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13. Surface Areas and Volumes MATHEMATICS class 9 in English Medium ncert book solutions Exercise 13.3
13. Surface Areas and Volumes Exercise 13.3 – Complete NCERT Book Solutions for Class 9 Mathematics (English Medium). Get all chapter explanations, extra questions, solved examples and additional practice questions for 13. Surface Areas and Volumes Exercise 13.3 to help you master concepts and score higher.
13. Surface Areas and Volumes MATHEMATICS class 9 in English Medium ncert book solutions Exercise 13.3
NCERT Solutions for Class 9 Mathematics play an important role in helping students understand the concepts of the chapter 13. Surface Areas and Volumes clearly. This chapter includes the topic Exercise 13.3, which is essential from both academic and examination point of view. The solutions provided here are prepared strictly according to the latest NCERT syllabus and follow the guidelines of CBSE to ensure accuracy and relevance. Each question is explained in a simple and student-friendly manner so that learners can grasp the concepts without confusion. These NCERT Solutions are useful for regular study, homework help, and exam preparation. All textbook questions are solved step by step to improve problem-solving skills and conceptual clarity. Students of Class 9 studying Mathematics can use these solutions to revise important topics, understand difficult questions, and practise effectively before examinations. The chapter 13. Surface Areas and Volumes is explained in a structured way, making it easier for students to connect the theory with the topic Exercise 13.3. By studying these updated NCERT Solutions for Class 9 Mathematics, students can build a strong foundation, boost their confidence, and score better marks in school and board exams.
13. Surface Areas and Volumes
Exercise 13.3
EXERCISE 13.3
Assume that π = 
, unless stated otherwise.
1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Solution:
Diameter = 10.5 cm ? radius = 5.25, height = 10 cm
Curved surface area of cone = πrl
? × 5.25 × 10
? × 52.5
? 165 cm2
2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Solution:
Slant height = 21 m , diameter = 24 m ? radius = 12 m
Total surface area of cone = πr(l + r)
3. Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find
(i) radius of the base and (ii) total surface area of the cone.
Solution:
CSA of cone = 308 cm2, height = 14 cm
(i) CSA of cone = 308 cm2
4. A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is 70.
Solution:
Height = 10 cm, radius = 24 m
l2 = h2 + r2
l2 = 102 + 242
l =
l = 26 m
(ii) curved surface area of cone = πrl
5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use π = 3.14).
Solution:
Wide = 3 m , height = 8 , radius = 6 m
l2 = r2 + h2
l2 = 62 + 82
l =
l = 10 m
Curved surface area of cone = πrl
? 3.14 × 6 × 10
? 31.4 × 6
? 188.4 m2
6. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of 210 per 100 m2.
Solution:
Length = 25 m, diameter = 14 m, radius = 7 m
Curved surface area of cone = πrl
7. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Solution:
Radius = 7 cm, height = 24 cm
l2 = r2 + h2
l2 = 72 + 242
8. A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is 12 per m2, what will be the cost of painting all these cones? (Use π = 3.14 )
Solution:
Diameter = 40 cm, radius = 20 cm, height = 1 m
l2 = r2 + h2
l2 = 0.22 + 12
See other sub-topics of this chapter:
1. Exercise 13.1 2. Exercise 13.2 3. Exercise 13.3 4. Exercise 13.4 5. Exercise 13.5 6. Exercise 13.6 7. Exercise 13.7 8. Exercise 13.8 9. Exercise 13.9