Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

NCERT Solutions for Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.1 – 2025-26

ffImage
banner

Step-by-Step Solutions for Class 10 Maths Chapter 12 Exercise 12.1 Surface Areas and Volumes

Exercise 12.1 Class 10 Maths NCERT Solutions focuses on developing your ability to calculate the surface areas and volumes of basic shapes, starting with cubes and cuboids. Understanding surface areas and volumes is not just an academic exercise; it has practical applications in real life. Whether you are packing a box, filling a tank with water, or designing a cylindrical container, these concepts are indispensable. Moreover, this chapter lays the groundwork for more advanced studies in geometry and calculus. By the end of Ex 12.1 Class 10, you will be proficient in handling problems related to surface areas and volumes, making you well-prepared for both academic exams and practical challenges.

toc-symbolTable of Content
toggle-arrow


Glance on NCERT Solutions Maths Chapter 12 Exercise 12.1 Class 10 | Vedantu

  • Chapter 12 Exercise 12.1 of your Class 10 Maths textbook likely deals with calculating the surface area and volume of composite solid figures. These figures are formed by joining together simpler shapes like cubes, cuboids, cones, cylinders, and hemispheres.

  • The key to solving these problems is to identify the individual shapes involved and then calculate their surface areas and volumes separately. Each shape, you will likely use formulas you have learned earlier in the chapter. 

  • Once you have the surface areas and volumes of the individual shapes, you can add them together to find the total surface area and volume of the composite figure.

Competitive Exams after 12th Science
tp-imag
bottom-arrow
tp-imag
bottom-arrow
tp-imag
bottom-arrow
tp-imag
bottom-arrow
tp-imag
bottom-arrow
tp-imag
bottom-arrow
Watch videos on

NCERT Solutions for Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.1 – 2025-26
Previous
Next
Vedantu 9&10
Subscribe
iconShare
Surface Areas and Volumes Full Chapter in One Shot | CBSE Class 10 Maths Chap 13 | Term 2 Solutions
8.1K likes
133.9K Views
3 years ago
Vedantu 9&10
Subscribe
iconShare
Surface Areas and Volumes L-1 [ Combination of Solids ] CBSE Class 10 Maths Ch13 | Term 2 Solutions
6K likes
159K Views
3 years ago

Access NCERT Solutions for Class 10th Exercise 12.1 Maths Chapter 12 – Surface Areas and Volumes

1. Two cubes each of volume $64\,\,c{{m}^{3}}$ are joined end to end. Find the surface area of the resulting cuboids.

Ans: Given: Volume of cubes \[=\text{ }64\text{ }c{{m}^{3}}\]


\[{{\left( Edge \right)}^{3}}=\text{ }64\]


Edge \[=\text{ }4\text{ }cm\]


the surface area of cuboids


If cubes are joined end to end, the dimensions of the resulting cuboid will be \[4\text{ }cm,4cm,8\text{ }cm\].


Surface area of cuboids:

$\Rightarrow 2\left( lb+bh+lh \right)$


$\Rightarrow 2\left( \left( 4 \right)\left( 4 \right)+\left( 4 \right)\left( 8 \right)+\left( 4 \right)\left( 8 \right) \right)$


$\Rightarrow 2\left( 16+32+32 \right)$


$\Rightarrow 2\left( 16+64 \right)$


$\Rightarrow 2\left( 80 \right)=160\,c{{m}^{2}}$


2. 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. $\left[ Use\,\,\pi =\dfrac{22}{7} \right]$.

Ans:


A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder


It can be observed that the radius \[\left( r \right)\]of the cylindrical part and the hemispherical part is the same (i.e.,\[7\text{ }cm\]).


Height of hemispherical part \[=Radius=7cm\]


Height of cylindrical part \[\left( h \right)=13-7=6\text{ }cm\]


Inner surface area of the vessel : \[\Rightarrow CSA\text{ }of\text{ }cylindrical\text{ }part+CSA\text{ }of\text{ }hemispherical\text{ }part\]


$\Rightarrow 2\pi rh+2\pi {{r}^{2}}$


$\Rightarrow \left[ 2\left( \dfrac{22}{7} \right)\left( 7 \right)\left( 6 \right) \right]+\left[ 2\left( \dfrac{22}{7} \right)\left( 7 \right)\left( 7 \right) \right]$


$\Rightarrow 44\left( 6+7 \right)$


$\Rightarrow 44\times 13=572\,c{{m}^{2}}$


3. A toy is in the form of a cone of radius $3.5cm$ 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. $\left[ Use\,\,\pi =\dfrac{22}{7} \right]$

Ans:


A toy is in the form of a cone of radius 3.5cm mounted on a hemisphere of same radius


It can be observed that the radius of the conical part and the hemispherical part is the same (i.e., 3.5 cm). 


Height of hemispherical part \[=Radius\left( r \right)=3.5=\dfrac{7}{2}\text{ }cm\]


Height of conical part \[\left( h \right)=5.5-3.5=12\text{ }cm\]


Slant height (l) of conical part:


$\Rightarrow \sqrt{{{r}^{2}}+{{h}^{2}}}$


$\Rightarrow \sqrt{{{\left( \dfrac{7}{2} \right)}^{2}}+{{\left( 12 \right)}^{2}}}=\sqrt{\dfrac{49}{4}+144}$


$\Rightarrow \sqrt{\dfrac{625}{4}}=\dfrac{25}{2}$


Total surface area of toy:

$\Rightarrow CSA\text{ }of\text{ }conical\text{ }part+CSA\text{ }of\text{ }hemispherical\text{ }part$


$\Rightarrow \pi rl+2\pi {{r}^{2}}$


$\Rightarrow \left[ \left( \frac{22}{7} \right)\left( \frac{7}{2} \right)\left( \frac{25}{2} \right) \right]+\left[ 2{{\left( \frac{7}{2} \right)}^{2}}\frac{22}{7} \right]$


$\Rightarrow 137.5+77=214.5\,c{{m}^{2}}$


4. 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. $\left[ Use\,\,\pi =\dfrac{22}{7} \right]$

Ans:


A cubical block of side 7cm is surmounted by a hemisphere

 

From the figure, it can be observed that the greatest diameter possible for such a hemisphere is equal to the cube’s edge, i.e.,\[7cm\].


Radius (r) of hemispherical part\[=\dfrac{7}{2}=3.5cm\]


Total surface area of solid:


Surface area of cubical part + CSA of hemispherical = Area of base of hemispherical part


$\Rightarrow 6{{\left( edge \right)}^{2}}+2\pi {{r}^{2}}-\pi {{r}^{2}}$


$\Rightarrow 6{{\left( edge \right)}^{2}}+\pi {{r}^{2}}$


\[\Rightarrow 6{{\left( 7 \right)}^{2}}+\left( \dfrac{22}{7} \right){{\left( \dfrac{7}{2} \right)}^{2}}\]


$\Rightarrow 294+38.5=332.5\,c{{m}^{2}}$


5. 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.

Ans:


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


Diameter of hemisphere \[=Edge\text{ }of\text{ }cube=L\] 


Radius of hemisphere \[=\dfrac{L}{2}\]


Total surface area of solid:


Surface area of cubical part + CSA of hemispherical part = Area of base of hemispherical part


$\Rightarrow 6{{\left( edge \right)}^{2}}+2\pi {{r}^{2}}-\pi {{r}^{2}}$


$\Rightarrow 6{{\left( edge \right)}^{2}}+\pi {{r}^{2}}$


$\Rightarrow 6\left( {{L}^{2}} \right)+\frac{\pi {{L}^{2}}}{4}$


$\Rightarrow 6{{L}^{2}}+\dfrac{\pi {{L}^{2}}}{4}$


$=\dfrac{1}{4}\left( 24+\pi  \right){{L}^{2}}$


6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its end (see the given figure). The length of the entire capsule is $14$ mm and the diameter of the capsule is 5 mm. Find its surface area [use $\pi =\frac{22}{7}$]


A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its end
 

Ans: It can be observed that,


surface area


Radius r of cylindrical part = Radius( r) of hemispherical part


$=\dfrac{Diameter\,of\,the\,capsule}{2}$


$=\dfrac{5}{2}$


Length of  cylindrical part ( h )=Length of the entire capsule$-2\times r$ 


$=14-5$


$=9\,mm$


Surface area of capsule:


2CSA of hemispherical part+CSA of cylindrical part


$\Rightarrow 2\left( 2\pi {{r}^{2}} \right)+2\pi rh$


$\Rightarrow 4\pi {{\left( \dfrac{5}{2} \right)}^{2}}+2\pi \left( \dfrac{5}{2} \right)\left( 9 \right)$

 

$\Rightarrow 25\pi +45\pi $


$\Rightarrow 70\left( \dfrac{22}{7} \right)$


$\Rightarrow 220\,\,m{{m}^{2}}$


7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are \[\mathbf{2}.\mathbf{1}\text{ }\mathbf{m}\]and \[\mathbf{4}\text{ }\mathbf{m}\] respectively, and the slant height of the top is\[\mathbf{2}.\mathbf{8}\text{ }\mathbf{m}\], 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 \[\mathbf{Rs}\text{ }\mathbf{500}\text{ }\mathbf{per}\text{ }{{\mathbf{m}}^{2}}\]. (Note that the base of the tent will not be covered with canvas.) $\left[ Use\,\,\pi =\dfrac{22}{7} \right]$

Ans:


A tent is in the shape of a cylinder surmounted by a conical top


Given that,


Height (h) of the cylindrical part \[=2.1m\]


Diameter of the cylindrical part \[=4m\]


Radius of the cylindrical part \[=2m\]


Slant height (l) of conical part \[=2.8m\]


Area of canvas used:


CSA of conical part+CSA of cylindrical part


$\Rightarrow \pi rl+2\pi rh$


$\Rightarrow \pi \left( 2 \right)\left( 2.8 \right)+2\pi \left( 2 \right)\left( 2.1 \right)$


$\Rightarrow 2\pi \left[ 2.8+2\left( 2.1 \right) \right]$ 


$\Rightarrow 2\pi \left[ 2.8+4.2 \right]=2\left( \dfrac{22}{7} \right)\left( 7 \right)$


$\Rightarrow 44{{m}^{2}}$


Cost of \[1\text{ }{{m}^{2}}\]canvas = \[Rs\text{ }500\]


Cost of \[44\text{ }{{m}^{2}}\]canvas \[=44\times 500=22000\]


Therefore, it will cost \[Rs\text{ }22000\]for making such a tent.


8. From a solid cylinder whose height is \[\mathbf{2}.\mathbf{4}\text{ }\mathbf{cm}\] and diameter \[\mathbf{1}.\mathbf{4}\text{ }\mathbf{cm}\], 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 \[\mathbf{c}{{\mathbf{m}}^{2}}\]. [ Use $\pi =\frac{22}{7}$ ]

Ans:


a solid cylinder


Given:


Height (h) of the conical part=Height (h) of the cylindrical part\[=2.4\text{ }cm\]


Diameter of the cylindrical part = 1.4 cm


Therefore, radius (r) of the cylindrical part \[=0.7cm\]


Slant height (l) of conical part:


$\Rightarrow \sqrt{{{r}^{2}}+{{h}^{2}}}$


$\Rightarrow \sqrt{{{\left( 0.7 \right)}^{2}}+{{\left( 2.4 \right)}^{2}}}$


$\Rightarrow \sqrt{0.49+5.76}$


$\Rightarrow \sqrt{0.49+5.76}$


$\Rightarrow \sqrt{6.25}=2.5$


Total surface area of the remaining solid will be:


CSA of cylindrical part+CSA of conical part +Area of cylindrical base $


$\Rightarrow 2\pi rh+\pi rl+\pi {{r}^{2}}$


$\Rightarrow \left( 2\times \dfrac{22}{7}\times 0.7\times 2.4 \right)+\left( \dfrac{22}{7}\times 0.7\times 2.5 \right)+\left( \dfrac{22}{7}\times 0.7\times 0.7 \right)$

$\Rightarrow \left( 4.4\times 2.4 \right)+\left( 2.2\times 2.5 \right)+\left( 2.2\times 0.7 \right)$


$\Rightarrow 10.56+5.50+1.54=17.60~\text{c}{{\text{m}}^{2}}$


The total surface area of the remaining solid to the nearest $\text{c}{{\text{m}}^{2}}$ is $18~\text{c}{{\text{m}}^{2}}$


9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the given figure. If the height of the cylinder is \[\mathbf{10}\text{ }\mathbf{cm}\], and its base is of radius \[\mathbf{3}.\mathbf{5}\text{ }\mathbf{cm}\], find the total surface area of the article. $\left[ Use\,\,\pi =\dfrac{22}{7} \right]$

Ans:


A wooden article was made by scooping out a hemisphere from each end of a solid cylinder


Given: Radius (r) of cylindrical part =Radius( r) of hemispherical part=3.5 cm


Height of cylindrical part (h) \[=10\text{ }cm\]


Surface area of article:


CSA of cylindrical part+2  CSA of hemispherical part


$\Rightarrow 2\pi rh+2\left( 2\pi {{r}^{2}} \right)$


$\Rightarrow 2\pi \left( 3.5 \right)\left( 10 \right)+2\left( 2\pi  \right){{\left( 3.5 \right)}^{2}}$


$\Rightarrow 70\pi +49\pi $


$\Rightarrow 119\pi $


$\Rightarrow 17\left( 22 \right)=374\,c{{m}^{2}}$

Conclusion

To sum up, it is imperative that you understand NCERT Solutions for Class 10 Maths Ex 12.1 Chapter 12 - Surface Areas and Volumes. Concentrate on comprehending the ideas behind surface area, volume, and cuboids. Get proficient at precisely applying formulas to solve issues quickly. Vedantu's all-inclusive solutions support complete preparation by offering precise explanations and step-by-step instructions. Class 10 Ex 12.1 will build a strong foundation for further mathematical concepts.


Class 10 Maths Chapter 12: Exercises Breakdown

Chapter 12 Surface Areas and Volumes All Exercises in PDF Format

Exercise 12.2

8 Questions and Solutions


CBSE Class 10 Maths Chapter 12 Other Study Materials


Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.



Study Resources for Class 10 Maths

For complete preparation of Maths for CBSE Class 10 board exams, check out the following links for different study materials available at Vedantu.

WhatsApp Banner
Best Seller - Grade 10
View More>
Previous
Next

FAQs on NCERT Solutions for Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.1 – 2025-26

1. What does NCERT Solutions for Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.1 focus on?

NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1 focus on stepwise problem-solving for calculating the surface areas and volumes of fundamental 3D shapes such as cubes, cuboids, cylinders, cones, spheres, and hemispheres, in line with the CBSE 2025–26 syllabus. These solutions emphasize applying formulas, visualizing composite solids, and strengthening conceptual understanding for CBSE board exam readiness.

2. How should students approach combined solid figures in Surface Areas and Volumes as per NCERT Solutions?

To solve problems involving composite solids in NCERT Solutions for Class 10 Maths Chapter 12, students should:

  • Identify each individual component (cube, cuboid, hemisphere, etc.).
  • Calculate surface area and volume for each component separately using the correct formulas.
  • Add or subtract the relevant areas or volumes according to how the solids are joined or cut.
  • Carefully note which surfaces are exposed or hidden in the final composite object as per the question.

3. What are the key formulas covered in Class 10 Maths Chapter 12 Surface Areas and Volumes NCERT Solutions?

Key formulas provided in NCERT Solutions for Class 10 Maths Chapter 12 include:

  • Surface Area (SA) and Volume (V) of basic solids: Cube, Cuboid, Sphere, Cylinder, Cone, Hemisphere.
  • Curved Surface Area (CSA), Total Surface Area (TSA) for each solid figure.
  • Volume conversions for composite and transformed solids.
  • Frustum of a cone surface area and volume formulas.

4. How many questions are in Exercise 12.1 of Class 10 Maths Chapter 12, and what types do they cover?

Exercise 12.1 of Class 10 Maths NCERT Chapter 12 contains nine questions (seven long and two short answer types). These cover direct application of surface area and volume formulas, as well as complex scenarios involving composite solids and real-life applications such as containers, vessels, toys, and practical geometrical arrangements.

5. What is a frustum of a cone as defined in the NCERT Solutions for Class 10 Maths Chapter 12?

A frustum of a cone is the part of a cone that remains after cutting through it with a plane parallel to its base and removing the smaller, top portion. The resulting solid has two parallel circular faces—one larger (base) and one smaller (top)—with the same axis as the original cone.

6. Why is understanding the practical application of surface area and volume important for Class 10 students?

Understanding surface area and volume is crucial because these concepts have direct real-world uses, such as:

  • Calculating material requirements (e.g., amount of canvas, tin, or paint needed to cover objects).
  • Solving problems related to packaging, construction, and manufacturing.
  • Forming a foundation for advanced topics in geometry, calculus, and engineering fields.

7. How do the NCERT Solutions ensure accuracy and CBSE compliance for Class 10 Maths Chapter 12?

Vedantu’s NCERT Solutions for Chapter 12 are prepared by CBSE subject experts. All solutions strictly follow the latest CBSE guidelines for 2025–26, use standard mathematical notation, and provide in-depth stepwise explanations. This ensures alignment with board expectation and facilitates high exam scores.

8. What is the most common mistake students make when solving surface area of composite solids?

The most frequent error is misidentifying which surfaces are actually exposed after combining or cutting solids. Students sometimes count surfaces twice or omit hidden faces, leading to incorrect answers. Always analyze the final structure carefully and calculate only the visible surface area required by the problem.

9. Can different composite solids have the same volume but different surface areas? Explain using an example.

Yes, two composite solids can have equal volumes but different surface areas. For example, a cube and a tall thin cuboid may both have a volume of 64 cm³, but the surface area will differ depending on edge lengths. Surface area depends on how the exposed faces are arranged; volume depends only on the space occupied.

10. What strategy should you use when a question asks for "area of canvas used" or "outer surface area covered"?

Focus on calculating the Curved Surface Area (CSA) or Total Surface Area (TSA) of the exposed part only. If the base is not required to be covered (e.g., tent or vessel), exclude its area. Always read the question to identify whether you need to include or omit certain faces in your calculation.

11. How is the slant height of a cone calculated in surface area questions in NCERT Solutions?

In Class 10 Maths Chapter 12, the slant height (l) of a cone is found using the formula: l = √(r² + h²), where r is the radius of the base and h is the vertical height. This value is crucial for calculating the curved surface area of conical shapes.

12. What are some real-life applications of NCERT Solutions for Class 10 Maths Chapter 12 concepts?

Important applications from this chapter include:

  • Designing tanks, vessels, capsules, and tents.
  • Planning packaging, shipping boxes, and storage containers.
  • Calculating the amount of paint or wrapping needed for curved surfaces.
  • Modeling and constructing solid toys and industrial parts with combined shapes.

13. How can Vedantu’s NCERT Solutions help students preparing for the Class 10 board exam in 2025–26?

Vedantu’s NCERT Solutions for Class 10 Maths Chapter 12 provide accurate, step-wise answers with explanations following CBSE 2025–26 patterns. This helps students quickly master concept application, avoid common errors, and build exam confidence, ensuring strong foundational understanding and improved performance in school and board exams.

14. What should a student do if they get stuck on a Surface Areas and Volumes problem in NCERT Ex 12.1?

  • Break the problem down into smaller parts: Identify each solid separately.
  • Write down and label all dimensions given in the question.
  • Draw a rough diagram to visualize the object.
  • Apply the correct NCERT formula for each step.
  • Check if any surfaces are hidden or not included in the calculation.
Using this approach, review Vedantu’s solution steps to find where you might have missed or misapplied a formula.

15. If a solid is transformed from one shape to another (like melting a cylinder to form a sphere), what remains constant and why?

The volume remains constant during such transformations. This is because the material used does not change in quantity; only the shape changes. The surface area often changes, but the total space occupied (volume) stays the same as per the law of conservation of matter in such problems.