Sector and segment are different circle regions.

59. Can area of a parallelogram be found from only side lengths?

No, parallelogram area cannot be found from only side lengths. The angle or height is also needed.

1. Same Sides: Different slants are possible.
2. Height: Changes with angle.
3. Area Formula: A = base × height.
4. Final Result: Side lengths alone do not fix parallelogram area.

60. Why does scaling a shape change area by square of scale?

Scaling a shape by k multiplies all lengths by k and area by k². Both dimensions scale.

1. Length Change: Multiplied by k.
2. Width Change: Multiplied by k.
3. Area Change: k × k = k².
4. Final Result: Area scales by k².