Class 10 Maths Chapter 11 Areas Related to Circles – Exercise 11.3 NCERT Solutions

Introduction

Exercise 11.3 applies the formulas of areas of circles, sectors, and segments to solve real‑life application problems. These include finding areas of flower beds, paths around circular fields, wheels, and other circular objects. This exercise strengthens your ability to connect geometry with practical situations.

Formula Used

1. Area of Circle:

$$A=\pi r^2$$

2. Circumference of Circle:

$$C=2\pi r$$

3. Area of Sector (θ in degrees):

$$A=\frac{\theta }{{360}^{\circ }}\cdot \pi r^2$$

4. Area of Segment:

$$A=\text{Area of sector}-\text{Area of triangle}$$

5. Area of Ring (Annulus):

$$A=\pi (R^2-r^2)$$

NCERT Questions with Solutions (10)

Q1. Find area of circular garden with radius 7 m.

Solution: $A=\pi r^2=\frac{22}{7}\times 7\times 7=154{\text{m}}^2$.

Q2. Find area of path 2 m wide around circular field of radius 10 m.

Solution: Outer radius = 12 m. Area = $\pi ({12}^2-{10}^2)=\pi (144-100)=44\pi =138.16{\text{m}}^2$.

Q3. Find area of ring formed between two concentric circles of radii 7 cm and 14 cm.

Solution: Area = $\pi ({14}^2-7^2)=\pi (196-49)=147\pi =462{\text{cm}}^2$.

Q4. Find area of wheel with radius 21 cm.

Solution: $A=\pi r^2=\frac{22}{7}\times 21\times 21=1386{\text{cm}}^2$.

Q5. Find area of semicircular protractor of radius 7 cm.

Solution: $A=\frac{1}{2}\pi r^2=\frac{1}{2}\times \frac{22}{7}\times 49=77{\text{cm}}^2$.

Q6. Find area of quadrant of circle with radius 14 cm.

Solution: $A=\frac{1}{4}\pi r^2=\frac{1}{4}\times \frac{22}{7}\times 196=154{\text{cm}}^2$.

Q7. Find area of circular path of width 3 m around circular park of radius 10 m.

Solution: Outer radius = 13 m. Area = $\pi ({13}^2-{10}^2)=\pi (169-100)=69\pi =216.77{\text{m}}^2$.

Q8. Find area of circular flower bed of radius 5 m.

Solution: $A=\pi r^2=\frac{22}{7}\times 25=78.5{\text{m}}^2$.

Q9. Find area of circular ring with outer radius 10 cm and inner radius 6 cm.

Solution: Area = $\pi ({10}^2-6^2)=\pi (100-36)=64\pi =201.06{\text{cm}}^2$.

Q10. Find area of circular path of width 2 m around circular pond of radius 7 m.

Solution: Outer radius = 9 m. Area = $\pi (9^2-7^2)=\pi (81-49)=32\pi =100.48{\text{m}}^2$.

FAQs (10 from NCERT)

1. Q: What is annulus? A: Ring‑shaped region between two concentric circles.

2. Q: What is sector? A: Region enclosed by two radii and arc.

3. Q: What is segment? A: Region enclosed by chord and arc.

4. Q: What is semicircle? A: Half of circle.

5. Q: What is quadrant? A: One‑fourth of circle.

6. Q: What is formula for ring area? A: $\pi (R^2-r^2)$.

7. Q: What is circumference formula? A: $2\pi r$.

8. Q: What is area formula? A: $\pi r^2$.

9. Q: Why are circular paths important? A: Used in parks, gardens, and construction.

10. Q: Why is this exercise important? A: It builds skills in solving real‑life problems using circle geometry.

Conclusion

Exercise 11.3 has 10 solved questions and 10 FAQs that strengthen your understanding of real‑life applications of circle areas. This builds the foundation for practical geometry in Class 10 Maths.

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New Syllabus-Class 10 Maths Chapter 11 Areas Related to Circles – Exercise 11.3 NCERT Solutions