SURFACE AREAS AND VOLUMES-Exercise 12.1

The solutions to the textbook exercises of NCERT Mathematics Class X – Surface Areas and Volumes are designed to provide students with a precise, logical, and concept-driven approach to solving numerical problems involving three-dimensional geometric figures. This chapter bridges abstract geometry with real-life measurement by exploring the surface areas and volumes of solids such as cubes, cuboids, cylinders, cones, spheres, hemispheres, frustums, and composite solids. These solutions follow a step-by-step methodology aligned strictly with the NCERT syllabus, ensuring conceptual clarity, correct formula application, and accurate unit handling. Each solution emphasizes reasoning, visualization, and systematic substitution, helping learners develop confidence in solving both routine and higher-order problems. Special attention is given to conversion of units, comparison of solids, and practical word problems, which are frequently tested in board examinations. This solution set serves as a reliable academic resource for revision, self-assessment, and exam preparation, enabling students to master spatial reasoning, improve numerical accuracy, and apply mathematical principles effectively in real-world contexts.

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December 17, 2025  |  By Academia Aeternum

SURFACE AREAS AND VOLUMES-Exercise 12.1

Maths - Exercise

Q1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.

Solution:

Volume of Cube = 64cm³
Side of Cube =a

$$\begin{aligned}a^{3}&=64\\ a&=\sqrt[3] {64}\\ &=\sqrt[3] {4^{3}}\\ a&=4\end{aligned}$$

On joining 2 cubes, the resulting cuboid dimensions will be 8,4,4
Hence surface Area of a cuboid

$$\begin{aligned}SA&=2\left( lb+bh+hl\right) \\ &=2\left( 8\times 4+4\times 4+4\times 8\right) \\ &=2\left( 32+16+32\right) \\ &=2\left( 64+16\right) \\ &=2\left( 80\right) \\ &=160\ \mathrm{cm^{2}}\end{aligned}$$
surface area of the resulting cuboid \(160\ \mathrm{cm^{2}}\)
Fig. 12.1.2-x
Fig. 12.1.2

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:

Diameter of hemisphere = 14 cm
Radius of hemisphere \((r)\) = 7 cm
Radius of cylinder will bethe same as the hemisphere = 7 cm
Height of Cylinder \((h)\) = 13-7 = 6 cm
Surface Area of the vessel will be the sum of curved surface Area of a hemisphere and a cylinder
Calculating CSA

$$\require{cancel}\begin{aligned}CSA&=2\pi rh+2\pi r^{2}\\ &=2\pi r\left( r+h\right) \\ &=2\times \dfrac{22}{\cancel{7}}\times \cancel{7}\times \left( 6+7\right) \\ &=2\times 22\times 13\\ &=44\times 13\\ &=572\ \mathrm{cm^2}\end{aligned}$$
Inner surface area of the vessel is \(572\ \mathrm{cm^2}\)
Fig. 12.1.3-x
Fig. 12.1.3

Q3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Solution:

Radius of the cone \((r)\) = 3.5 cm
Height of the toy = 15.5 cm
Height of the cone \((h)\)= 15.5-3.5 = 12 cm
Total surface Area of the toy = CSA of the cone + CSA of the hemisphere

$$\begin{aligned}TCSA&=\pi rl+2\pi r^{2}\\ &=\pi r\left( l+2r\right) \\\\ l&=\sqrt{r^{2}+h^{2}}\\ &=\sqrt{\left( 3.5\right) ^{2}+12^{2}}\\ &=\sqrt{12.25+144}\\ &=12.50\\\\ TCSA&=\pi \times 3.5\left( 12\cdot 5+2\times 3.5\right) \\ &=\dfrac{22}{7}\times 3.5\left( 12.5+7\right) \\ &=22\times 0.5\left( 19.5\right) \\ &=11\times 19.5\\ &=214.5\ \mathrm{cm^{2}}\end{aligned}$$
Total surface area of the toy is \(214.5\ \mathrm{cm^{2}}\)
Fig. 12.1.4-x
Fig. 12.1.4

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:

Side of cubical block \((a)\) = 7 cm
Greatest diameter of the hemisphere will be 7 cm \(\Rightarrow r=3.5\)
Surface Area of the solid = Surface Area of the cuboid + curved surface area of the hemisphere - Base Area of hemisphere

$$\require{cancel}\begin{aligned}SA&=6a^{2}+2\pi r^{2}-\pi r^{2}\\ &=6a^{2}+\pi r^{2}\\ &=6\cdot 7^{2}+\dfrac{22}{\cancel{7}}\times \cancelto{0.5}{3.5} \times 3.5\\ &=6\times 49+\times 22\times 3.5\times 0.5\\ &=294+11\times 3.5\\ &=294+38.5\\ &=332.5\ \mathrm{cm^{2}}\end{aligned}$$
Surface area of the solid is \(332.5\ \mathrm{cm^{2}}\)
Fig. 12.1.5-x
Fig. 12.1.5

Q5. A hemispherical depression is cut out from one face of a cubic 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:

Diameter of the hemisphere = \(l\)
Radius of the hemisphere \((r)\)= \(l/2\)
Side of cube = \(l\)
Surface Area of the remaining solid = surface Area of cubic block + CSA of hemisphere- Area of base of hemisphere

$$\begin{aligned}SA&=6l^{2}+2\pi \left( \frac{l}{2}\right) ^{2}-\pi \left( \frac{l}{2}\right) ^{2}\\\\ &=6l^{2}+A\left( \frac{l}{2}\right) ^{2}\\\\ &=6l^{2}+\dfrac{\pi l^{2}}{4}\\\\ &=\dfrac{24l^{2}+\pi l^{2}}{4}\\\\ &=\dfrac{l}{4}\left[ \pi +24\right] \end{aligned}$$
Surface area of the remaining solid \(\dfrac{l}{4}\left[ \pi +24\right] \)
Fig. 12.10-x
Fig. 12.10 (Source: NCERT)

Q6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig. 12.10). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Solution:

Diameter of the capsule = 5 mm
Radius of Capsule \((r)\)= 2.5 mm
Entire length of capsule = 14 mm
length of Cylinder \((h)\) = 14-(2.5+2.5) = 9 mm
SA of the capsule = CSA of Cylinder + 2 (CSA of hemisphere )

$$\require{cancel}\begin{aligned}TSA&=2\pi rh+4\pi r^{2}\\ &=2\pi rh+4\pi r^{2}\\ &=2\pi r\left( h+2r\right) \\ &=2\times \dfrac{22}{7}\times 2.5\times \left( 9+5\right) \\ &=2\times \dfrac{22}{\cancel{7}}\times 2.5\times \cancelto{2}{14} \\ &=2\times 22\times 2.5\times 2\\ &=220\ \mathrm{mm^{2}}\end{aligned}$$
Surface area of Capsule is \(=220\ \mathrm{mm^{2}}\)
Fig. 12.1.7-x
Fig. 12.1.7

Q7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical parts 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 ₹ 500 per m2. (Note that the base of the tent will not be covered with canvas.)

Solution:

Height of cylindrical part \((h)\) 2.1 m
Diameter of cylindrical part = 4 m
Radius of cylindrical part \((r)\) = 4/2 =2 m
Slant-height \((l)\) of conical past =2.8 m
canvas required for the tent = CSA of Cylindrical part + CSA of conical part

$$\begin{aligned}SA&=2\pi rh+\pi rl\\ &=\pi r\left( 2h+l\right) \\ &=\dfrac{22}{7}\times 2\left( 2\times 2.1+2.8\right) \\ &=\dfrac{22}{7}\times 2\left( 4.2+2.8\right) \\ &=\dfrac{22}{7}\times 2\times 7\\ &=44\ \mathrm{m^{2}}\end{aligned}$$

Rate of canvas = ₹500/m² therefore cost of canvas

$$\begin{aligned}&=44\times 500\\ &=\mathbb{\text{₹ }}22000\end{aligned}$$
Cost of the canvas of the tent is \(\mathbb{\text{₹ }}22000\)
Fig. 12.1.8-x
Fig. 12.1.8

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 \(\mathrm{cm^2}\).

Solution:

Height of cylinder \((h)\) = 2.4 cm
Diameter of cylinder = 1.4 cm
Radius of Cylinder \((r)\) = 1.4/2 = 0.7 cm
Total Surface Area = CSA of cylinder + CSA of Cone + Base Area of cylinder

$$\begin{aligned}TSA&=2\pi rh+\pi rl+\pi r^{2}\\\\ &=\pi r\left[ 2h+l+r\right] \\\\ l&=\sqrt{r^{2}+h^{2}}\\\\ &=\sqrt{\left( 0.7\right) ^{2}+\left( 0.7\right) ^{2}}\\\\ &=\sqrt{0.49+\left( 2.4\right) ^{2}}\\\\ &=2.5\\\\ TSA&=\scriptsize\dfrac{22}{7}\times 0.7\left[ \left( 2\times 2.4\right) +2.5+0.7\right] \\\\ &=2.2\left[ 4.8+3.2\right] \\\\ &=2.2\times 8\\\\ &=17.6\\\\ &\approx 18\ \mathrm{cm^{2}}\end{aligned}$$
Total surface area of the remaining solid is \(\approx 18\ \mathrm{cm^{2}}\)
Fig. 12.11-x
Fig. 12.11 (Source: NCERT)

Q9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 12.11. 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:

Height of cylinder \(h\) = 10 cm
Radius of the base \(r\)= 3.5 cm
Total surface Area = CSA of Cylinder + 2 (CSA of hemisphere)

$$\begin{aligned}TSA&=2\pi rh+2\left( 2\pi r^{2}\right) \\ &=2\pi rh+4\pi r^{2}\\ &=2\pi r\left[ h+2r\right] \\ &=2\times \dfrac{22}{7}\times 3.5\left( 10+2\times 3\cdot 5\right) \\ &=2\times 22\times 0.5\left( 10+7\right) \\ &=22\times 17\\ &=374\ \mathrm{cm^{2}}\end{aligned}$$
Total surface area of the article is \(374\ \mathrm{cm^{2}}\)

Frequently Asked Questions

The surface area of a solid is the total area covered by all its outer faces. It represents the amount of material required to cover the solid from the outside.

Volume is the measure of space occupied by a solid object. It indicates the capacity of the solid to hold material such as liquid or gas.

Curved surface area is the area of only the curved part of a solid, excluding any flat circular or polygonal faces.

Total surface area is the sum of the curved surface area and the areas of all flat faces of a solid.

The chapter includes cube, cuboid, right circular cylinder, right circular cone, sphere, hemisphere, hollow solids, and combinations of these solids.

The total surface area of a cube is given by \(6a^2\), where \(a\) is the length of one edge.

The volume of a cuboid is calculated using the formula \(l \times b \times h\), where \(l\), \(b\), and \(h\) are length, breadth, and height respectively.

The curved surface area of a cylinder is \(2\pi rh\), where \(r\) is the radius and \(h\) is the height.

The total surface area of a cylinder is \(2\pi r(h + r)\), which includes the curved surface and both circular ends.

Slant height is the distance from the centre of the base of a cone to a point on the curved surface along the side. It is denoted by \(l\).

Slant height is calculated using \(l = \sqrt{r^2 + h^2}\), where \(r\) is radius and \(h\) is height of the cone.

The volume of a cone is \(\frac{1}{3}\pi r^2 h\).

The surface area of a sphere is \(4\pi r^2\), where \(r\) is the radius.

The volume of a sphere is \(\frac{4}{3}\pi r^3\).

A hemisphere is exactly half of a sphere, having one flat circular face and one curved surface.

The curved surface area of a hemisphere is \(2\pi r^2\).

The total surface area of a hemisphere is \(3\pi r^2\), including the circular base.

Combination of solids refers to objects formed by joining two or more basic solids such as cone and cylinder or sphere and cylinder.

The total volume is obtained by adding or subtracting the volumes of the individual solids depending on the structure.

These problems involve melting or reshaping a solid into another solid without loss of material, so volume remains constant.

The principle of conservation of volume is used, which states that volume before and after transformation remains the same.

Hollow solids have thickness and empty space inside, requiring subtraction of inner volume or surface area from the outer one.

A hollow cylinder has an outer radius, inner radius, and thickness, commonly used in pipes and tubes.

Volume is calculated as \(\pi h(R^2 - r^2)\), where \(R\) is outer radius and \(r\) is inner radius.

All dimensions must be in the same unit to avoid incorrect results in surface area or volume calculations.

Applications include water tanks, packaging boxes, ice-cream cones, pipes, spherical balls, containers, and construction materials.

Such problems require calculation of surface area since only the outer surface is coated.

Capacity problems involve volume, while surface area problems involve covering or coating material.

Formula-based numericals, word problems, combination of solids, recasting solids, and application-based questions are common.

By clearly identifying the solid, choosing the correct formula, maintaining unit consistency, and writing steps systematically.

The formulas are fixed and questions are predictable, making it easier to score high with proper practice.

Students should revise formulas, practice mixed numericals, and focus on real-life application problems.

It enhances spatial reasoning, numerical accuracy, logical thinking, and real-world problem interpretation.

It directly applies geometry to real objects and measurements encountered in everyday situations.

p is used because many solids like cylinders, cones, and spheres involve circular bases or curved surfaces.

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