Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.3 Solutions

Area of a circle is the space covered inside its boundary, and it is calculated using πr².
Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.3 connects this formula with sectors, quadrants, segments, wipers and inscribed polygons.

Measuring Space Perimeter and Area Class 9 moves from perimeter and triangle area to circular area in Exercise 6.3. Students use area of circle Class 9, area of sector Class 9, area of quadrant Class 9, and area of segment Class 9 in one exercise. Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.3 Solutions cover 10 questions, including minute hand sweep, wiper blade sweep, minor and major sectors, minor and major segments, and area ratios for an equilateral triangle, square and hexagon inside a circle. The textbook asks students to use 22/7 for π unless another value is given.

Key Takeaways

• Sector Area: A sector with angle θ° has area πr² × θ/360.
• Quadrant Area: A quadrant is one-fourth of a circle.
• Segment Area: Minor segment area equals sector area minus triangle area.
• Inscribed Polygons: Area ratios compare polygon area with πr².

Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.3 Solutions Structure 2026

Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.3 Solutions for Sector Area

A sector is the part of a circle enclosed by two radii and the arc between them. Its area depends on the central angle because the sector occupies θ/360 of the full circle.

Q1. Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.

The area of the sector is 77/3 cm².

Given:

Radius = 7 cm

Sector angle = 60°

Use:

Area of sector = πr² × θ/360

π = 22/7

Substitute values:

Area = 22/7 × 7² × 60/360

Area = 22/7 × 49 × 1/6

Area = 154/6

Area = 77/3 cm²

Answer:

The area of the sector is 77/3 cm².

Ganita Manjari Class 9 Chapter 6 Exercise 6.3: Area of Quadrant Class 9

A quadrant is formed when a circle is divided into four equal parts by two perpendicular diameters. Its area is one-fourth of the circle’s area.

Q2. Find the area of a quadrant of a circle whose circumference is 44 cm.

The area of the quadrant is 77/2 cm².

Given:

Circumference = 44 cm

Use:

Circumference = 2πr

π = 22/7

Find radius:

44 = 2 × 22/7 × r

44 = 44r/7

r = 7 cm

Now find area of quadrant:

Area of quadrant = 1/4 × πr²

Area of quadrant = 1/4 × 22/7 × 7²

Area of quadrant = 1/4 × 154

Area of quadrant = 77/2 cm²

Answer:

The area of the quadrant is 77/2 cm².

Class 9 Maths Chapter 6 Exercise 6.3 Solutions for Area Swept by Minute Hand

A minute hand sweeps a sector as it moves around the clock. In 10 minutes, it covers 60° because the full 360° rotation takes 60 minutes.

Q3. The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.

The area swept by the minute hand in 10 minutes is 77/3 cm².

Given:

Length of minute hand = 7 cm

Radius = 7 cm

Time = 10 minutes

Full rotation time = 60 minutes

Angle swept:

Angle = 10/60 × 360°

Angle = 60°

Use:

Area swept = area of sector

Area = πr² × θ/360

Area = 22/7 × 7² × 60/360

Area = 22/7 × 49 × 1/6

Area = 77/3 cm²

Answer:

The area swept by the minute hand is 77/3 cm².

Class 9 Maths Ganita Manjari Chapter 6 Solutions for Minor and Major Sectors

A minor sector has the smaller central angle, and a major sector has the larger central angle. In Question 4, the minor sector is 90° and the major sector is 270°.

Q4. A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the corresponding: (i) minor sector and (ii) major sector. Use π ≈ 3.14.

The minor sector area is 78.5 cm², and the major sector area is 235.5 cm².

Given:

Radius = 10 cm

Minor sector angle = 90°

Major sector angle = 270°

π = 3.14

Q4(i). Minor sector

Area of minor sector = πr² × θ/360

Area = 3.14 × 10² × 90/360

Area = 3.14 × 100 × 1/4

Area = 78.5 cm²

Answer:

The area of the minor sector is 78.5 cm².

Q4(ii). Major sector

Area of major sector = πr² × θ/360

Area = 3.14 × 10² × 270/360

Area = 3.14 × 100 × 3/4

Area = 235.5 cm²

Answer:

The area of the major sector is 235.5 cm².

Class 9 Maths Chapter 6 Exercise 6.3 Solutions for Area of Segment Class 9

A circular segment is the region between a chord and its corresponding arc. A minor segment is found by subtracting the triangle area from the minor sector area.

Q5. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. Use π ≈ 3.14 and √3 ≈ 1.73.

The minor segment area is 20.44 cm², and the major segment area is 686.06 cm².

Given:

Radius = 15 cm

Central angle = 60°

π = 3.14

√3 = 1.73

The triangle formed by the two radii and chord is equilateral.

So:

Side of equilateral triangle = 15 cm

First find minor sector area:

Minor sector area = πr² × θ/360

Minor sector area = 3.14 × 15² × 60/360

Minor sector area = 3.14 × 225 × 1/6

Minor sector area = 117.75 cm²

Now find area of equilateral triangle:

Area of equilateral triangle = √3/4 × side²

Area = 1.73/4 × 15²

Area = 1.73/4 × 225

Area = 97.3125 cm²

Minor segment area:

Minor segment area = minor sector area - triangle area

Minor segment area = 117.75 - 97.3125

Minor segment area = 20.4375 cm²

Minor segment area ≈ 20.44 cm²

Full circle area:

Circle area = πr²

Circle area = 3.14 × 225

Circle area = 706.5 cm²

Major segment area:

Major segment area = circle area - minor segment area

Major segment area = 706.5 - 20.4375

Major segment area = 686.0625 cm²

Major segment area ≈ 686.06 cm²

Answer:

Minor segment area = 20.44 cm²

Major segment area = 686.06 cm²

Measuring Space Perimeter and Area Class 9: Area Cleaned by Wiper Blades

A wiper blade sweeps a sector because one end stays fixed and the blade rotates through an angle. Two wipers clean two equal sectors when their blades have equal length and do not overlap.

Q6. A car has two wipers which do not overlap. Each wiper has a blade of length 28 cm and sweeps through an angle of 120°. Find the total area cleaned at each sweep of the blades.

The total area cleaned at each sweep is 4928/3 cm².

Given:

Number of wipers = 2

Blade length = 28 cm

Radius = 28 cm

Angle swept = 120°

Use:

Area cleaned by one wiper = area of sector

Area of one sector = πr² × θ/360

π = 22/7

Area of one sector = 22/7 × 28² × 120/360

Area of one sector = 22/7 × 784 × 1/3

Area of one sector = 2464/3 cm²

Total area cleaned:

Total area = 2 × 2464/3

Total area = 4928/3 cm²

Answer:

The total area cleaned at each sweep is 4928/3 cm².

Class 9 Maths Area Solutions: Minor Segment Formula Proof

A 60° sector with radius r contains an equilateral triangle of side r. This gives the minor segment formula by subtracting the equilateral triangle area from the sector area.

Q7. A chord of a circle of radius r subtends an angle of 60° at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to r²(π/6 - √3/4).

The minor segment area is r²(π/6 - √3/4).

Given:

Radius = r

Central angle = 60°

The triangle formed by the two radii and the chord is equilateral.

So:

Side of triangle = r

Area of sector:

Sector area = πr² × 60/360

Sector area = πr²/6

Area of equilateral triangle:

Triangle area = √3/4 × r²

Triangle area = √3r²/4

Minor segment area:

Minor segment area = sector area - triangle area

Minor segment area = πr²/6 - √3r²/4

Minor segment area = r²(π/6 - √3/4)

Answer:

The area of the minor segment is r²(π/6 - √3/4).

Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.3 Solutions for Inscribed Polygon Area Ratio

An inscribed polygon has all vertices on the circle. Its area ratio is found by comparing the polygon area with the circle area πr².

Q8. An equilateral triangle is inscribed in a circle of radius r. Show that the ratio of the area of the triangle to the area of the circle is equal to 3√3/(4π) ≈ 0.413.

The required ratio is 3√3/(4π).

In an equilateral triangle inscribed in a circle:

Circumradius = r

Side of equilateral triangle = √3r

Area of equilateral triangle:

Area = √3/4 × side²

Area = √3/4 × (√3r)²

Area = √3/4 × 3r²

Area = 3√3r²/4

Area of circle:

Circle area = πr²

Ratio:

Area of triangle : Area of circle = 3√3r²/4 : πr²

Ratio = 3√3/(4π)

Approximate value:

Ratio ≈ 0.413

Answer:

The ratio is 3√3/(4π), or about 0.413.

Q9. A square is inscribed in a circle of radius r. Show that the ratio of the area of the square to the area of the circle is equal to 2/π ≈ 0.637.

The required ratio is 2/π.

In a square inscribed in a circle:

Diagonal of square = diameter of circle

So:

Diagonal = 2r

Let side of square be a.

Using Baudhāyana-Pythagoras theorem:

a² + a² = (2r)²

2a² = 4r²

a² = 2r²

Area of square:

Area = a²

Area = 2r²

Area of circle:

Circle area = πr²

Ratio:

Area of square : Area of circle = 2r² : πr²

Ratio = 2/π

Approximate value:

Ratio ≈ 0.637

Answer:

The ratio is 2/π, or about 0.637.

Q10. A hexagon is inscribed in a circle of radius r. Show that the ratio of the area of the hexagon to the area of the circle is equal to 3√3/(2π) ≈ 0.827. Can you see why the answer is exactly twice the answer to Question 8?

The required ratio is 3√3/(2π), and it is twice the answer to Question 8.

In a regular hexagon inscribed in a circle:

Side of hexagon = radius

So:

Side = r

A regular hexagon can be divided into 6 equilateral triangles of side r.

Area of one equilateral triangle:

Area = √3/4 × r²

Area of hexagon:

Area = 6 × √3r²/4

Area = 3√3r²/2

Area of circle:

Circle area = πr²

Ratio:

Area of hexagon : Area of circle = 3√3r²/2 : πr²

Ratio = 3√3/(2π)

Approximate value:

Ratio ≈ 0.827

Comparison with Question 8:

Question 8 ratio = 3√3/(4π)

Question 10 ratio = 3√3/(2π)

So:

3√3/(2π) = 2 × 3√3/(4π)

Answer:

The ratio is 3√3/(2π), which is exactly twice the ratio in Question 8.

Measuring Space Perimeter and Area Class 9 Concepts Used in Exercise 6.3

Exercise 6.3 uses circular area formulas after the chapter has already covered circumference and arc length. The main focus is the area enclosed by sectors, segments and inscribed polygons.

Area of Circle Class 9

The area of a circle is the space inside its boundary.

Copy-friendly formula:

Area of circle = πr²

Here:

r = radius

Area of Sector Class 9

A sector takes the same fraction of the circle as its angle takes from 360°.

Copy-friendly formula:

Area of sector = πr² × θ/360

Here:

θ = angle at centre

Area of Quadrant Class 9

A quadrant is one-fourth of a circle.

Copy-friendly formula:

Area of quadrant = πr²/4

Area of Segment Class 9

A minor segment is found by subtracting the triangle area from the sector area.

Copy-friendly formula:

Area of minor segment = area of sector - area of triangle

For 60°:

Area of minor segment = r²(π/6 - √3/4)

Area Swept by Minute Hand

The minute hand sweeps a sector of the clock face.

Copy-friendly formula:

Area swept = πr² × θ/360

For 10 minutes:

θ = 60°

Area Cleaned by Wiper Blades

Each wiper blade sweeps a sector.

Copy-friendly formula:

Area cleaned by one wiper = πr² × θ/360

For two equal non-overlapping wipers:

Total area = 2 × πr² × θ/360

Quick Formula Table for Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.3 Solutions

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 6