Class 9 Maths Chapter 6 Measuring Space: Perimeter and Area Exercise 6.3 Solutions 2026-27

Think and Reflect

Why were human beings so fond of using circular shapes? Was this only for practical reasons, or were there other reasons too? What kinds of uses have human beings found for the circular shape?

Answer:

Human beings have used circular shapes for both practical and cultural reasons. A circle has no corners, is symmetrical in every direction, and allows smooth rotation. These properties made circular forms especially useful for objects that roll, rotate, or need to maintain an equal distance from a centre.

Circular shapes appear in wheels, gears, clocks, coins, plates, pots, wells, domes, and circular buildings. The wheel, in particular, made transportation and many mechanical systems easier because it can rotate smoothly around its centre.

Circles also appear in art, architecture, decoration, and symbols. Their balanced, continuous shape has made them useful not only in mathematics and engineering but also in various forms of human design and expression.

Exercise Set 6.3

Unless stated otherwise, use the approximation 22/7 for π.

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

Solution:

Given:

Radius, r = 7 cm

Angle of the sector, θ = 60°

Area of a sector:

= θ/360° × πr²

= 60°/360° × 22/7 × 7²

= 1/6 × 22/7 × 49

= 77/3 cm²

= 25⅔ cm²

Hence, the area of the sector is 25⅔ cm².

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

Solution:

Given:

Circumference of the circle = 44 cm

We know:

Circumference = 2πr

Therefore,

2 × 22/7 × r = 44

r = 7 cm

A quadrant is one-fourth of a circle.

Area of the quadrant:

= 1/4 × πr²

= 1/4 × 22/7 × 7²

= 77/2 cm²

= 38.5 cm²

Hence, the area of the quadrant is 38.5 cm².

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

Solution:

Length of the minute hand = 7 cm

Therefore,

r = 7 cm

The minute hand completes 360° in 60 minutes.

Angle swept in 10 minutes:

θ = 10/60 × 360°

= 60°

Area swept by the minute hand:

= Area of sector

= θ/360° × πr²

= 60°/360° × 22/7 × 7²

= 77/3 cm²

= 25⅔ cm²

Hence, the area swept by the minute hand in 10 minutes is 25⅔ cm².

4. A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the corresponding: (i) minor sector (that subtends 90° at the centre), and (ii) major sector (that subtends 270° at the centre). (Use π ≈ 3.14.)

Solution:

Given:

Radius, r = 10 cm

(i) Area of the minor sector

Angle of the minor sector:

θ = 90°

Area of sector:

= θ/360° × πr²

= 90°/360° × 3.14 × 10²

= 1/4 × 3.14 × 100

= 78.5 cm²

Therefore, the area of the minor sector is 78.5 cm².

(ii) Area of the major sector

Angle of the major sector:

θ = 270°

Area:

= 270°/360° × 3.14 × 10²

= 3/4 × 314

= 235.5 cm²

Therefore, the area of the major sector is 235.5 cm².

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

Solution:

Given:

Radius, r = 15 cm

Angle at the centre, θ = 60°

The triangle formed by joining the chord endpoints to the centre has two sides equal to the radius.

Since the angle between them is 60°, the triangle is equilateral.

Therefore, each side of the triangle = 15 cm.

Area of the minor sector

= 60°/360° × πr²

= 1/6 × 3.14 × 15²

= 117.75 cm²

Area of the equilateral triangle

= √3/4 × a²

= 1.73/4 × 15²

= 1.73/4 × 225

= 97.3125 cm²

Area of the minor segment

= Area of minor sector − Area of triangle

= 117.75 − 97.3125

= 20.4375 cm²

≈ 20.44 cm²

Therefore, the area of the minor segment is approximately 20.44 cm².

Area of the major segment

Area of circle:

= πr²

= 3.14 × 15²

= 706.5 cm²

Area of major segment:

= Area of circle − Area of minor segment

= 706.5 − 20.44

= 686.06 cm²

Therefore, the area of the major segment is approximately 686.06 cm².

6. 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.

Solution:

Given:

Length of each wiper blade, r = 28 cm

Angle swept, θ = 120°

Area cleaned by one wiper:

= θ/360° × πr²

= 120°/360° × 22/7 × 28²

= 1/3 × 22/7 × 784

= 2464/3 cm²

= 821⅓ cm²

There are two wipers, and their swept areas do not overlap.

Therefore,

Total area cleaned:

= 2 × 821⅓

= 1642⅔ cm²

Hence, the total area cleaned at each sweep of the blades is 1642⅔ cm².

7. 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 πr²(1/6 − √3/4π).

Solution:

Given:

Radius = r

Angle subtended at the centre = 60°

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

Area of the 60° sector:

= 60°/360° × πr²

= πr²/6

Area of the equilateral triangle:

= √3/4 × r²

Therefore,

Area of minor segment:

= Area of sector − Area of triangle

= πr²/6 − √3r²/4

Taking πr² common:

= πr²(1/6 − √3/4π)

Hence,

Area of the corresponding minor segment = πr²(1/6 − √3/4π).

Hence proved.

8. 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.

Solution:

Let ABC be an equilateral triangle inscribed in a circle of radius r.

Let the side of the equilateral triangle be a.

For an equilateral triangle, the circumradius is:

r = a/√3

Therefore,

a = √3r

Area of the equilateral triangle:

= √3/4 × a²

= √3/4 × (√3r)²

= 3√3/4 × r²

Area of the circle:

= πr²

Therefore,

Area of triangle / Area of circle

= (3√3r²/4)/(πr²)

= 3√3/4π

Using √3 ≈ 1.73 and π ≈ 3.14:

3√3/4π ≈ 0.413

Hence,

Area of triangle: Area of circle = 3√3/4π ≈ 0.413.

Hence proved.

9. 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.

Solution:

Let the side of the square be a.

Since the square is inscribed in the circle:

Diagonal of the square = Diameter of the circle

Therefore,

a√2 = 2r

So,

a = √2r

Area of the square:

= a²

= (√2r)²

= 2r²

Area of the circle:

= πr²

Therefore,

Area of square / Area of circle

= 2r²/πr²

= 2/π

Using π ≈ 3.14:

2/π ≈ 2/3.14

≈ 0.637

Hence,

Area of square: Area of circle = 2/π ≈ 0.637.

Hence proved.

10. 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?

Solution:

A regular hexagon inscribed in a circle can be divided into 6 equilateral triangles, each having side r.

Area of one equilateral triangle:

= √3/4 × r²

Therefore,

Area of hexagon:

= 6 × √3/4 × r²

= 3√3/2 × r²

Area of circle:

= πr²

Therefore,

Area of hexagon / Area of circle

= (3√3r²/2)/(πr²)

= 3√3/2π

Using √3 ≈ 1.73 and π ≈ 3.14:

3√3/2π ≈ 0.827

Hence,

Area of hexagon: Area of circle = 3√3/2π ≈ 0.827.

The answer is exactly twice the answer to Question 8 because the inscribed equilateral triangle has area:

3√3r²/4

while the regular hexagon has area:

3√3r²/2

Therefore,

Area of hexagon = 2 × Area of equilateral triangle

So, their ratios with the area of the same circle also differ by a factor of 2.

Hence proved.

Understand Exercise 6.3 with Step-by-Step NCERT Solutions

The Maths Class 9 Chapter 6 Exercise 6.3 Solutions help students strengthen their understanding of perimeter and area by applying concepts to textbook problems. While solving the exercise, students should carefully identify the given measurements, choose the appropriate area relationship or formula, and show each calculation clearly. Attempting the questions before checking the NCERT solutions can make revision more effective and help students become more confident with the concepts covered in Chapter 6.

Students can also download the Class 9 Maths Chapter 6 Exercise 6.3 Solutions PDF for convenient practice and revision. Students can use it to revisit solutions whenever they need help with the exercise.

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