Class 10 Maths Chapter 12: Surface Areas and Volumes Notes & Formulas

Exam Weightage & Blueprint

Total: 5-6 Marks

This chapter focuses on Combination of Solids (Surface Area & Volume) and Conversion of Solids. Calculation accuracy is key.

Question Type	Marks	Frequency	Focus Topic
MCQ	1	High	Formulas, Ratios of Vol/Area
Short Answer	2 or 3	Medium	Combined Solids (Toy, Capsule)
Long Answer	4 or 5	Very High	Volume Conversion, Embankment, Flow of Water

CBSE 2026–27 Syllabus Status: Surface Areas and Volumes

Unit VI: Mensuration • 5–6 Marks

This chapter is strictly structured according to the CBSE Rationalised Mathematics Syllabus. Focus your board preparation only on prescribed topics and avoid studying deleted sections:

In 2026–27 Board Syllabus:

Deleted / Not in Board Exam:

⏰ Last 24-Hour Checklist

Formula Sheet: Memorize Cylinder, Cone, Sphere, Hemisphere formulas.
Slant Height: $l = \sqrt{h^2 + r^2}$ for Cone.
TSA of Combination: Add CSA of visible parts. Don't add base areas!

Volume of Combination: Simply add volumes of individual solids.
Conversion: Volume remains constant (Vol 1 = Vol 2).
Units: Ensure all units are same (cm vs m).

📐 Formulas Recap

Solid	CSA	TSA	Volume
Cylinder	$2\pi rh$	$2\pi r(r+h)$	$\pi r^2h$
Cone	$\pi rl$	$\pi r(l+r)$	$\frac{1}{3}\pi r^2h$
Hemisphere	$2\pi r^2$	$3\pi r^2$	$\frac{2}{3}\pi r^3$
Sphere	$4\pi r^2$	$4\pi r^2$	$\frac{4}{3}\pi r^3$

Concept: 📐 Combination & Conversion

Surface Area (Combination)

Logic: We calculate the area of the visible surface only.

Example: A toy (Cone on Hemisphere).
TSA = CSA(Cone) + CSA(Hemisphere). (Base areas hidden).

Volume (Combination)

Logic: Volume is the total space occupied. We simply add the volumes.

Example: A Gulab Jamun.
Total Vol = Vol(Cylinder) + 2 - Vol(Hemisphere).

Conversion of Solids

Logic: When a solid is melted and recast, Volume remains constant.

Formula: $V_1 = V_2 or n \times v = V$ (if recast into $n$ smaller pieces).

⚡ High-Yield Conceptual Rules (Board MCQs & A-R):

Surface Area of Scooped-Out Solids: Scooping out a cavity (e.g. hemisphere or cone from a cylinder) INCREASES total surface area because new inner surfaces are created: $$\text{TSA of scooped cylinder} = \text{Outer CSA} + \text{Base Area} + \text{Inner Cavity CSA}$$ Rule: Surface area is ADDED for cavity, but Volume is SUBTRACTED.
Liquid Rate of Flow Rule: $$\text{Volume of fluid discharged} = \text{Cross-sectional Area of Pipe} \times (\text{Speed} \times \text{Time})$$
Hollow Cylinder / Embankment Volume: $$V = \pi (R^2 - r^2) h = \pi (R - r)(R + r) h \quad \text{where } R = r + \text{width}$$

Key Combinations in CBSE: Joined base areas are inside and NOT included in Total Surface Area!

Solved Examples (Board Marking Scheme)

Q1. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area. (3 Marks)

Step 1: Identify Dimensions 0.5 Mark

$r = 3.5$ cm. Total height = 15.5 cm.

Height of cone ($h$) = $15.5 - 3.5 = 12$ cm.

Step 2: Slant Height 1 Mark

$l = \sqrt{r^2 + h^2} = \sqrt{3.5^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5$ cm.

Step 3: Calculate Area 1.5 Marks

TSA = CSA(Cone) + CSA(Hemisphere) = $\pi rl + 2\pi r^2$

$= \frac{22}{7} \times 3.5 (12.5 + 2 \times 3.5) = 11 \times 19.5 = \mathbf{214.5}$ cm-.

Q2. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder. (2 Marks)

Step 1: Concept 0.5 Mark

Volume of Sphere = Volume of Cylinder (Since it is melted & recast).

Step 2: Equation 1 Mark

$\frac{4}{3}\pi R^3 = \pi r^2 h$

$\frac{4}{3} \times (4.2)^3 = (6)^2 \times h$

Step 3: Solve 0.5 Mark

$h = \frac{4 × 4.2 × 4.2 × 4.2}{3 × 6 × 6} = 1.4 × 1.4 × 1.4 × 4 / 4$

$h = \frac{4 \times 74.088}{108} = 2.74$ cm.

Q3. A gulab jamun contains sugar syrup up to about 30% of its volume. Find approximate syrup in 45 gulab jamuns, shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm. (4 Marks)

Step 1: Dimensions 1 Mark

Diameter = 2.8 cm $\Rightarrow r = 1.4$ cm. Length of cylindrical part $h = 5 - (1.4+1.4) = 2.2$ cm.

Step 2: Volume of 1 Piece 1.5 Marks

Vol = Vol(Cyl) + 2 $ \times $ Vol(Hemi) = $\pi r^2 h + \frac{4}{3}\pi r^3$

$= π r^2 (h + \frac{4}{3}r) = \frac{22}{7} × 1.4 × 1.4 (2.2 + 1.87) ≈ 25.05$ cm-.

Step 3: Total Syrup 1.5 Marks

Vol of 45 pieces = $45 \times 25.05$.

Syrup = $30%$ of Total Vol = $\frac{30}{100} \times 45 \times 25.05 ≈ \mathbf{338}$ cm-.

Previous Year Questions (PYQs)

2023 (3 Marks): A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of $\pi$.
Ans: Vol = Vol(Cone) + Vol(Hemi) = $$\frac{1}{3}$π (1)^2(1) + $\frac{2}{3}$π (1)^3 = π$ cm-.

2020 (5 Marks): A vessel is in the form of an inverted cone. Its height is 8 cm and radius of its top is 5 cm. It is filled with water. When lead shots (spheres of radius 0.5 cm) are dropped, one-fourth of water flows out. Find number of lead shots.
Ans: $n \times $ Vol(Sphere) = $\frac{1}{4} \times $ Vol(Cone). Answer: $n = 100$.

Exam Strategy & Mistake Bank

⚠️ Mistake Bank

Adding Bases: In TSA of combinations (e.g., cone on cylinder), do NOT add the area of the base where they join. Only add visible CSA.

Diameter vs Radius: Always check if $d$ or $r$ is given. Questions often give diameter to trap you.

Calculation: Don't multiply $\pi$ ($\frac{22}{7}$) early. Keep it common and cancel it out at the end if possible.

💡 Scoring Tips

Draw Diagrams: Always draw a rough sketch with dimensions labeled (radius, height). It gets marks.

Step Marks: Even if calculation is wrong, writing correct formula gets you marks.

Units: Write units (cm-, m-) in the final answer. Missing units = -0.5 marks.

📝 More Solved Board Questions

Q4. A well of diameter 3 m is dug 14 m deep. The earth taken out is spread evenly around it to form an embankment of width 4 m. Find the height of the embankment. 5 Marks

Sol. Vol of earth dug out = Vol of Cylinder = $\pi r^2 h = \pi (1.5)^2 \times 14 = 31.5\pi$ m³.

Outer radius of embankment ($R$) = $1.5 + 4 = 5.5$ m.

Inner radius ($r$) = 1.5 m.

Area of embankment = $\pi (R^2 - r^2) = \pi (5.5^2 - 1.5^2) = \pi (30.25 - 2.25) = 28\pi$ m².

Height of embankment ($H$) = $\frac{Volume of earth}{Area of embankment} = \frac{31.5\pi}{28\pi} = 1.125$ m.

Answer: 1.125 m

🎯 Board Pattern (2018–2025): Embankment and "Flow of water through a pipe" are the most common 5-mark questions. Remember: Area of embankment is a Ring (Outer Area - Inner Area).

📋 Board Revision Checklist

✅ Cube: $V = a^3$; $TSA = 6a^2$
✅ Cylinder: $V = \pi r^2h$; $CSA = 2\pi rh$
✅ Cone: $V = (1/3)\pi r^2h$; $CSA = \pi rl$ (where $l = \sqrt{h^2+r^2}$)
✅ Sphere: $V = (4/3)\pi r^3$; $SA = 4\pi r^2$
✅ Hemisphere: $V = (2/3)\pi r^3$; $TSA = 3\pi r^2$
✅ For Combined TSA: Only add the surfaces you can touch (visible CSA)
✅ Scooped-out cavity: TSA increases (add cavity CSA); Volume decreases (subtract cavity volume)
✅ Rate of Flow: Volume $= \text{Pipe Area} \times \text{Speed} \times \text{Time}$
✅ Embankment volume $= \pi(R^2 - r^2)h$ where $R = r + \text{embankment width}$
✅ Melting/Recasting: Volume $V_{old} = V_{new}$
✅ Deleted Topic: Frustum of a cone is NOT in syllabus

💡 Exam Tip:
Keep $\pi$ as it is in the middle steps. Many times $\pi$ from both sides of the equation will cancel out, saving you tedious calculations.

CBSE Section E • 4 Marks

Case Study: Pharmaceutical Capsule Design & Dosing

Scenario: A pharmaceutical manufacturer produces antibiotic capsules. Each medicine capsule is in the shape of a right circular cylinder with two hemispheres stuck to each of its ends. The overall length of the entire capsule is 14 mm and the outer diameter of the capsule is 5 mm.

Q1: Find the radius ($r$) and the height ($h$) of the cylindrical middle section. 1 Mark

Q2: Calculate the total outer surface area of the capsule (use $\pi = 22/7$). 1 Mark

Q3: Find the maximum internal volume of medicine the capsule can contain. (OR) If 10,000 capsules are packaged, find the total surface area to be coated with gelatin. 2 Marks

📝 10-Mark Self-Assessment Test

Time: 25 Mins

Solve these high-frequency CBSE Board questions on paper, then reveal step-by-step solutions to check your marks.

Q1 (1 Mark - MCQ): Two cubes each of volume $27 cm^3$ are joined end to end. The surface area of the resulting cuboid is:
(a) $90 cm^2$ (b) $81 cm^2$ (c) $54 cm^2$ (d) $72 cm^2$

Q2 (2 Marks - VSA): A solid metallic sphere of radius $10.5 cm$ is melted and recast into a number of smaller cones, each of radius $3.5 cm$ and height $3 cm$. Find the number of cones so formed.

Q3 (3 Marks - SA): A decorative block is made of two solids — a cube and a hemisphere. The base of the block is a cube with edge $5 cm$, and the hemisphere fixed on the top has a diameter of $4.2 cm$. Find the total surface area of the block. [Take $\pi = 22/7$]

Q4 (4 Marks - LA): A toy is in the form of a cone of radius $3.5 cm$ mounted on a hemisphere of same radius. The total height of the toy is $15.5 cm$. Find the total surface area and volume of the toy. [Take $\pi = 22/7$]

Sol 1: Edge of cube $a = \sqrt[3]{27} = 3 cm$.
When joined end to end: $l = 3 + 3 = 6 cm, b = 3 cm, h = 3 cm$.
$TSA = 2(lb + bh + hl) = 2(18 + 9 + 18) = 2(45) = 90 cm^2$.
Correct Option: (a) $90 cm^2$ [1 Mark]

Sol 2: Let number of cones be $n$.
$$n \times Vol(Cone) = Vol(Sphere) \implies n \times (\frac{1}{3}\pi r^2 h) = \frac{4}{3}\pi R^3$$ $$n \times (3.5)^2 \times 3 = 4 \times (10.5)^3$$ $$n \times 12.25 \times 3 = 4 \times 1157.625 \implies n = \frac{4630.5}{36.75} = 126$$ Answer: Number of cones $= 126$ [2 Marks]

Sol 3: Edge $a = 5 cm$, hemisphere radius $r = 2.1 cm$.
$TSA of block = TSA of cube - Base area of hemisphere + CSA of hemisphere$ [1 Mark]
$$= 6a^2 - \pi r^2 + 2\pi r^2 = 6a^2 + \pi r^2 [1 Mark]$$ $$= 6(5)^2 + \frac{22}{7} \times 2.1 \times 2.1 = 150 + 22 \times 0.3 \times 2.1 = 150 + 13.86 = 163.86 cm^2 [1 Mark]$$ Answer: $163.86 cm^2$ [3 Marks]

Sol 4: Radius $r = 3.5 cm$. Height of cone $h = 15.5 - 3.5 = 12 cm$.
Slant height $l = \sqrt{h^2 + r^2} = \sqrt{12^2 + 3.5^2} = \sqrt{144 + 12.25} = \sqrt{156.25} = 12.5 cm$ [1 Mark].
(i) Total Surface Area:
$TSA = CSA of cone + CSA of hemisphere = \pi rl + 2\pi r^2 = \pi r(l + 2r)$
$$= \frac{22}{7} \times 3.5 \times (12.5 + 7) = 11 \times 19.5 = 214.5 cm^2 [1.5 Marks]$$ (ii) Volume:
$Vol = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{1}{3}\pi r^2 (h + 2r) = \frac{1}{3} \times \frac{22}{7} \times (3.5)^2 \times (12 + 7)$
$$= \frac{1}{3} \times \frac{22}{7} \times 12.25 \times 19 = \frac{1}{3} \times 38.5 \times 19 = \frac{731.5}{3} ≈ 243.83 cm^3 [1.5 Marks]$$ Answer: $TSA = 214.5 cm^2$, $Volume = 243.83 cm^3$ [4 Marks]

Concept Mastery Quiz 🎯

Test your readiness for the board exam.

1. A solid piece of iron in the form of a cuboid of dimensions $49cm \times 33cm \times 24cm$, is moulded to form a solid sphere. The radius of the sphere is:

2. If two solid hemispheres of same base radius $r$ are joined together along their bases, then curved surface area of this new solid is:

3. The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is:

4. During conversion of a solid from one shape to another, the volume of the new shape will:

5. A metallic spherical shell of internal and external diameters 4 cm and 8 cm is melted and recast into the form of a cone of base diameter 8 cm. The height of the cone is:

Frequently Asked Questions: Class 10 Maths Chapter 12

How many marks does Surface Areas and Volumes carry in the board exam?

Surface Areas and Volumes carries 5 to 6 marks in Unit 6: Mensuration, often appearing as a 5-mark long answer problem or a 4-mark case study question on combined 3D solids.

Is Frustum of a Cone included in the Class 10 syllabus?

No, Frustum of a Cone has been completely removed from the rationalized CBSE Class 10 syllabus. The syllabus covers cubes, cuboids, spheres, hemispheres, right circular cylinders, and right circular cones.

What is the rule for finding the surface area of a combined solid?

When two solids are joined, only the exposed visible surfaces are added. The contacting base surfaces become internal and must NOT be included in the total surface area calculation.

How do you handle volume conservation in reshaping problems?

When one solid is melted and recast into another solid (or multiple small solids), total volume remains constant: Volume of original solid = n * Volume of each new solid.

Chapter 12 Study Modules & Practice Resources

Complete Notes & Formulas • NCERT Solutions • Previous Year Questions • Chapter Tests & Quizzes • Practice Worksheets

Chapter 12: Surface Areas and Volumes

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