What is a frustum of a cone?

A frustum of a cone is the solid shape that remains when a cone is cut by a plane parallel to its base. It has two circular ends — a larger base with radius R and a smaller top with radius r — connected by a curved surface. The slant height of the frustum is l = sqrt(h squared + (R - r) squared). Common real-life examples include buckets, lampshades, and drinking glasses. The frustum formulas are new in Class 10 and are frequently tested in CBSE board exams.

What is the key principle for conversion of solids problems?

When a solid is melted and recast into a different shape (or multiple smaller shapes), the volume remains constant. This is the fundamental principle: Volume of original solid = Volume of new solid(s). Surface area does NOT remain the same — only volume is conserved. To find the number of smaller objects that can be made, divide the volume of the original solid by the volume of one smaller object.

How do you find the surface area of a combination of solids?

For the surface area of combined solids, you add the visible (exposed) surface areas of each component and subtract the areas that are hidden at the joints. For example, if a cone sits on top of a cylinder with the same radius, the TSA = CSA of cylinder + base area of cylinder + CSA of cone. The top of the cylinder and the base of the cone are hidden at the joint and must not be counted. Always draw a sketch to identify which surfaces are exposed.

What is the relationship between the volumes of a cone, hemisphere, and cylinder?

For a cone, hemisphere, and cylinder all having the same radius r and height equal to r, their volumes are in the ratio 1:2:3. Specifically, Volume of cone = (1/3) pi r cubed, Volume of hemisphere = (2/3) pi r cubed, Volume of cylinder = pi r cubed. This elegant ratio is worth remembering as it sometimes appears as a direct question in board exams.

How many marks does Surface Areas and Volumes carry in CBSE Class 10?

Surface Areas and Volumes typically carries 6 to 8 marks in the CBSE Class 10 board exam. Questions range from 1-mark MCQs on direct formula application to 4-5 mark long-answer problems involving combination of solids, conversion problems, or frustum calculations. This chapter is considered highly scoring because most problems follow predictable formula-based patterns. Thorough practice of NCERT exercises and previous year papers is the best preparation strategy.

How can SparkEd help me prepare for Surface Areas and Volumes?

SparkEd provides structured practice on its Surface Areas and Volumes topic page with problems sorted by difficulty level. The AI Math Solver gives instant step-by-step solutions when you're stuck on frustum or combination-of-solids problems. The AI Coach identifies your weak areas — whether it's unit conversion, slant height calculation, or formula selection — and recommends targeted practice. You can access all these tools at sparkedmaths.com.

What is a frustum of a cone?

A frustum of a cone is the solid shape that remains when a cone is cut by a plane parallel to its base. It has two circular ends — a larger base with radius R and a smaller top with radius r — connected by a curved surface. The slant height of the frustum is l = sqrt(h squared + (R - r) squared). Common real-life examples include buckets, lampshades, and drinking glasses. The frustum formulas are new in Class 10 and are frequently tested in CBSE board exams.

What is the key principle for conversion of solids problems?

When a solid is melted and recast into a different shape (or multiple smaller shapes), the volume remains constant. This is the fundamental principle: Volume of original solid = Volume of new solid(s). Surface area does NOT remain the same — only volume is conserved. To find the number of smaller objects that can be made, divide the volume of the original solid by the volume of one smaller object.

How do you find the surface area of a combination of solids?

For the surface area of combined solids, you add the visible (exposed) surface areas of each component and subtract the areas that are hidden at the joints. For example, if a cone sits on top of a cylinder with the same radius, the TSA = CSA of cylinder + base area of cylinder + CSA of cone. The top of the cylinder and the base of the cone are hidden at the joint and must not be counted. Always draw a sketch to identify which surfaces are exposed.

What is the relationship between the volumes of a cone, hemisphere, and cylinder?

For a cone, hemisphere, and cylinder all having the same radius r and height equal to r, their volumes are in the ratio 1:2:3. Specifically, Volume of cone = (1/3) pi r cubed, Volume of hemisphere = (2/3) pi r cubed, Volume of cylinder = pi r cubed. This elegant ratio is worth remembering as it sometimes appears as a direct question in board exams.

How many marks does Surface Areas and Volumes carry in CBSE Class 10?

Surface Areas and Volumes typically carries 6 to 8 marks in the CBSE Class 10 board exam. Questions range from 1-mark MCQs on direct formula application to 4-5 mark long-answer problems involving combination of solids, conversion problems, or frustum calculations. This chapter is considered highly scoring because most problems follow predictable formula-based patterns. Thorough practice of NCERT exercises and previous year papers is the best preparation strategy.

How can SparkEd help me prepare for Surface Areas and Volumes?

SparkEd provides structured practice on its Surface Areas and Volumes topic page with problems sorted by difficulty level. The AI Math Solver gives instant step-by-step solutions when you're stuck on frustum or combination-of-solids problems. The AI Coach identifies your weak areas — whether it's unit conversion, slant height calculation, or formula selection — and recommends targeted practice. You can access all these tools at sparkedmaths.com.

Study Guide

Surface Areas & Volumes Class 10: Formulas, Conversions & Solved Examples

Your complete formula bank and problem-solving guide for combination of solids, frustums, and conversion problems!

CBSEClass 10

The SparkEd Authors (IITian & Googler)15 March 202650 min read

Why Surface Areas & Volumes Is a High-Scoring Chapter

$Surface Areas and Volumes (NCERT Chapter 12) is one of the most formula-driven chapters in Class 10 Maths. It typically carries 6-8 marks in the CBSE board exam, and the best part? Almost every question is a direct application of formulas. No tricky proofs, no abstract reasoning — just plug the right values into the right formula and compute.$

$This chapter builds on what you learned in Class 9 about basic solids (cuboid, cylinder, cone, sphere) and takes it further with three new problem types: 1. Combination of solids — finding surface area/volume when two or more basic solids are joined together. 2. Conversion of solids — when one solid is melted and recast into another shape. 3. Frustum of a cone — the shape you get when you cut a cone with a plane parallel to the base.$

$Let's start with a quick revision of all the formulas you need, then dive into each problem type with solved examples.$

Complete Formula Bank: All Solids at a Glance

$Here is every formula you need for this chapter. Bookmark this section for quick revision!$

$r$

\text{CSA} = 2\pi r h

\text{TSA} = 2\pi r(r + h)

\text{Volume} = \pi r^2 h

$r$

\text{CSA} = \pi r l

\text{TSA} = \pi r(r + l)

\text{Volume} = \frac{1}{3}\pi r^2 h

$r$

\text{Surface Area} = 4\pi r^2

\text{Volume} = \frac{4}{3}\pi r^3

$r$

\text{CSA} = 2\pi r^2

\text{TSA} = 3\pi r^2

\text{Volume} = \frac{2}{3}\pi r^3

$r$

\text{CSA} = \pi(R + r)l

\text{TSA} = \pi\left[R^2 + r^2 + (R + r)l\right]

\text{Volume} = \frac{1}{3}\pi h(R^2 + r^2 + Rr)

$l^2 = r^2 + h^2$

Combination of Solids: Surface Area

$When two or more basic solids are combined to form a new shape, finding the surface area requires careful thinking about which surfaces are visible and which are hidden.$

$Key Principle: When two solids are joined, the surfaces at the joint are hidden and must NOT be included in the total surface area.$

$For example, if a cone is placed on top of a cylinder (like an ice cream cone on a cylindrical cup), the total surface area is:$

\text{TSA} = \text{CSA of cylinder} + \text{Area of base of cylinder} + \text{CSA of cone}

$Notice: the top circle of the cylinder and the base circle of the cone are the same — they're hidden at the joint, so we don't count them.$

Solved Example 1: Tent Problem

$Problem: A tent is in the shape of a cylinder of diameter 20 m and height 2.5 m, surmounted by a cone of the same diameter with a slant height of 8 m. Find the area of canvas used to make the tent.$

$r = 10$

$The canvas covers the CSA of the cylinder + CSA of the cone (no base needed for a tent):$

\text{Area} = 2\pi r h + \pi r l

= \pi r(2h + l)

= \pi \times 10 \times (2 \times 2.5 + 8)

= \pi \times 10 \times 13

= 130\pi

= 130 \times \frac{22}{7} = \frac{2860}{7} \approx 408.57 \text{ m}^2

$\approx 408.57$

Solved Example 2: Toy (Hemisphere + Cone)

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

$r = 3.5$

$-$

$Slant height of cone:$

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

$TSA of toy = CSA of cone + CSA of hemisphere:$

= \pi r l + 2\pi r^2

= \pi r(l + 2r)

= \frac{22}{7} \times 3.5 \times (12.5 + 7)

= \frac{22}{7} \times 3.5 \times 19.5

= 11 \times 19.5 = 214.5 \text{ cm}^2

$214.5$

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Combination of Solids: Volume

$Finding the volume of a combination of solids is generally simpler than finding the surface area, because volumes simply add up.$

$Key Principle: The volume of a combined solid = sum of volumes of individual parts. No subtraction needed (unlike surface area where joint areas are hidden).$

$This works because volume measures the space inside, and combining two solids means the total internal space is the sum of individual spaces.$

Solved Example 3: Gulab Jamun Problem

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

$r = 1.4$

\text{Volume} = \text{Volume of cylinder} + 2 \times \text{Volume of hemisphere}

= \pi r^2 h + 2 \times \frac{2}{3}\pi r^3

= \pi r^2\left(h + \frac{4r}{3}\right)

= \frac{22}{7} \times (1.4)^2 \times \left(5 + \frac{4 \times 1.4}{3}\right)

= \frac{22}{7} \times 1.96 \times \left(5 + \frac{5.6}{3}\right)

= \frac{22}{7} \times 1.96 \times \frac{20.6}{3}

= \frac{22 \times 1.96 \times 20.6}{7 \times 3}

= \frac{888.272}{21} \approx 42.30 \text{ cm}^3

$\approx 42.30$

Solved Example 4: Silo Problem

$Problem: A solid is in the form of a cylinder with hemispherical ends. If the total height is 19 cm and the diameter of the cylinder is 7 cm, find the volume and total surface area of the solid.$

$r = 3.5$

$Volume:$

V = \pi r^2 h + 2 \times \frac{2}{3}\pi r^3 = \pi r^2\left(h + \frac{4r}{3}\right)

= \frac{22}{7} \times 3.5^2 \times \left(12 + \frac{14}{3}\right) = \frac{22}{7} \times 12.25 \times \frac{50}{3}

= \frac{22 \times 12.25 \times 50}{21} = \frac{13475}{21} \approx 641.67 \text{ cm}^3

$TSA:$

\text{TSA} = 2\pi rh + 2 \times 2\pi r^2 = 2\pi r(h + 2r)

= 2 \times \frac{22}{7} \times 3.5 \times (12 + 7) = 22 \times 19 = 418 \text{ cm}^2

$\approx 641.67$

Conversion of Solids

$When a solid made of metal (or clay, wax, etc.) is melted and recast into a different shape, the volume remains the same . This is the single most important principle for conversion problems.$

\text{Volume of original solid} = \text{Volume of new solid(s)}

$If the original solid is recast into many smaller solids:$

\text{Volume of original} = n \times \text{Volume of one small solid}

where

n

is the number of smaller solids.

$These problems are very popular in board exams because they test both your formula knowledge and your algebraic skills.$

Solved Example 5: Sphere to Cones

$Problem: How many cones with radius 1 cm and height 3 cm can be made by melting a metallic sphere of radius 3 cm?$

$Solution: Volume of sphere:$

V_{\text{sphere}} = \frac{4}{3}\pi (3)^3 = \frac{4}{3}\pi \times 27 = 36\pi \text{ cm}^3

$Volume of one cone:$

V_{\text{cone}} = \frac{1}{3}\pi (1)^2 \times 3 = \pi \text{ cm}^3

$Number of cones:$

n = \frac{V_{\text{sphere}}}{V_{\text{cone}}} = \frac{36\pi}{\pi} = 36

$Answer: 36 cones can be made.$

Solved Example 6: Water Rising in Cylinder

$Problem: A sphere of diameter 6 cm is dropped into a cylindrical vessel partly filled with water. The diameter of the vessel is 12 cm. If the sphere is completely submerged, find the rise in the water level.$

$Solution: When the sphere is submerged, it displaces water equal to its own volume. This displaced water causes the water level to rise.$

$h$

\text{Volume of sphere} = \text{Volume of water displaced (cylinder of height } h\text{)}

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

\frac{4}{3} \times 27 = 36h

36 = 36h

h = 1 \text{ cm}

$1$

Solved Example 7: Well and Embankment

$Problem: A well of diameter 3 m is dug 14 m deep. The earth taken out is spread evenly all around the well to form a 4 m wide embankment. Find the height of the embankment.$

$= 1.5$

\text{Volume of earth dug} = \text{Volume of embankment}

\pi (1.5)^2 \times 14 = \pi \left[(5.5)^2 - (1.5)^2\right] \times h

2.25 \times 14 = (30.25 - 2.25) \times h

31.5 = 28h

h = \frac{31.5}{28} = 1.125 \text{ m}

$= 1.125$

Frustum of a Cone

$When a cone is cut by a plane parallel to its base, the part between the base and the cutting plane is called a frustum of the cone. Think of it as a cone with the top chopped off.$

$R$

$Slant height of frustum:$

l = \sqrt{h^2 + (R - r)^2}

$Formulas:$

\text{CSA} = \pi(R + r)l

\text{TSA} = \pi\left[R^2 + r^2 + (R+r)l\right]

\text{Volume} = \frac{1}{3}\pi h(R^2 + r^2 + Rr)

$H$

Solved Example 8: Frustum (Bucket Problem)

$Problem: A bucket is in the form of a frustum of a cone. Its depth is 15 cm and the diameters of the top and bottom are 56 cm and 42 cm. Find the capacity and cost of tin sheet used at the rate of Rs 1.50 per sq cm.$

$R = 28$

$Capacity (Volume):$

V = \frac{1}{3}\pi h(R^2 + r^2 + Rr)

= \frac{1}{3} \times \frac{22}{7} \times 15 \times (784 + 441 + 588)

= \frac{22}{7} \times 5 \times 1813

= \frac{22 \times 5 \times 1813}{7} = \frac{199430}{7} \approx 28490 \text{ cm}^3

$Slant height:$

l = \sqrt{15^2 + (28-21)^2} = \sqrt{225 + 49} = \sqrt{274} \approx 16.55 \text{ cm}

$Tin sheet area = CSA of frustum + area of bottom circle:$

= \pi(R+r)l + \pi r^2

= \frac{22}{7}\left[(28+21) \times 16.55 + 21^2\right]

= \frac{22}{7}\left[811.05 + 441\right]

= \frac{22}{7} \times 1252.05 \approx 3934.30 \text{ cm}^2

$= 3934.30 \times 1.50 \approx$

$\approx 28490$

Solved Example 9: Frustum as Part of a Cone

$Problem: The radii of the circular ends of a frustum are 6 cm and 14 cm. If its slant height is 10 cm, find the total surface area.$

$r = 6$

\text{TSA} = \pi\left[R^2 + r^2 + (R+r)l\right]

= \pi\left[196 + 36 + 20 \times 10\right]

= \pi \times [196 + 36 + 200]

= 432\pi

= 432 \times \frac{22}{7} \approx 1357.71 \text{ cm}^2

$\approx 1357.71$

Common Mistakes Students Make

$Watch out for these pitfalls — they're responsible for most marks lost in this chapter:$

$1. Confusing CSA and TSA: * Mistake: Using TSA when the problem asks for CSA, or vice versa. * Fix: CSA = only the curved/lateral surface. TSA = CSA + area of all flat faces. Read the question carefully.$

$h$

$3. Diameter vs. Radius: * Mistake: Using diameter directly in formulas instead of dividing by 2. * Fix: The very first step should be: read the given dimensions and convert diameters to radii.$

$4. Wrong Surfaces in Combination Problems: * Mistake: Including hidden surfaces when solids are joined (e.g., counting the top of a cylinder when a cone sits on it). * Fix: Sketch the combined solid. Identify which surfaces are exposed and which are internal.$

$5. Forgetting Volume Conservation in Conversion: * Mistake: Equating surface areas instead of volumes when a solid is melted and recast. * Fix: When melting/recasting, volume stays constant, NOT surface area.$

$6. Unit Conversion Errors: * Mistake: Mixing cm and m in the same calculation. * Fix: Convert all measurements to the same unit before starting any calculation.$

$R$

Board Exam Strategy for Surface Areas & Volumes

$Weightage: This chapter typically carries 6-8 marks in the board exam.$

$Question Patterns:$

$* 1-2 Marks (MCQ/VSA): Direct formula application — find the volume of a cone, the TSA of a hemisphere, or the slant height of a frustum.$

$* 3 Marks (SA): Combination of two solids (like a toy = cone + hemisphere), or a simple conversion problem (sphere melted into cylinders).$

$* 4-5 Marks (LA): Multi-step problems involving frustum calculations, well-and-embankment type problems, or complex combination of three solids.$

$Pro Tips:$

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

$1$

$Practice extensively on SparkEd's Surface Areas & Volumes page to build speed and accuracy.$

Quick Revision: Formula Comparison Table

$Here's a handy comparison to help you see the relationships between formulas:$

$= \frac{1}{3} \times$

$Remember these ratios for a cylinder, hemisphere, and cone with the same radius and height = radius:$

V_{\text{cone}} : V_{\text{hemisphere}} : V_{\text{cylinder}} = 1 : 2 : 3

$This is a beautiful relationship and is sometimes asked as a direct question.$

$= 2\pi rh$

$1$

Ace This Chapter with SparkEd

$Surface Areas and Volumes is one of those chapters where practice directly translates to marks. The more problems you solve, the faster you get at identifying which formula to use and avoiding calculation errors.$

$Here's how SparkEd helps you master this chapter:$

$* Graded Practice: Our Surface Areas & Volumes page has problems from easy formula-application questions to challenging multi-step board-level problems.$

$* AI Math Solver: Stuck on a tricky frustum problem or a combination of solids? Paste it into the AI Solver for step-by-step solutions showing every substitution and simplification.$

$* AI Coach: Get personalized insights on where you're making errors — whether it's unit conversions, slant height calculations, or forgetting to exclude hidden surfaces.$

$* Related Topics: Since this chapter uses circles extensively (for cross-sections), make sure to also practice Areas Related to Circles .$

$Visit sparkedmaths.com and start converting your formula knowledge into full marks!$

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