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13. Surface Areas and Volumes Mathematics Exercise - 13.4 class 9 Maths in English - CBSE Study

13. Surface Areas and Volumes Mathematics Class 9 exercise - 13.4 class 9 Maths cbse board school study materials like cbse notes in English medium, all chapters and exercises are covered the ncert latest syllabus 2026 - 27.

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13. Surface Areas and Volumes Mathematics Exercise - 13.4 class 9 Maths in English - CBSE Study

NCERT Solutions for Class 9 Mathematics are carefully prepared according to the latest CBSE syllabus and NCERT textbooks to help students understand every concept clearly. These solutions cover all important 13. Surface Areas and Volumes with detailed explanations and step-by-step answers for better exam preparation. Each Exercise 13.4 is explained in simple language so that students can easily grasp the fundamentals and improve their academic performance. The study material is designed to support daily homework, revision practice, and final exam preparation for Class 9 students. With accurate answers, concept clarity, and structured content, these NCERT solutions help learners build confidence and score higher marks in their examinations. Whether you are revising a specific topic or preparing an entire chapter, this resource provides reliable and syllabus-based guidance for complete success in Mathematics.

Class 9 English Medium Mathematics All Chapters:

13. Surface Areas and Volumes

4. Exercise 13.4

Exercise 13.4


Assume π = 22/7 , unless stated otherwise.

Q1. Find the surface area of a sphere of radius:

(i) 10.5 cm                  (ii) 5.6 cm                (iii) 14 cm

Solution:

Q2. Find the surface area of a sphere of diameter:

(i) 14 cm             (ii) 21 cm                 (iii) 3.5 m

Solution:

(i) Surface area of sphere = 4πr2

Q3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)

Solution:

Total surface area of hemisphere = 3πr2

⇒ 3 × 3.14 × 10 × 10

⇒ 3 × 314

⇒ 942 cm2

Q4. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Solution:

Surface area of sphere = 4πr2

Q5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of 16 per 100 cm2.

Solution:

Curved surface area of hemisphere = 2πr2  

Q6. Find the radius of a sphere whose surface area is 154 cm2.

Solution:

Area = 154 cm2

Surface area of sphere = 4πr2

Q7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Solution:

Let the diameter of earth = x

Q8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Solution:

Inner radius = 5 cm,    width = 0.25 cm,   radius = 5.25 cm

Curved surface area of hemisphere = 2πr2

⇒ 44 × 0.75 × 5.25

⇒ 173.25

Q9. A right circular cylinder just encloses a sphere of

radius (see Fig. 13.22). Find

(i) surface area of the sphere,

(ii) curved surface area of the cylinder,

(iii) ratio of the areas obtained in (i) and (ii).  

Solution:

Radius of sphere = r,   radius of cylinder = r + r = 2r

(i) surface area of sphere = 4πr2

(ii) curved surface area of cylinder = 2πrh

⇒ 2πr(2r)

 4πr2

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