Aakash Repeater Courses
Take Aakash iACST and get instant scholarship on coaching programs.
Imagine you're designing a water tank for your home. You need to calculate how much water it can hold (volume) and how much material will be needed to paint the outside of this water tank(surface area). The answer to these questions lies in understanding surface areas and volumes. This chapter helps us to understand how to calculate the area related to covering a solid object and the space it occupies. It also deals with calculating the surface area and volume of different solids like cubes, cylinders, cones, and spheres, and the formulas for surface area and volume. NCERT Solutions for Class 10 can help the students understand these concepts and will make them more efficient in solving problems involving Surface Areas and Volumes.
This Story also Contains
These NCERT Solutions for class 10 Maths are designed by our experienced subject experts to offer a structured approach and step-by-step solutions for students to prepare for the board exams. These Solutions cover all exercises in the NCERT textbook and are easily available for download. The step-by-step explanations for different types of problems help students gain a strong understanding of surface area and volume. Many toppers rely on NCERT Solutions since they are designed as per the latest syllabus. Get NCERT solved questions, NCERT syllabus, and NCERT free PDF directly from this NCERT article.
The NCERT Solutions for Class 10 Maths Chapter 12 have been prepared by Careers360 experts to make learning simpler and to help you score better in exams. You can also download the solutions in PDF format.
Class 10 Maths Chapter 12 Solutions Exercise: 12.1 Page number: 166-167 Total questions: 9 |
Take Aakash iACST and get instant scholarship on coaching programs.
Q1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Answer:
We are given that the volume of the cube $=\ 64\ cm^3$
Also, the volume of a cube is given by $=\ a^3$ ( here $a$ is the edge of the cube)
Thus : $a^3\ =\ 64$
$a\ =\ 4\ cm$
Now, according to the question, we have combined the two cubes the edge lengths of the formed cuboid are 4 cm, 4 cm, and 8 cm.
The surface area of a cuboid is : $=\ 2\left ( lb\ +\ bh\ +\ hl \right )$
$=\ 2\left ( 8\times 4\ +4\times 4\ +\ 4\times 8 \right )$
$=\ 2\left ( 80 \right )$
$=\ 160\ cm^2$
Thus, the area of the formed cuboid is 160 cm2.
Answer:
Since the vessel consists of a hemisphere and cylinder, its area is given by :
Area of vessel = Inner area of the cylinder(curved) + Inner area of hemisphere
The inner surface area of the hemisphere $=\ 2\pi r^2$
$=\ 2\times \left ( \frac{22}{7} \right ) \times 7^2$
$=\ 308\ cm^2$
The surface area of the cylinder $=\ 2\pi rh$
$=\ 2\times \frac{22}{7}\times 7\times 6$
$=\ 264\ cm^2$
Thus, the inner surface area of the vessel is $= 308 + 264 = 572 \:cm^2.$
Answer:
The required surface area of the toy is given by :
Area of the toy = Surface area of the hemisphere + Surface area of the cone
Firstly, consider the hemisphere :
The surface area of a hemisphere $=\ 2 \pi r^2$
$=\ 2 \times \frac{22}{7}\times \left ( 3.5 \right )^2$
$=\ 77\ cm^2$
Now for cone, we have :
The surface area of a cone $=\ \pi rl$
Thus, we need to calculate the slant of the cone.
We know that :
$l^2\ =\ h^2\ +\ r^2$
$=\ 12^2\ +\ 3.5^2$
$=\ \frac{625}{4}$
$l\ =\ \frac{25}{2}\ =\ 12.5\ cm$
Thus, the surface area of a cone $=\ \pi rl$
$=\ \frac{22}{7}\times 3.5\times 12.5$
$=\ 137.5\ cm^2$
Hence, the total surface area of the toy $=\ 77\ +\ 137.5\ =\ 214.5\ cm^2.$
Answer:
It is given that the hemisphere is mounted on the cuboid, thus, the hemisphere can take on complete as its diameter (which is maximum).
Thus, the greatest diameter of the hemisphere is 7 cm.
Now, the total surface area of the solid = Surface area of the cube + Surface area of the hemisphere - Area of the base of a hemisphere (as this is counted on one side of the cube)
The surface area of the cube $=\ 6a^3$
$=\ 6\times 7^3\ =\ 294\ cm^2$
Now, the area of a hemisphere $=\ 2\pi r^2$
$=\ 2\times \frac{22}{7}\times \left ( \frac{7}{2} \right )^2\ =\ 77\ cm^2$
The area of the base of a hemisphere $=\ \pi r^2\ =\ \frac{22}{7}\times \left ( \frac{7}{2} \right )^2\ =\ 38.5\ cm^2$
Hence, the surface area of the solid is $= 294 + 77 - 38.5 = 332.5 \:cm^2$.
Answer:
It is given that the diameter of the hemisphere is equal to the edge length of the cube.
The total surface area of the solid is given by :
The surface area of solid = Surface area of cube + Surface area of the hemisphere - Area of the base of the hemisphere
The surface area of the cube $=\ 6l^2$
The surface area of the hemisphere:
$=\ 2\pi r^2\ =\ 2\pi \left ( \frac{l}{2} \right )^2$
Area of base of the hemisphere:
$=\ \pi r^2\ =\ \pi \left ( \frac{l}{2} \right )^2$
Thus, the area of the solid is:
$=\ 6l^2\ +\ \pi \left ( \frac{l}{2} \right )^2\ unit^2.$
Answer:
It is clear from the figure that the capsule has a hemisphere and cylinder structure.
The surface area of the capsule = 2 (Area of the hemisphere) + Area of the cylindrical part
Area of hemisphere $=\ 2\pi r^2$
$=\ 2\pi\times \left ( \frac{5}{2} \right )^2$
$=\ \frac{25}{2} \pi\ mm^2$
The area of the cylinder $=\ 2\pi rh$
$=\ 2\pi \times \frac{5}{2}\times 9$
$=\ 45 \pi\ mm^2$
Thus, the area of the solid $=\ 2\left ( \frac{25}{2} \right )\pi\ +\ 45 \pi$
$=\ 25\pi\ +\ 45 \pi$
$=\ 70 \pi$
$=\ 220\ mm^2.$
Answer:
The canvas will cover the cylindrical part as well as the conical part.
So, the area of the canvas = Area of the cylindrical part (curved) + Area of the conical part
Now, the area of the cylindrical part $=\ 2\pi rh$
$=\ 2\pi \times 2\times 2.1$
$=\ 8.4 \pi\ m^2$
The area of the cone is $=\ \pi rl$
$=\ \pi \times 2\times 2.8$
$=\ 5.6\pi\ m^2$
Thus, the area of the canvas $=\ 8.4\pi\ +\ 5.6\pi$
$=\ 14\pi\ =\ 44\ m^2$
Further, it is given that the rate of canvas per m2 is Rs. 500.
Thus, the required money is $=500 \times 44=$ Rs. 22,000.
Answer:
Firstly, we need to calculate the slant height of the cone :
$l^2\ =\ r^2\ +\ h^2$
$=\ (0.7)^2\ +\ (2.4)^2$
or $l^2\ =\ 6.25$
or $l\ =\ 2.5\ cm$
Now, the total surface area of a solid can be calculated as :
The surface area of the solid = Surface area of the cylinder + Surface area of the cone + Area of the base of the cylinder
The surface area of the cylinder is $=\ 2 \pi rh$
$=\ 2 \pi \times 0.7\times 2.4$
$=\ 10.56\ cm^2$
Now, the surface area of a cone $=\ \pi rl$
$=\ \pi \times 0.7\times 2.5$
$=\ 5.50\ cm^2$
And the area of the base of the cylinder is $=\ \pi r^2$
$=\ \pi \times 0.7\times 0.7$
$=\ 1.54\ cm^2$
Thus, the required area of the solid = 10. 56 + 5.50 + 1.54 = 17.60 cm2.
Thus, the total surface area of the remaining solid to the nearest cm2 is 18 cm2.
Answer:
The required surface area is given by :
The surface area of the article = Surface area of cylindrical part + 2 (Surface area of the hemisphere)
Now, the area of the cylinder $=\ 2\pi rh$
$=\ 2\pi \times 3.5\times 10$
$=\ 70\pi \ cm^2$
The surface area of the hemisphere $=\ 2\pi r^2$
$=\ 2\pi \times 3.5\times 3.5$
$=\ 24.5\pi\ cm^2$
Thus the required area $=70 \pi +2\times 24.5 \pi = 374 cm^2.$
Class 10 Maths Chapter 12 Solutions Exercise: 12.2 Page number: 169-170 Total questions: 8 |
Answer:
The volume of the solid is given by :
The volume of solid = Volume of cone + Volume of a hemisphere
The volume of the cone $=\ \frac{1}{3} \pi r^2h$
$=\ \frac{1}{3} \pi \times 1^2\times 1$
$=\ \frac{\pi}{3}\ cm^3$
The volume of the hemisphere $=\ \frac{2}{3}\pi r^3$
$=\ \frac{2}{3}\pi \times 1^3$
$=\ \frac{2\pi}{3}\ cm^3$
Hence, the volume of so lithe d is :
$=\ \frac{\pi}{3}\ +\ \frac{2\pi}{3}\ =\ \pi\ cm^3.$
Answer:
The volume of air present = Volume of cylinder + 2 (Volume of a cone)
Now, the volume of a cylinder: $=\ \pi r^2h$
$=\ \pi \left ( \frac{3}{2} \right )^2\times 8$
$=\ 18\pi \ cm^3$
The volume of a cone is $=\ \frac{1}{3} \pi r^2h$
$=\ \frac{1}{3} \pi \times \left ( \frac{3}{2} \right )^2\times 2$
$=\ \frac{3}{2} \pi \ cm^3$
Thus, the volume of air is $=\ 18 \pi\ +\ 2\times \frac{3}{2} \pi \ =\ 21\pi$
$=\ 66\ cm^3$
Answer:
It is clear from the figure that Gulab Jamun has one cylindrical part and two hemispherical parts.
Thus, the volume of the gulab jamun is = Volume of cylindrical part + 2 (Volume of the hemisphere )
Now, the volume of the cylinder is $=\ \pi r^2h$
$=\ \pi\times 1.4^2\times 2.2$
$=\ 13.55\ cm^3$
The volume of a hemisphere is :
$\\=\ \frac{2}{3}\pi r^3\\\\=\ \frac{2}{3}\times \pi \times (1.4)^3\\\\=\ 5.75\ cm^3$
Thus, the volume of 1 gulab jamun is $= 13.55 + 2 (5.75) = 25.05 cm^3.$
Hence, the volume of 45 gulab jamun $=\ 45(25.05)\ cm^3\ =\ 1127.25\ cm^3$
Further, it is given that one gulab jamun contains sugar syrup up to 30%.
So, the total volume of sugar present $=\ \frac{30}{100}\times 1127.25\ =\ 338\ cm^3.$
Answer:
The volume of wood is given by = volume of the cuboid - the volume of four cones.
Firstly, the volume of the cuboid $=\ lbh$
$=\ 15\times 10\times 3.5$
$=\ 525\ cm^3$
And the volume of the cone $=\frac13 \pi r^2h=\frac13 \pi \times 0.5^2 \times 1.4= 0.3665 \ cm^3$
Thus, the volume of wood is $= 525 - 4 (0.3665) = 523.53 \:cm^3.$
Answer:
According to the question :
Water spilled from the container = Volume of lead balls.
Let us assume the number of lead balls to be n.
Thus, the equation becomes :
$\frac{1}{4}\times Volume_{cone}\ =\ n\times \frac{4}{3} \pi r^3$
or $\frac{1}{4}\times \frac{1}{3}\pi\times5^2\times 8\ =\ n\times \frac{4}{3} \pi\times 0.5^3$
or $n\ =\ \frac{25\times 8}{16\times \left ( \frac{1}{2} \right )^3}$
or $n\ =\ 100$
Hence, the number of lead shots dropped is 100.
Answer:
The pole can be divided into one large cylinder and one small cylinder.
Thus, the volume of the pole = volume of a large cylinder + volume of a small cylinder
$=\ \pi r_l^2h_l\ +\ \pi r_s^2h_s$
$=\ \pi \times 12^2\times 220\ +\ \pi \times 8^2\times 60$
$=\ \pi \times \left ( 144\times 220\ +\ 64\times 60 \right )$
$=\ 3.14\times 35520$
$=\ 111532.5\ cm^3$
Now, according to the question, the mass of the pole is :
$=\ 8\times 111532.5$
$=\ 892262.4\ g\ =\ 892.262\ Kg.$
Answer:
It is clear from the question that the required volume is :
The volume of water (left) =Volume of a cylinder - Volume of solid
Now the volume of the cylinder is $=\ \pi r^2h$
or $=\ \pi\times (60)^2\times 180\ cm^3$
The volume of the solid is :
$\\=\ \frac{1}{3} \pi r^2h\ +\ \frac{2}{3}\pi r^3\\\\=\ \frac{1}{3} \times \pi \times (60)^2\times 120\ +\ \frac{2}{3}\times \pi \times (60)^3\\\\=\ \pi (60)^2\times 80\ cm^3$
Thus, the volume of water left :
$\\=\ \pi (60)^2\times 180\ -\ \pi (60)^2\times 80\\\\=\ \pi (60)^2\times 100\\\\=\ 1131428.57\ cm^3\\\\=\ 1.131\ m^3.$
Answer:
The volume of the vessel is given by :
The volume of the vessel = Volume of the sphere + Volume of the cylindrical part
Now, the volume of the sphere is :
$\\=\ \frac{4}{3}\pi r^3\\\\=\ \frac{4}{3}\pi \left ( \frac{8.5}{2} \right )^3\\\\=\ 321.55\ cm^3$
The volume of the cylinder is:-
$\\=\ \pi r^2h\\\\=\ \pi \times (1)^2\times 8\\\\=\ 25.13\ cm^3$
Thus, the volume of the vessel is $= 321.55 + 25.13 = 346.68\: cm^3.$
Exercise-wise NCERT Solutions of Surface Areas and Volumes Class 10 Maths Chapter 12 are provided in the link below.
Question:
A cylinder of height 8 cm and radius 6 cm is melted and converted into three cones of the same radius and height as the cylinder. Determine the total curved surface area of cones.
Answer:
The radius ($r$) of each cone = radius of the cylinder = 6 cm
The height ($h$) of each cone = height of the cylinder = 8 cm
The curved surface area of a cone, where $l$ is the slant height of the cone $= \pi r l$
Now, $l = \sqrt{r^2 + h^2}= \sqrt{(6)^2 + (8)^2} = 10 \text{ cm}$
The curved surface area of a cone, where $l$ is the slant height of the cone $= \pi r l$
So, the curved surface area of each cone,
$= \pi (6) (10) = 60\pi \text{ cm}^2$
Since there are 3 cones, the total curved surface area,
$= 3 \times 60\pi = 180\pi \text{ cm}^2$
Hence, the correct answer is $180\pi \text{ cm}^2$.
Total Surface Area, denoted as TSA, signifies the entire expanse covered by an object's surface. This encompasses the collective area of all its external facets. Here are some prominent geometrical figures and their corresponding TSA formulas.
The concept of Curved Surface Area (CSA) or Lateral Surface Area (LSA) comes into focus, particularly in forms such as cylinders, cones, and pyramids. CSA represents the area of the curved component or sides, excluding the top and bottom facets. LSA, on the other hand, pertains to the lateral area of various shapes. The following are some key examples of these surface area measures:
Volume emerges as a fundamental attribute, denoting the spatial extent occupied by an object or substance, quantified in cubic units. The following volume formulas unravel the essence of these geometric measurements:
Topics you will learn in NCERT Class 10 Maths Chapter 12 Surface Areas and Volumes include:
Access all NCERT Class 10 Maths solutions from one place using the links below.
Also, read,
Given below are the subject-wise exemplar solutions of class 10 NCERT:
Here are some useful links for NCERT books and the NCERT syllabus for class 10:
Frequently Asked Questions (FAQs)
The concepts of surface areas and volumes are applied widely in real life. In construction, architects and engineers use these principles to calculate the amount of material needed for building structures, such as the walls or roofs of buildings. In the packaging industry, companies use surface area formulas to design packaging materials that minimize waste while ensuring the container fits the product properly.
To find the surface area of a combination of solids we require breaking the complex solid into simpler shapes. Like, if we have a cone placed on top of a hemisphere, we first calculate the surface area of the cone(excluding its base) and then calculate the surface area of the hemisphere. Once we have the surface areas of the individual solids, we can simply add them together to find the total surface area.We have to carefully exclude any overlapping areas, like the base of the cone that touches the hemisphere, to avoid double-counting.
In board exams, questions from Chapter 12 typically focus on calculating the surface area and volume of solids. These can include both direct calculation problems, where you simply apply the formulas, and more complex word problems that involve real-life situations, such as finding the amount of material needed to cover a tank or determining the volume of water a container can hold. There may also be questions involving the combination of solids, where you need to calculate the surface area or volume of an object made up of more than one solid. These questions test both your conceptual understanding and problem-solving skills.
Class 10 Chapter 12, Surface Areas and Volumes, focuses on calculating the surface area and volume of various solids. Key formulas include the surface area and volume of shapes like cubes, spheres, cylinders, and cones.
The surface area of a cube is given by $6a^2$ (where a is the side length), and the volume is $a^3$.
The surface area of a sphere is $4\pi r^{2}$, and its volume is $\frac{4}{3}\pi r^{3}$, where r is the radius.
For a cylinder, the surface area is $2πr(r+h)$, and the volume is $πr^2h$, where r is the radius and h is the height.
The volume of a cone is $\frac{1}{3}\pi r^{2}h$, and the volume of a frustum of a cone is $\frac{1}{3}\pi h({r_1}^2 +r_1 r_2 +{r_2}^2)$, where $r_1$ and $r_2$ are the radii of the two circular ends, and h is the height.
$\text{Volume} = \frac{1}{3} \pi h (r_1^2 + r_1r_2 + r_2^2)$
Where, $r_1$ and $r_2$ are the radii of the top and bottom bases of the frustum, and h is the vertical height of the frustum. This formula calculates the volume of the frustum by taking into account both the radii of the circular ends and the height of the truncated cone, providing an accurate measure of the space inside the frustum.
On Question asked by student community
Hello,
If you want to get your 10th marksheet online, you just need to visit an official website like https://www.cbse.gov.in/ or https://results.cbse.nic.in/ for the CBSE board, and for the state board, you can check their website and provide your roll number, security PIN provided by the school, and school code to download the result.
I hope it will clear your query!!
Hello, if you are searching for Class 10 books for exam preparation, the right study material can make a big difference. Standard textbooks recommended by the board should be your first priority as they cover the syllabus completely. Along with that, reference books and guides can help in practicing extra questions and understanding concepts in detail. You can check the recommended books for exam preparation from the link I am sharing here.
https://school.careers360.com/ncert/ncert-books-for-class-10
https://school.careers360.com/boards/cbse/cbse-best-reference-books-for-cbse-class-10-exam
Hello
You asked about Class 10 sample paper board exam and most important questions. Practicing sample papers and previous year questions is one of the best ways to prepare for the board exam because it gives a clear idea of the exam pattern and types of questions asked. Schools and teachers usually recommend students to solve at least the last five years question papers along with model papers released by the board.
For Class 10 board exams, the most important areas are Mathematics, Science, Social Science, English, and Hindi or regional language. In Mathematics, questions from Algebra, Linear Equations, Geometry, Trigonometry, Statistics, and Probability are repeatedly seen. For Science, the key chapters are Chemical Reactions, Acids Bases and Salts, Metals and Non metals, Life Processes, Heredity, Light and Electricity. In Social Science, priority should be given to Nationalism, Resources and Development, Agriculture, Power Sharing, Democratic Politics, and Economics related topics. In English, focus on unseen passages, grammar exercises, and important writing tasks like letter writing and essays.
Follow these steps to access the SQPs and marking schemes:
Step 1: Visit https://cbseacademic.nic.in/
Step 2: Click on the link titled “CBSE Sample Papers 2026”
Step 3: A PDF will open with links to Class 10 and 12 sample papers
Step 4: Select your class (Class 10 or Class 12)
Step 5: Choose your subject
Step 6: Download both the sample paper and its marking scheme
If you are looking for Class 10 previous year question papers for 2026 preparation, you can easily access them through the links I’ll be attaching. These papers are very helpful because they give you a clear idea about the exam pattern, marking scheme, and the type of questions usually asked in board exams. Practicing these will not only improve your time management but also help you identify important chapters and commonly repeated questions.
https://school.careers360.com/boards/cbse/cbse-previous-year-question-papers-class-10
https://school.careers360.com/boards/cbse/cbse-previous-year-question-papers
Hello,
Yes, you can give the CBSE board exam in 2027.
If your date of birth is 25.05.2013, then in 2027 you will be around 14 years old, which is the right age for Class 10 as per CBSE rules. So, there is no problem.
Hope it helps !
Take Aakash iACST and get instant scholarship on coaching programs.
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
As per latest syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE