Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.1 Solutions

Perimeter is the total distance around the boundary of a shape, and the perimeter of a circle is called circumference.
Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.1 connects this idea with arc length, sector perimeter, curved shapes, tyre revolutions and circle ratios.

Chapter 6, Measuring Space: Perimeter and Area Class 9, begins with the 4 × 100 m relay track, where athletes in outer lanes start ahead because curved lanes have larger radii. This idea leads to circumference, arc length and curved boundaries. Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.1 Solutions cover radius from circumference, circumference for given radii, arc length, sector perimeter, mixed curved-shape perimeters, car tyre revolutions, flower-petal perimeters and ratio of radii. The textbook asks students to use 22/7 for π unless stated otherwise.

Key Takeaways

• Circumference: The perimeter of a circle is 2πr or πd.
• Arc Length: An arc with centre angle θ° has length 2πr × θ/360.
• Sector Perimeter: A sector perimeter includes the arc length and two radii.
• Tyre Revolution: One complete revolution covers one circumference.

Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.1 Solutions Structure 2026

Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.1 Solutions for Circumference

Exercise 6.1 begins with the circumference of a circle. The main formulas are circumference = 2πr and circumference = πd.

Q1. The perimeter of a circle is 44 cm. What is its radius?

The radius of the circle is 7 cm.

Given:

Perimeter of circle = 44 cm

Use:

Circumference = 2πr

π = 22/7

So:

44 = 2 × 22/7 × r

44 = 44r/7

r = 44 × 7 / 44

r = 7 cm

Answer:

The radius of the circle is 7 cm.

Q2. Calculate, correct to 3 significant figures, the circumference of a circle with: (i) radius 7 cm (ii) radius 10 cm (iii) radius 12 cm.

The circumferences are 44.0 cm, 62.9 cm and 75.4 cm to 3 significant figures.

Use:

Circumference = 2πr

π = 22/7

Q2(i). Radius = 7 cm

Circumference = 2 × 22/7 × 7

Circumference = 44 cm

To 3 significant figures:

Circumference = 44.0 cm

Q2(ii). Radius = 10 cm

Circumference = 2 × 22/7 × 10

Circumference = 440/7

Circumference = 62.857...

To 3 significant figures:

Circumference = 62.9 cm

Q2(iii). Radius = 12 cm

Circumference = 2 × 22/7 × 12

Circumference = 528/7

Circumference = 75.428...

To 3 significant figures:

Circumference = 75.4 cm

Answer:

(i) 44.0 cm
(ii) 62.9 cm
(iii) 75.4 cm

Ganita Manjari Class 9 Chapter 6 Exercise 6.1: Arc Length Class 9

Arc length is the length of a part of a circle’s circumference. If an arc subtends θ° at the centre, it covers θ/360 of the full circumference.

Q3. Calculate the length of the arc of a circle if: (i) the radius is 3.5 cm and the angle at the centre is 60°, and (ii) the radius is 6.3 m and the angle at the centre is 120°.

The arc lengths are 11/3 cm and 13.2 m.

Use:

Arc length = 2πr × θ/360

Q3(i). Radius = 3.5 cm, angle = 60°

Arc length = 2 × 22/7 × 3.5 × 60/360

Arc length = 22 × 1/6

Arc length = 22/6

Arc length = 11/3 cm

Answer:

Arc length = 11/3 cm

Q3(ii). Radius = 6.3 m, angle = 120°

Arc length = 2 × 22/7 × 6.3 × 120/360

Arc length = 2 × 22/7 × 6.3 × 1/3

Arc length = 13.2 m

Answer:

Arc length = 13.2 m

Class 9 Maths Chapter 6 Exercise 6.1 Solutions for Sector Perimeter

A sector perimeter includes one curved arc and two straight radii. This is the main extra step after finding the arc length.

Q4. Find the perimeter of a sector of a circle of radius 14 cm and sector angle 75°.

The perimeter of the sector is 139/3 cm.

Given:

Radius = 14 cm

Sector angle = 75°

Use:

Sector perimeter = arc length + 2r

First find arc length:

Arc length = 2πr × θ/360

Arc length = 2 × 22/7 × 14 × 75/360

Arc length = 88 × 75/360

Arc length = 55/3 cm

Now add the two radii:

Sector perimeter = 55/3 + 14 + 14

Sector perimeter = 55/3 + 28

Sector perimeter = 55/3 + 84/3

Sector perimeter = 139/3 cm

Answer:

The perimeter of the sector is 139/3 cm.

Class 9 Maths Ganita Manjari Chapter 6 Solutions: Perimeters of Curved Shapes

Question 5 uses semicircles, quarter circles and three-quarter circles from Fig. 6.14. The method is to split each outer boundary into known arcs and straight parts.

Q5. Find the perimeters of the following shapes in Fig. 6.14.

Use π = 22/7.

Q5(i)

The perimeter is 2440/7 m, or 348.6 m approximately.

The shape has two straight portions of 80 m each and two semicircles of diameter 60 m.

Two semicircles make one full circle.

Perimeter = 80 + 80 + π × 60

Perimeter = 160 + 22/7 × 60

Perimeter = 160 + 1320/7

Perimeter = 1120/7 + 1320/7

Perimeter = 2440/7 m

Perimeter = 348.57 m

Answer:

Perimeter = 348.6 m

Q5(ii)

The perimeter is 248/7 cm.

The shape has an outer semicircle of diameter 12 cm, an inner semicircle of diameter 8 cm and two straight base parts.

Outer semicircle length = 1/2 × π × 12

Outer semicircle length = 6π

Inner semicircle length = 1/2 × π × 8

Inner semicircle length = 4π

Total straight parts = 12 - 8

Total straight parts = 4 cm

Perimeter = 6π + 4π + 4

Perimeter = 10π + 4

Perimeter = 10 × 22/7 + 4

Perimeter = 220/7 + 28/7

Perimeter = 248/7 cm

Answer:

Perimeter = 248/7 cm

Q5(iii)

The perimeter is 440/7 cm.

The boundary is made of four semicircles, each with diameter 10 cm.

One semicircle length = 1/2 × π × 10

One semicircle length = 5π

Perimeter = 4 × 5π

Perimeter = 20π

Perimeter = 20 × 22/7

Perimeter = 440/7 cm

Answer:

Perimeter = 440/7 cm

Q5(iv)

The perimeter is 396/7 cm.

The boundary is made of three semicircular arcs, each with diameter 12 cm.

One semicircle length = 1/2 × π × 12

One semicircle length = 6π

Perimeter = 3 × 6π

Perimeter = 18π

Perimeter = 18 × 22/7

Perimeter = 396/7 cm

Answer:

Perimeter = 396/7 cm

Q5(v)

The perimeter is 88 cm.

The boundary is made of four semicircles, each with diameter 14 cm.

One semicircle length = 1/2 × π × 14

One semicircle length = 7π

Perimeter = 4 × 7π

Perimeter = 28π

Perimeter = 28 × 22/7

Perimeter = 88 cm

Answer:

Perimeter = 88 cm

Q5(vi)

The perimeter is 80 cm.

The boundary has one large semicircle of diameter 28 cm and two smaller semicircular arcs. The remaining boundary includes two straight parts.

Large semicircle length = 1/2 × π × 28

Large semicircle length = 14π

Two small semicircles together have arc length:

7π

Straight parts together:

14 cm

Perimeter = 14π + 7π + 14

Perimeter = 21π + 14

Perimeter = 21 × 22/7 + 14

Perimeter = 66 + 14

Perimeter = 80 cm

Answer:

Perimeter = 80 cm

Q5(vii)

The perimeter is 264/7 cm.

The figure uses semicircles on the three sides of a right triangle with sides 6 cm, 8 cm and 10 cm.

Hypotenuse:

10 cm

Perimeter = 1/2 × π × 6 + 1/2 × π × 8 + 1/2 × π × 10

Perimeter = 3π + 4π + 5π

Perimeter = 12π

Perimeter = 12 × 22/7

Perimeter = 264/7 cm

Answer:

Perimeter = 264/7 cm

Q5(viii)

The perimeter is 264/7 cm.

The large semicircle has diameter 12 cm.

The three smaller semicircles each have diameter 4 cm.

Large semicircle length = 1/2 × π × 12

Large semicircle length = 6π

Three small semicircles:

3 × 1/2 × π × 4 = 6π

Perimeter = 6π + 6π

Perimeter = 12π

Perimeter = 264/7 cm

Answer:

Perimeter = 264/7 cm

Q5(ix)

The perimeter is 440/7 cm.

The figure has one large semicircle of diameter 20 cm and two smaller semicircular arcs of diameter 10 cm each.

Large semicircle length = 1/2 × π × 20

Large semicircle length = 10π

Two smaller semicircles:

2 × 1/2 × π × 10 = 10π

Perimeter = 10π + 10π

Perimeter = 20π

Perimeter = 440/7 cm

Answer:

Perimeter = 440/7 cm

Measuring Space Perimeter and Area Class 9: Car Tyre Revolution Problem

A wheel covers a distance equal to its circumference in one complete revolution. This connects circumference with real wheel and tyre movement.

Q6. If the diameter of a car tyre is 56 cm, then: (i) How far does the car need to travel for the tyre to complete one revolution? (ii) How many revolutions does the tyre make if the car travels 10 km?

The car travels 176 cm in one revolution, and the tyre makes 62500/11 revolutions in 10 km.

Given:

Diameter = 56 cm

Use:

Distance in one revolution = circumference

Circumference = πd

Circumference = 22/7 × 56

Circumference = 176 cm

Q6(i)

Answer:

The car travels 176 cm in one revolution.

Q6(ii)

Convert 10 km into cm:

10 km = 10 × 1000 m

10 km = 10000 m

10000 m = 10000 × 100 cm

10000 m = 1000000 cm

Number of revolutions:

Revolutions = total distance ÷ distance in one revolution

Revolutions = 1000000 ÷ 176

Revolutions = 1000000/176

Revolutions = 62500/11

Revolutions = 5681.8 approximately

Answer:

The tyre makes 62500/11 revolutions, or about 5682 revolutions.

Class 9 Maths Perimeter Solutions: Flower Petal Perimeters

Question 7 asks for the total perimeter of all petals. Each petal boundary is made of circular arcs.

Q7. Find the total perimeter of all the petals in each of the given flowers.

Q7(i)

The total perimeter of all petals is 88 cm.

The square has side 14 cm.

Each petal boundary is made of two quarter-circle arcs of radius 7 cm.

There are 4 petals.

Total arcs = 4 × 2

Total arcs = 8 quarter-circle arcs

One quarter-circle arc length:

Arc length = 1/4 × 2πr

Arc length = πr/2

Arc length = 7π/2

Total perimeter of all petals:

8 × 7π/2 = 28π

28π = 28 × 22/7

28π = 88 cm

Answer:

The total perimeter of all petals is 88 cm.

Q7(ii)

The total perimeter of all petals is 528 cm.

The flower has 6 petals inside a regular hexagon of side 42 cm.

Each petal has two arcs.

Total arcs:

6 × 2 = 12 arcs

Each arc has radius 42 cm and centre angle 60°.

Arc length = 2πr × 60/360

Arc length = 2π × 42 × 1/6

Arc length = 14π

Total perimeter:

12 × 14π = 168π

168π = 168 × 22/7

168π = 528 cm

Answer:

The total perimeter of all petals is 528 cm.

Class 9 Maths Chapter 6 Exercise 6.1 Solutions for Ratio of Perimeters

The circumference of a circle is directly proportional to its radius. This means the ratio of perimeters is the same as the ratio of radii.

Q8. The ratio of the perimeters of two circles is 5:4. What is the ratio of their radii?

The ratio of their radii is 5:4.

Let the radii be r1 and r2.

Their perimeters are:

2πr1 and 2πr2

Given:

2πr1 : 2πr2 = 5 : 4

Cancel 2π:

r1 : r2 = 5 : 4

Answer:

The ratio of their radii is 5:4.

Measuring Space Perimeter and Area Class 9 Concepts Used in Exercise 6.1

Exercise 6.1 focuses on the perimeter of circular and curved shapes. Students use circumference, arc length and sector perimeter before the chapter moves into area formulas for rectangles, parallelograms and triangles.

Perimeter of Circle Class 9

The perimeter of a circle is called circumference.

Copy-friendly formulas:

Circumference = 2πr

Circumference = πd

Here:

r = radius

d = diameter

Circumference of Circle Class 9

Circumference is the distance covered by one complete round along a circle.

In a tyre question:

Distance in one revolution = circumference

For diameter 56 cm:

Circumference = πd

Circumference = 22/7 × 56

Circumference = 176 cm

Arc Length Class 9

Arc length depends on the angle at the centre.

Copy-friendly formula:

Arc length = 2πr × θ/360

Here:

θ = angle at centre

Perimeter of Sector Class 9

A sector has one curved arc and two radii.

Copy-friendly formula:

Perimeter of sector = arc length + 2r

So:

Perimeter of sector = 2πr × θ/360 + 2r

Ratio of Perimeters of Circles

The ratio of perimeters of circles equals the ratio of their radii.

Copy-friendly result:

2πr1 : 2πr2 = r1 : r2

So:

If perimeters are 5 : 4, radii are 5 : 4.

Quick Formula Table for Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.1 Solutions

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 6