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13. Surface Areas and Volumes MATHEMATICS class 10 in English Medium ncert book solutions Exercise 13.1
13. Surface Areas and Volumes Exercise 13.1 – 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 13.1 to help you master concepts and score higher.
13. Surface Areas and Volumes MATHEMATICS class 10 in English Medium ncert book solutions Exercise 13.1
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 13.1, 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 13.1. 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
Exercise 13.1
Chapter 13. Surface Areas and Volumes Exercise 13.1 solved solution
Q1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Solution:
The volume of cube = 64 cm3
Joining two surfaces;
l = 4 + 4 = 8 cm
b = 4 cm
h = 4 cm
The Surface Area of such formed cuboid = 2(lb + bh + lh)
= 2(8×4 + 4×4 + 8×4)
= 2(32 + 16 + 32)
= 2×80
= 160 cm2
Therefore, The area obtained from this cuboid is 160 cm2
Q2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution:
Q3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:
The radius of hemisphererr = 3.5 cm
The radius of conical partr = 3.5 cm
The height of conical partrh = 15.5 – 3.5 = 12 cm
Q4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.
Solution:
The edge of cubical block = 7 cm
The maximum diameter of the blockd = 7 cm
The total surface area of block = The Area of cuboidal block + The Area of hemisphere – Area of a circle covered with hemisphere
= 294 + 38.5 = 332.5 cm2
Therefore, total surface area of block = 332.5 cm2
Q5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Solution:
Let the diameter of hemisphered = l unit
And the edge of cubea = l unit
(Hence, diameter of hemisphere and the edge of cube are same)
Total surface Area of remaining solid = Area of cuboidal block + Area of hemisphere – Area of a circle covered with hemisphere
Q6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig. 13.10). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Solution:
The height of cylinder = length of capsule – 2(2.5 mm)
= 14 – 5 [Hence diameter = 2.5 mm]
= 9 mm
Total Surface area of capsule = 2(curve surface area of hemisphere) + curve surface area of cylinder
Q7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m2. (Note that the base of the tent will not be covered with canvas.)
Solution:
The diameter of the cylindrical part of tent = 4 m
So, radius = 2 m
The height of cylindrical part = 2.1 m
The slant height of cone = 2.8 m
And radius = 2 m
The area of required canvas = the curved surface area of cylindrical part + curved surface area of conical part
= 2πrh + πrl
Q8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.
Solution:
The height of cylinder h = 2.4 cmcm
The diameter of cylinder = 1.4 cm
So, radius of cylinder = 0.7 cm
The height of hollowed out = 2.4 cm
Radius = 0.7 cm
Total surface area of remaining solid = Curved surface area of cylinder + curved surface area of cone + Area of the bottom of solid
= 2πrh + πrl + πr2
= πr(2h + l + r)
The total surface area of the remaining solid to the nearest cm2 = 18 cm2 Answer
Q9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 13.11. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.
Solution:
The height of cylinder = 10 cm
The radius (r) of base = 3.5 cm
The radius (r) of hemisphere = 3.5 cm
Total surface area = Curved surface area of cylinder + Curved surface area of top of hemisphere + Curved surface area of bottom of hemisphere
= 2πrh + 2πr2 + 2πr2
= 2πr(h h + rr + rr )
= 2πr(hh + 2rr)
Total surface area is 374 cm2
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