Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.3 Solutions – Measuring Space: Perimeter and Area
NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.3 help students understand and solve questions based on Measuring Space: Perimeter and Area. This exercise provides practice in applying concepts and formulas related to the perimeter and area of different plane figures. The step-by-step solutions are explained in a simple and clear manner to help students understand the approach used for solving each question.
Students can use these Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.3 Solutions for regular practice, revision, and exam preparation. A printable PDF of the solutions is also available for students who want to revise the exercise conveniently.
Class 9 Maths Ganita Manjari Chapter 6 Exercise 6.3 Solutions
NCERT Solutions for Class 9 Maths Chapter 6 Exercise Set 6.3
Exercise Set 6.3 focuses on area of sectors, quadrants, minor segments, major segments and circular motion-based area problems.
Question 1. Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.
Answer:
Area of sector = πr² × θ/360°
= 22/7 × 7² × 60/360
= 22/7 × 49 × 1/6
= 77/3 cm²
Final answer:
77/3 cm²
Question 2. Find the area of a quadrant of a circle whose circumference is 44 cm.
Answer:
Circumference = 44 cm
2Ï€r = 44
2 × 22/7 × r = 44
r = 7 cm
Area of quadrant = 1/4 × πr²
= 1/4 × 22/7 × 49
= 77/2 cm²
Final answer:
77/2 cm² or 38.5 cm²
Question 3. The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.
Answer:
In 60 minutes, the minute hand sweeps 360°.
In 10 minutes, angle swept:
= 360° × 10/60
= 60°
Area swept = sector area
= πr² × θ/360°
= 22/7 × 7² × 60/360
= 77/3 cm²
Final answer:
77/3 cm²
Question 4. A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the minor sector and major sector. Use π = 3.14.
Answer:
Radius = 10 cm
Area of circle = πr²
= 3.14 × 100
= 314 cm²
Minor sector angle = 90°
Minor sector area = 314 × 90/360
= 314/4
= 78.5 cm²
Major sector angle = 270°
Major sector area = 314 × 270/360
= 314 × 3/4
= 235.5 cm²
Final answer:
Minor sector = 78.5 cm², major sector = 235.5 cm²
Question 5. A chord of a circle of radius 15 cm subtends 60° at the centre. Find the areas of the corresponding minor and major segments.
Answer:
Radius = 15 cm
Angle = 60°
Ï€ = 3.14
√3 = 1.73
Area of minor sector:
= πr² × 60/360
= 3.14 × 225 × 1/6
= 117.75 cm²
The triangle formed by the two radii and chord is equilateral because the central angle is 60°.
Area of equilateral triangle with side 15 cm:
= √3/4 × 15²
= 1.73/4 × 225
= 97.3125 cm²
Area of minor segment:
= sector area - triangle area
= 117.75 - 97.3125
= 20.4375 cm²
Area of circle:
= 3.14 × 225
= 706.5 cm²
Area of major segment:
= 706.5 - 20.4375
= 686.0625 cm²
Final answer:
Minor segment ≈ 20.44 cm², major segment ≈ 686.06 cm²
Question 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.
Answer:
Area cleaned by one wiper:
= πr² × θ/360°
= 22/7 × 28² × 120/360
= 22/7 × 784 × 1/3
= 2464/3 cm²
For two wipers:
Total area = 2 × 2464/3
= 4928/3 cm²
Final answer:
4928/3 cm²
Question 7. A chord of a circle of radius r subtends 60° at the centre. Show that the area of the corresponding minor segment is πr²/6 - √3r²/4.
Answer:
Area of sector with angle 60°:
= πr² × 60/360
= πr²/6
The triangle formed by two radii and the chord is equilateral with side r.
Area of equilateral triangle:
= √3/4 r²
Area of minor segment:
= sector area - triangle area
= πr²/6 - √3r²/4
Final answer:
Area of minor segment = πr²/6 - √3r²/4
Question 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 3√3/4π.
Answer:
For an equilateral triangle inscribed in a circle of radius r, side length is:
a = √3r
Area of equilateral triangle:
= √3/4 × a²
= √3/4 × 3r²
= 3√3r²/4
Area of circle:
= πr²
Ratio:
= (3√3r²/4)/(πr²)
= 3√3/4π
Final answer:
Ratio = 3√3/4π
Question 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 2/Ï€.
Answer:
For a square inscribed in a circle, the diagonal of the square is the diameter of the circle.
Diagonal = 2r
If side = a, then:
a√2 = 2r
a = √2r
Area of square:
= a²
= 2r²
Area of circle:
= πr²
Ratio:
= 2r²/πr²
= 2/Ï€
Final answer:
Ratio = 2/Ï€
Question 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 3√3/2π.
Answer:
A regular hexagon inscribed in a circle of radius r has side length r.
It consists of 6 equilateral triangles of side r.
Area of one equilateral triangle:
= √3/4 r²
Area of hexagon:
= 6 × √3/4 r²
= 3√3r²/2
Area of circle:
= πr²
Ratio:
= (3√3r²/2)/(πr²)
= 3√3/2π
Final answer:
Ratio = 3√3/2π
This is twice the answer to Question 8 because the regular hexagon consists of 6 equilateral triangles, while the inscribed equilateral triangle covers half as many equivalent triangular sectors.