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# NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1- Surface Areas and Volumes

Last updated date: 04th Aug 2024
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## NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1 - FREE PDF Download

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.

Table of Content
1. NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 12 Exercise 12.1 Class 10 | Vedantu
3. Access NCERT Solutions for Class 10th Exercise 12.1 Maths Chapter 12 – Surface Areas and Volumes
4. Class 10 Maths Chapter 12: Exercises Breakdown
5. CBSE Class 10 Maths Chapter 12 Other Study Materials
6. Chapter-Specific NCERT Solutions for Class 10 Maths
FAQs

## 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.

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## 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 $=\text{ }4\text{ }cm$

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:

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:

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:

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:

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 $\pi =\frac{22}{7}$]

Ans: It can be observed that,

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

$=\dfrac{5}{2}$

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:

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:

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: 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

## 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.

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

1. What are the topics and sub-topics of the Class 10 Maths Chapter 12?

There are quite a few topics and sub-topics of this chapter. They are- 1) Surface Areas and Volumes, 2) Introduction, 3) Surface Area of a Combination Of Solid, 4) Volume of a combination of Solids, 5) Conversion of Solid From One Shape to Another, 6) Frustum of a Cone.

2. How many questions are there in Class 10 Maths Chapter 12 Exercise 12.1 of the NCERT textbook?

Chapter 12 Exercise 12.1 of Class 10 Mathematics NCERT book contains a total of seven long and two short questions. We, at Vedantu, provide the answers for the same as well in our solutions PDF, which you will find on our website and mobile application. You can download these solutions without spending a penny and use them for better exam preparation.

3. What is a frustum of a cone?

Here goes the definition- when we cut through a cone or cone-shaped object with a plane parallel to the base and remove the cone that is formed on one side of the plane, now the part which is now leftover on the other side of the plane is called as a frustum of a cone.

4. How is Vedantu’s NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1 prepared?

Class 10 students who are looking for a better NCERT Solutions for Mathematics should definitely look into the solutions PDF provided by Vedantu, India’s no.1 ed-tech company. All the answers provided in these solutions are 100% accurate. The problems on this exercise have been answered by our Maths expertise, to make students understand the theories in a better way. Topics covered in these materials are according to NCERT syllabus and guidelines. Therefore, solutions for Maths class 10, are very helpful for students who are doing the preparation for the exams and to score good marks.