Overview of Class 10 Maths Chapter 12: Surface Areas and Volumes
The NCERT Solutions for Class 10 Maths Chapter 12 Surface Areas and Volumes offer students an easy-to-access solution that enables them to study this critical chapter of their curriculum. The chapter teaches students to determine surface area and volume measurements for various three-dimensional objects which include cones and cylinders and spheres and hemispheres.
Students learn to transform one solid object into another while determining surface area and volume measurements for the resulting object. The exam questions require students to demonstrate their ability to apply formulas through precise execution of each procedural step.
The solutions present information in an organized format which enables students to comprehend 3D figure concepts. This chapter provides students with explicit explanations which present complete procedures that enable them to develop essential skills in mensuration while enhancing their abilities to solve problems.
The NCERT Solutions for Class 10 Maths provided by ToppersSky help students prepare confidently for their board exams in advance. The structured solutions of this system provide a complete solution which helps students solve their conceptual problems while building their knowledge of all subjects. The explanation presents each step clearly which enables students to develop their problem-solving abilities while they establish a solid knowledge base.
The solutions have been developed according to current NCERT syllabus standards to assist students with their exam preparation needs. ToppersSky ensures that all important concepts and commonly asked question types are covered so students can practice effectively and perform well in their exams.
Class 10 Maths Chapter 12 – Surface Areas and Volumes
Chapter 12 of the Class 10 Maths curriculum includes essential mathematical content through its study of surface areas and volumes. The board examination assigns major marks to this chapter. The chapter will have multiple questions which test students on surface area and volume and solid shape conversions. Students must prepare this chapter because it has both academic weightage and real-life significance.
Topics Covered in Chapter 12
- Surface area of a combination of solids
- Volume of a combination of solids
- Conversion of one solid shape into another
- Frustum of a cone
These topics help students understand how to calculate measurements of 3D shapes using formulas and logical reasoning.
Why This Chapter Is Important
Surface Areas and Volumes are not only important for exams but also useful in real life. Many objects around us such as rectangular boxes, gas cylinders, footballs, cones, and storage tanks are three-dimensional shapes. To calculate how much space they occupy or how much material is needed to cover them, we use formulas of surface area and volume.
Students learn formulas for spheres, cylinders, cones, cuboids, hemispheres, and combinations of solids. The students who practice solved examples will successfully handle both textbook questions and application-based problems which are part of their exams.
ToppersSky provides structured NCERT Solutions for Class 10 Maths Chapter 12 with clear step-by-step explanations. The solutions assist students in understanding proper formula usage while helping them gain confidence for their application.
Key Features of ToppersSky NCERT Solutions for Chapter 12
- Complete step-by-step solutions to all textbook exercise questions
- Clear explanation of tricky numerical problems
- Helpful reference material for exam preparation
- Structured format for quick revision
- Focus on formula understanding and application
- Prepared according to the latest NCERT syllabus guidelines
Students can use the Animation learning videos together with chapter notes and mind maps and practice sets from the topperssky app to achieve exam success through better understanding of their concepts.
The practice of these solutions will help students achieve high marks in Surface Areas and Volumes while developing essential skills needed for mensuration.
NCERT Class 10 Maths Ch 13 Q&A – Surface Areas & Volumes
Q.1 Two cubes, each having a volume of 64 cm³, are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Volume of each cube = 64 cm³
Let the side of each cube be a.
So,
a³ = 64
a = 4 cm
When two cubes are joined end to end:
Length of cuboid = 4 + 4 = 8 cm
Breadth = 4 cm
Height = 4 cm
Surface area of cuboid = 2(lb + bh + lh)
= 2(8 × 4 + 4 × 4 + 4 × 8)
= 2(32 + 16 + 32)
= 2 × 80
= 160 cm²
Therefore, the surface area of the cuboid is 160 cm².
Q.2 A vessel is in the form of a hollow hemisphere mounted on a hollow cylinder. The diameter of the hemisphere is 14 cm, and the total height is 13 cm. Find the inner surface area of the vessel.
Solution:
Diameter of hemisphere = 14 cm
Radius = 7 cm
Height of cylinder = 13 − 7
= 6 cm
Inner surface area = Curved surface area of cylinder + Curved surface area of hemisphere
= 2πrh + 2πr²
= 2πr(h + r)
= 2 × (22/7) × 7 × (6 + 7)
= 2 × 22 × 13
= 572 cm²
Therefore, the inner surface area of the vessel is 572 cm².
Q.3 A toy is in the form of a cone mounted on a hemisphere. The radius is 3.5 cm and total height is 15.5 cm. Find the total surface area.
Solution:
Radius (r) = 3.5 cm
Height of cone = 15.5 − 3.5
= 12 cm
Slant height of cone:
l = √(r² + h²)
= √(3.5² + 12²)
= √(12.25 + 144)
= √156.25
= 12.5 cm
Curved surface area of cone = πrl
= (22/7) × 3.5 × 12.5
= 275/2 cm²
Curved surface area of hemisphere = 2πr²
= 2 × (22/7) × (3.5)²
= 77 cm²
Total surface area = 275/2 + 77
= 429/2
= 214.5 cm²
Therefore, the total surface area of the toy is 214.5 cm².
Q.4 A cube of side 7 cm is surmounted by a hemisphere. Find the greatest diameter of the hemisphere and the surface area of the solid.
Solution:
Side of cube = 7 cm
Maximum diameter of hemisphere = 7 cm
Radius = 7/2 cm
Total surface area = Surface area of cube + Curved surface area of hemisphere − Area of base of hemisphere
= 6 × 7² + 2πr² − πr²
= 6 × 49 + πr²
= 294 + (22/7) × (7/2)²
= 294 + 38.5
= 332.5 cm²
Therefore, the surface area of the solid is 332.5 cm².
Q.5 A hemispherical depression is cut from one face of a cubical block. The diameter of the hemisphere is equal to the edge of the cube. Find the surface area of the remaining solid.
Solution:
Edge of cube = l
Radius of hemisphere = l/2
Surface area of remaining solid = Surface area of cube + Curved surface area of hemisphere − Area of base
= 6l² + 2πr² − πr²
= 6l² + πr²
= 6l² + π(l/2)²
= 6l² + πl²/4
= (l²/4)(24 + π)
Therefore, the surface area of the remaining solid is (l²/4)(24 + π) square units.
Q.6 A medicine capsule consists of a cylinder with two hemispheres at both ends. Length = 14 mm, diameter = 5 mm. Find the surface area.
Solution:
Radius = 2.5 mm
Length of cylinder = 14 − (2.5 + 2.5)
= 9 mm
Surface area of one hemisphere = 2πr²
= 2 × (22/7) × 2.5 × 2.5
= 275/7 mm²
Surface area of cylinder = 2πrh
= 2 × (22/7) × 2.5 × 9
= 990/7 mm²
Total surface area = 2 × (275/7) + 990/7
= 550/7 + 990/7
= 1540/7
= 220 mm²
Therefore, the surface area of the capsule is 220 mm².
Q.7 A tent is in the shape of a cylinder with a conical top. Diameter = 4 m, height of cylinder = 2.1 m, slant height of cone = 2.8 m. Find the area of canvas and its cost at ₹500 per m².
Solution:
Radius = 4/2 = 2 m
Surface area required = Curved surface area of cone + Curved surface area of cylinder
= πrl + 2πrh
= πr(l + 2h)
= (22/7) × 2 × (2.8 + 2 × 2.1)
= (44/7) × (2.8 + 4.2)
= (44/7) × 7
= 44 m²
Cost of canvas = 44 × 500
= ₹22,000
Therefore, the total cost of the canvas is ₹22,000.
Q.8 Rachel made a model in the shape of a cylinder with two cones attached at both ends using a thin aluminium sheet. The diameter of the model is 3 cm, and its total length is 12 cm. Each cone has a height of 2 cm. Find the volume of air contained inside the model.(Assume outer and inner dimensions are nearly the same.)
Solution:
Diameter of the model = 3 cm
Radius (r) = 1.5 cm
Height of each cone = 2 cm
Since there are two cones, total height of cones = 2 + 2 = 4 cm
Height of cylindrical part = 12 − 4
= 8 cm
Now,
Volume of cylinder = πr²h
= π × (1.5)² × 8
= π × 2.25 × 8
= 18π cm³
Volume of one cone = (1/3)πr²h
= (1/3)π × (1.5)² × 2
= (1/3)π × 2.25 × 2
= 1.5π cm³
Volume of two cones = 2 × 1.5π
= 3π cm³
Total volume = Volume of cylinder + Volume of two cones
= 18π + 3π
= 21π cm³
Taking π = 22/7,
Total volume = 66 cm³
Therefore, the volume of air inside the model is 66 cm³.
Q.9 A gulab jamun contains sugar syrup equal to 30% of its volume. Each gulab jamun is shaped like a cylinder with two hemispherical ends. The total length is 5 cm, and the diameter is 2.8 cm. Find approximately how much syrup is contained in 45 such gulab jamuns.
Solution:
Diameter = 2.8 cm
Radius (r) = 1.4 cm
Total length = 5 cm
Height of cylindrical part = 5 − (1.4 + 1.4)
= 2.2 cm
Volume of cylinder = πr²h
= π × (1.4)² × 2.2
= 4.312π cm³
Volume of two hemispheres = Volume of one sphere
= (4/3)πr³
= (4/3)π × (1.4)³
= (10.976/3)π cm³
Total volume of one gulab jamun = 25.05 cm³
Since syrup is 30% of total volume:
Syrup in one gulab jamun = 30% of 25.05
= 7.515 cm³
Syrup in 45 gulab jamuns:
= 45 × 7.515
= 338.18 cm³
Therefore, approximately 338.18 cm³ of syrup is present in 45 gulab jamuns.
Q.10 A pen stand made of wood is shaped like a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm × 10 cm × 3.5 cm. Each conical depression has a radius of 0.5 cm and a depth of 1.4 cm. Find the volume of wood in the stand.
Solution:
Volume of cuboid = length × width × height
= 15 × 10 × 3.5
= 525 cm³
Now, depressions are in the shape of cones.
Volume of one cone = (1/3)πr²h
= (1/3)π × (0.5)² × 1.4
= (1/3)π × 0.25 × 1.4
= 0.3667π cm³
Volume of 4 cones = 4 × 0.3667π
≈ 1.46 cm³
Volume of wood remaining = Volume of cuboid − Volume of 4 cones
= 525 − 1.46
= 523.54 cm³
Therefore, the volume of wood in the pen stand is 523.54 cm³.
FAQs
1. How can I score full marks in Class 10 Maths Chapter 12 tests?
The solutions explain how to solve problems through their step-by-step demonstrations. The students will achieve exact test results when they practice numerical problems while reviewing their study material for academic tests and board examinations.
2. How are NCERT Solutions for Class 10 Maths Chapter 12 helpful in exams?
The solutions present complete explanations which follow the current NCERT syllabus requirements. The practice of these questions enables students to learn various question formats which include application-based problems and solid shape conversion and shape combination. This process creates complete understanding of concepts which enhances students’ exam confidence.
3. What are the main concepts covered in Chapter 12?
The important topics covered in this chapter include:
- The surface area measurements for solid material combinations
- The complete volume measurement for solid material combinations
- The process of transforming one solid object into a different solid object
- Frustum of a cone
Students need to understand essential formulas for these concepts because they appear frequently in exams.
4. Why is Chapter 12 important in real life?
People use surface area and volume measurements to determine tank capacities and design containers and calculate storage volumes and create 3D models of cylinders and cones and spheres. The chapter enables students to use mathematics in their real-world activities.
5. What is the best way to prepare for Surface Areas and Volumes?
Students should study through three activities which include learning essential mathematical formulas, solving diagrammatic problems, and working through combination challenges while maintaining a regular study schedule. Students should focus on understanding the logic behind each formula instead of just memorizing it.
