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13. Surface Areas and Volumes MATHEMATICS class 10 in English Medium ncert book solutions Exercise 13.4

13. Surface Areas and Volumes Exercise 13.4 – Complete NCERT Book Solutions for Class 10 Mathematics (English Medium). Get all chapter explanations, extra questions, solved examples and additional practice questions for 13. Surface Areas and Volumes Exercise 13.4 to help you master concepts and score higher.

13. Surface Areas and Volumes MATHEMATICS class 10 in English Medium ncert book solutions Exercise 13.4

13. Surface Areas and Volumes MATHEMATICS class 10 in English Medium ncert book solutions Exercise 13.4

NCERT Solutions for Class 10 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 13.4, 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 10 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 13.4. By studying these updated NCERT Solutions for Class 10 Mathematics, students can build a strong foundation, boost their confidence, and score better marks in school and board exams.

13. Surface Areas and Volumes

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Exercise 13.4

Last Update On: 06 March 2026

 

13. Surface Areas and Volumes Exercise 13.4


Q1. A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

Solution: 

Diameter of upper end D = 4 cm

Radius of upper end R = 2 cm

Diameter of lower end d = 2 cm

Radius of lower end r = 1 cm

Q2. The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.

Solution: 

Slant height (l ) of the frustum of the cone = 4 cm

Perimeter of upper end = 18 cm

Thus, the curved surface area of the frustum is 48 cm2.              

Q3. A fez, the cap used by the Turks, is shaped like the frustum of a cone (see Fig. 13.24). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

Solution: 

Slant height (l) of the cap = 15 cm

Radius of open end (R) = 10 cm

Radius of upper end (r) = 4 cm

Area of material used for making it = CSA of Frustum + Area of upper part


Q4. A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs 8 per 100 cm2. (Take π = 3.14)

Solution: 

Height of the container (h) = 16 cm

Radius (R) of the top of the container = 20 cm

Radius of the lower end of the container (r) = 8 cm

Capacity in litre= 10.45 litre (Approx)

The cost of milk = 20 × 10.45 = 209.00

Area of used sheet = πl (R­ + r) + πr2

= 3.14 × 20 (20­ + 8) + 3.14 × 8 × 8

= 3.14 × 20 (28) + 3.14 × 64

= 3.14 (560 + 64)

= 3.14 (624)

= 3.14 (624)

= 1959.36 cm2 

The cost of sheet at the rate of ₹ 8 per 100 cm2

Hence, the cost of the metal sheet is 156.75.

Q5. A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.

Solution:   

AD = 20 cm

Then, AG = 10 cm (cut into two parts at the middle)

∠BAC = 60

AD bisects ∠BAC

Thus, ∠CAD = 30o 

In right angled triangle ΔAGF

Similarly, in right angled triangle DADC,

Lets the length of wire be H  

Volume of the wire = Volume of the frustum obtained

H = 7964.44 m

Hence, the length of the wire is 7964.44 m.

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