Class 10 Maths Chapter 12 Surface Areas and Volumes – Exercise 12.4 NCERT Solutions

Introduction

Exercise 12.4 applies the formulas of surface areas and volumes of solids to solve real‑life application problems. These include finding areas and volumes of circular paths, rings, wheels, flower beds, and other practical designs. This exercise strengthens your ability to connect geometry with everyday situations.

Formula Used

• Cube:

$$\text{TSA}=6a^2,V=a^3$$

• Cuboid:

$$\text{TSA}=2(lb+bh+hl),V=l\cdot b\cdot h$$

• Cylinder:

$$\text{CSA}=2\pi rh,\text{TSA}=2\pi r(h+r),V=\pi r^{2}h$$

• Sphere:

$$\text{Surface Area}=4\pi r^2,V=\frac{4}{3}\pi r^3$$

• Hemisphere:

$$\text{CSA}=2\pi r^2,\text{TSA}=3\pi r^2,V=\frac{2}{3}\pi r^3$$

• Ring (Annulus):

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

NCERT Questions with Solutions (10)

Q1. Find area of circular path of width 2 m around circular park of radius 7 m.

$$R=9\text{m},r=7\text{m}$$

$$A=\pi (R^2-r^2)=\pi (81-49)=32\pi =100.48{\text{m}}^2$$

Q2. Find area of ring formed between two concentric circles of radii 14 cm and 7 cm.

$$A=\pi ({14}^2-7^2)=\pi (196-49)=147\pi =462{\text{cm}}^2$$

Q3. Find area of semicircular protractor of radius 10 cm.

$$A=\frac{1}{2}\pi r^2=\frac{1}{2}\times \pi \times 100=157.08{\text{cm}}^2$$

Q4. Find area of quadrant of circle with radius 14 cm.

$$A=\frac{1}{4}\pi r^2=\frac{1}{4}\times \pi \times 196=154{\text{cm}}^2$$

Q5. Find area of circular flower bed of radius 5 m.

$$A=\pi r^2=\pi \times 25=78.5{\text{m}}^2$$

Q6. Find area of wheel with radius 21 cm.

$$A=\pi r^2=\pi \times 441=1386{\text{cm}}^2$$

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

$$R=13\text{m},r=10\text{m}$$

$$A=\pi (R^2-r^2)=\pi (169-100)=69\pi =216.77{\text{m}}^2$$

Q8. Find area of ring with outer radius 10 cm and inner radius 6 cm.

$$A=\pi ({10}^2-6^2)=\pi (100-36)=64\pi =201.06{\text{cm}}^2$$

Q9. Find area of circular path of width 2 m around circular pond of radius 12 m.

$$R=14\text{m},r=12\text{m}$$

$$A=\pi (R^2-r^2)=\pi (196-144)=52\pi =163.36{\text{m}}^2$$

Q10. Find area of semicircle with radius 7 cm.

$$A=\frac{1}{2}\pi r^2=\frac{1}{2}\times \pi \times 49=77{\text{cm}}^2$$

FAQs (10)

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

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

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

4. What is semicircle? Half of circle.

5. What is quadrant? One‑fourth of circle.

6. Formula for ring area?

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

7. Formula for circumference?

$$C=2\pi r$$

8. Formula for circle area?

$$A=\pi r^2$$

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

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

Conclusion

Exercise 12.4 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.

Visit: www.fuzymathacademy.com

New Syllabus-Class 10 Maths Chapter 12 Surface Areas and Volumes – Exercise 12.4 NCERT Solutions