Class 10 Surface Areas and Volumes Complete Guide 2027 — Combined Solids, Conversions + Tricks

In short: Surface Areas and Volumes carries 6 marks in CBSE Class 10 Maths 2027 under the Mensuration unit. The chapter tests your ability to combine solids (a toy shaped as a cone on a hemisphere), convert between shapes (melting a cylinder into spheres), and work with frustums. You need the formulas from Class 9 plus three new ones. The problems are visual and diagram-based — draw first, solve second.

Surface Areas and Volumes is Chapter 12 in CBSE Class 10 Maths. It sits inside the Mensuration unit (10 marks total), alongside Areas Related to Circles. This chapter specifically carries about 6 marks and it builds directly on the solid geometry you learned in Class 9. The difference is that Class 10 introduces combined solids and conversions — shapes joined together or reshaped into different forms.

The good news: the formulas are not new. You already know the surface area and volume of a cylinder, cone, sphere, and hemisphere from Class 9. What is new is combining them. For all 14 chapters and their marks, see the CBSE Class 10 chapter-wise weightage.

The Formula Sheet You Must Know

These formulas from Class 9 are the foundation. You cannot solve a single problem without them.

Cylinder

• CSA = 2πrh
• TSA = 2πr(r + h)
• Volume = πr²h

Cone

• Slant height l = √(r² + h²)
• CSA = πrl
• TSA = πr(r + l)
• Volume = ⅓πr²h

Sphere

• SA = 4πr²
• Volume = ⁴⁄₃πr³

Hemisphere

• CSA = 2πr²
• TSA = 3πr² (curved + flat circular base)
• Volume = ⅔πr³

Frustum of a Cone (NEW in Class 10)

• CSA = π(r₁ + r₂)l, where l = √[h² + (r₁ − r₂)²]
• TSA = π(r₁ + r₂)l + πr₁² + πr₂²
• Volume = ⅓πh(r₁² + r₂² + r₁r₂)

Memory trick for frustum volume: "One-third π h times the sum of squares plus their product." Think of it as the cone volume formula's big brother — it has r₁², r₂², AND r₁r₂.

Memory trick for all volumes: Cylinder = πr²h (full), Cone = ⅓ of cylinder, Hemisphere = ⅔ of cylinder (with h = r). The fractions go ⅓, ⅔, 1 — cone is the smallest, hemisphere in the middle, cylinder is the largest for the same r and h.

The Three Problem Types on Board Exams

Type 1: Combined Solids — Surface Area (3-5 marks)

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

Step-by-step:

1. Draw the figure. Cone sits on top of hemisphere. Both share a circular face where they join.
2. Identify dimensions. r = 3.5 cm, total height = 15.5 cm, so cone height h = 15.5 − 3.5 = 12 cm (subtract hemisphere radius)
3. Find slant height of cone. l = √(3.5² + 12²) = √(12.25 + 144) = √156.25 = 12.5 cm
4. Surface area = CSA of cone + CSA of hemisphere (NOT TSA — the circular base where they join is internal, not exposed)
5. = πrl + 2πr² = π(3.5)(12.5) + 2π(3.5)² = 43.75π + 24.5π = 68.25π = 214.5 cm²

The critical insight: When two solids are joined, the surface where they meet is NOT counted. Only exposed surfaces contribute to the total surface area. This is the #1 concept the board tests.

Type 2: Conversion of Solids — Volume (3-5 marks)

"A metallic sphere of radius 4.2 cm is melted and recast into the form of a cylinder of radius 6 cm. Find the height of the cylinder."

Step-by-step:

1. Key principle: When a solid is melted and recast, the volume stays the same.
2. Volume of sphere = Volume of cylinder
3. ⁴⁄₃π(4.2)³ = π(6)²h
4. ⁴⁄₃ × 74.088 = 36h
5. 98.784 = 36h
6. h = 2.744 cm

Variations: sphere melted into cones, cylinder melted into smaller spheres (find how many), cone melted into smaller cones. The principle is always: total volume before = total volume after.

Type 3: Frustum Problems (3 marks)

"A bucket is in the form of a frustum of a cone whose height is 30 cm and radii of the top and bottom are 21 cm and 7 cm. Find the capacity of the bucket."

1. h = 30, r₁ = 21, r₂ = 7
2. Volume = ⅓πh(r₁² + r₂² + r₁r₂) = ⅓π(30)(441 + 49 + 147) = 10π(637) = 6370π = 20,020 cm³
3. In litres: 20,020/1000 = 20.02 litres (1 litre = 1000 cm³)

Common Mistakes That Cost Marks

1. Counting the joined surface in combined solids. When a cone sits on a hemisphere, the flat circle where they meet is INSIDE the toy — it has zero surface area. Students add it anyway and get the wrong answer. Rule: if you cannot see a surface from outside, do not count it.

2. Confusing height and slant height in cones. The "height" of a cone is the perpendicular distance from base to tip. The "slant height" is the distance along the slope. They are connected by l = √(r² + h²). Board questions deliberately give one and ask you to find the other first.

3. Using diameter instead of radius. If the question says "diameter 14 cm," the radius is 7 cm. This sounds obvious but under exam pressure, students substitute 14 directly into the formula. Always write "r = d/2 = ..." as your first step.

4. Forgetting to subtract the hemisphere radius from total height. In a toy shaped as cone + hemisphere, the total height includes the hemisphere's radius. The cone height = total height − radius of hemisphere. Missing this gives the wrong cone dimensions.

5. Unit conversion errors. 1 m³ = 1,000,000 cm³. 1 litre = 1,000 cm³. 1 m³ = 1,000 litres. Board problems love mixing units. Convert everything to the same unit before calculating.

6. Not simplifying π. Unless the question says "take π = 22/7" or "take π = 3.14," leave your answer in terms of π. If a specific value is given, use it and compute the decimal answer. Read the instruction line at the top of the paper carefully.

4-Week Practice Plan

Week 1: Revise Class 9 Formulas

• Write all formulas (cylinder, cone, sphere, hemisphere) from memory
• Solve 10 straightforward problems: find SA or volume of a single solid
• Make sure you can switch between height and slant height for cones instantly
• Refer to the Class 10 Maths formulas list for a quick cross-check

Week 2: Combined Solids

• Solve NCERT Exercise 12.1 (surface areas of combinations)
• For every problem, draw the figure first and mark which surfaces are exposed
• Practice identifying the shared surface — this is the key skill
• Solve 5 additional combined solid SA problems from sample papers

Week 3: Conversion of Solids + Frustum

• Solve NCERT Exercise 12.2 (volumes and conversions)
• Solve NCERT Exercise 12.3 (frustum problems)
• Master the "melted and recast" logic: volume before = volume after
• Solve 5 frustum problems until the formula feels automatic

Week 4: Board Paper Practice

• Solve this chapter's questions from the last 5 years of CBSE board papers. See how to use previous year papers.
• Time yourself: 5-mark questions should take 8-10 minutes
• Solve the CBSE 2027 sample paper's mensuration questions
• Final check: can you write all formulas including frustum from memory in under 2 minutes?

Quick Revision Card for Exam Day

• Cylinder: V = πr²h, CSA = 2πrh, TSA = 2πr(r+h)
• Cone: V = ⅓πr²h, CSA = πrl, l = √(r²+h²)
• Sphere: V = ⁴⁄₃πr³, SA = 4πr²
• Hemisphere: V = ⅔πr³, CSA = 2πr², TSA = 3πr²
• Frustum: V = ⅓πh(r₁²+r₂²+r₁r₂), l = √[h²+(r₁−r₂)²]
• Combined solids: DON'T count the joined surface
• Conversions: volume before = volume after
• Always use radius, not diameter
• Cone height ≠ slant height — check which one is given
• 1 litre = 1000 cm³, 1 m³ = 10⁶ cm³

For visual, step-by-step practice with combined solid problems, try Padhaao's AI tutor — it walks you through each calculation and catches errors before they cost you marks.

Frequently Asked Questions

How many marks does Surface Areas and Volumes carry in CBSE Class 10 2027?

Surface Areas and Volumes carries approximately 6 marks in the CBSE Class 10 Maths board exam 2027. It falls under the Mensuration unit (10 marks total, shared with Areas Related to Circles). Typically, one 3-mark and one 3-mark or one 5-mark question appear from this chapter, often involving combined solids or conversion problems.

What is the most important concept in this chapter for board exams?

The most important concept is combined solids — calculating the surface area or volume when two or more basic solids are joined together. The key rule is: when two solids are joined, the surface where they meet is not counted in the total surface area, but both volumes are added. Nearly every board paper includes one combined solid question.

How do I solve "melted and recast" problems?

When a solid is melted and recast into a different shape, the total volume remains the same. Set up the equation: volume of original shape = volume of new shape(s). If it is recast into multiple smaller shapes, use: volume of original = number × volume of one small shape. Solve for the unknown (height, radius, or number of pieces).

What is a frustum and how is it different from a cone?

A frustum is the shape you get when you slice a cone with a plane parallel to its base and remove the smaller cone on top. It looks like a bucket — wider at one end, narrower at the other. Unlike a full cone, a frustum has two circular faces (top and bottom) with different radii r₁ and r₂. Its volume formula is ⅓πh(r₁² + r₂² + r₁r₂).

Should I memorise all the formulas or can I derive them during the exam?

Memorise them. Deriving formulas during the exam wastes precious time and risks errors. There are about 12-15 formulas total (cylinder, cone, sphere, hemisphere, frustum — each with SA and volume). Write them on a rough sheet at the start of the exam so you can refer to them quickly. The frustum formulas especially should be memorised since they are harder to derive under pressure.

Surface Areas and Volumes rewards students who draw diagrams, use the correct formulas, and pay attention to whether they are computing surface area or volume. Draw first, identify exposed surfaces, check your units, and those 6 marks are reliable.