Surface Areas and Volumes Class 10 Worksheet with Solutions
Strengthen your mensuration skills with this Surface Areas and Volumes Class 10 worksheet with solutions. Revise the formulas, understand combinations of solids and practise questions involving exposed surfaces, capacity and everyday objects.
How much material is needed to make a toy? How much water can a container hold? How do you calculate the volume of a sweet with rounded ends? Surface Areas and Volumes Class 10 connects familiar three-dimensional shapes with these practical questions. The key is to recognise the cylinders, cones, spheres, hemispheres, cubes and cuboids that make up an object.
This surface area and volume Class 10 worksheet supports revision through a formula chart, step-by-step examples, extra questions and FAQs. Before checking a solution, draw a simple labelled sketch and decide whether the question concerns the outside surface or the space occupied by the solid. That decision determines which formula you need.
Surface Area and Volume Class 10 Notes
Surface area measures the area of a solid’s boundary, while volume measures the space it occupies. Painting, polishing and covering problems usually involve surface area. Capacity, filling and material-quantity problems usually involve volume.
- Curved surface area: The curved part of a cylinder, cone or hemisphere, excluding its flat circular faces.
- Lateral surface area: The side faces of a cube or cuboid, excluding the top and bottom.
- Total surface area: All boundary surfaces of an individual solid.
- Exposed surface area: The visible outer surfaces of a combined object; joined faces are excluded.
- Capacity: The internal volume a container can hold. Use its internal dimensions when wall thickness matters.
Surface Area and Volume Class 10 All Formulas Chart
In this formula chart, r is the radius, h is the perpendicular height, s is the slant height of a cone, a is the side of a cube, and L, B and H are the length, breadth and height of a cuboid.
| Solid | Curved or Lateral Surface Area | Total Surface Area | Volume |
|---|---|---|---|
| Cube | 4a2 | 6a2 | a3 |
| Cuboid | 2H(L + B) | 2(LB + BH + HL) | LBH |
| Cylinder | 2πrh | 2πr(h + r) | πr2h |
| Right circular cone | πrs | πr(s + r) | ⅓πr2h |
| Sphere | 4πr2 | 4πr2 | ⁴⁄₃πr3 |
| Hemisphere | 2πr2 | 3πr2 | ⅔πr3 |
Height, Slant Height and Units
For a right circular cone, s = √(r2 + h2). Use slant height for its curved surface area and perpendicular height for its volume. Write surface area in square units and volume in cubic units. Remember: 1 cm3 = 1 mL and 1000 cm3 = 1 litre.
How to Solve Combinations of Solids
Many surface area and volume Class 10 important questions involve two or more joined solids. A toy may contain a cone and a hemisphere; a capsule may contain a cylinder and two hemispheres. Separate the object into familiar shapes before calculating.
- Identify each solid and label its dimensions.
- Convert diameters to radii and make the measurement units consistent.
- Find the height of each part rather than using the overall height automatically.
- For surface area, include only the exposed faces.
- For volume, add non-overlapping parts or subtract cavities.
- Use the specified value of π and round at the end.
Example 1: A Cone Joined to a Hemisphere
Question: A solid toy consists of a cone mounted on a hemisphere. Both have radius 3 cm, and the cone has height 4 cm. Find the exposed surface area and total volume.
Solution:
Cone slant height = √(32 + 42)
Cone slant height = 5 cm
Exposed area = cone curved area + hemisphere curved area
Exposed area = π × 3 × 5 + 2π × 32
Exposed area = 33π cm2
Total volume = cone volume + hemisphere volume
Total volume = ⅓π × 32 × 4 + ⅔π × 33
Total volume = 12π + 18π
Total volume = 30π cm3
The common circular base is inside the toy, so it is excluded from the exposed surface area.
Example 2: Capacity of an Open Cylindrical Container
Question: An open cylindrical container has internal radius 7 cm and height 10 cm. Find its capacity. Also find the sheet area needed for its curved wall and bottom, assuming negligible thickness. Take π = 22/7.
Solution:
Capacity = πr2h
Capacity = (22 ÷ 7) × 72 × 10
Capacity = 1540 cm3
Capacity = 1.54 litres
Sheet area = 2πrh + πr2
Sheet area = 440 + 154
Sheet area = 594 cm2
Surface Area and Volume Class 10 Gulab Jamun Question
The gulab jamun question tests both combinations of solids and percentages. Model the sweet as a cylinder with a hemisphere at each end. The two hemispheres together form a sphere, but their lengths must first be removed from the overall length to obtain the cylinder’s height.
Solved Practice Example: Syrup in Rounded Sweets
Question: Each of 20 sweets has a total length of 7 cm and diameter 3 cm. Each sweet is shaped like a cylinder with two hemispherical ends. If syrup occupies 30% of its volume, find the total syrup volume in terms of π.
Solution:
Radius = 3 ÷ 2 = 1.5 cm
Cylindrical height = 7 − 2 × 1.5
Cylindrical height = 4 cm
Volume of one sweet = πr2h + ⁴⁄₃πr3
Volume of one sweet = π × 1.52 × 4 + ⁴⁄₃π × 1.53
Volume of one sweet = 9π + 4.5π
Volume of one sweet = 13.5π cm3
Volume of 20 sweets = 20 × 13.5π
Volume of 20 sweets = 270π cm3
Total syrup volume = (30 ÷ 100) × 270π
Total syrup volume = 81π cm3
Surface Area and Volume Class 10 Extra Questions with Answers
Attempt these questions independently to check your formula selection and calculations. Leave answers in terms of π where appropriate.
-
Find the total surface area and volume of a cube of side 5 cm.
Answer: 150 cm2 and 125 cm3. -
Find the curved surface area and volume of a cylinder with radius
3 cm and height 7 cm.
Answer: 42π cm2 and 63π cm3. -
A cone has radius 5 cm and height 12 cm. Find its slant height
and curved surface area.
Answer: 13 cm and 65π cm2. -
Find the total surface area and volume of a hemisphere of radius 3 cm.
Answer: 27π cm2 and 18π cm3. -
A cylindrical solid has radius 3 cm and height 8 cm. A conical cavity
of the same radius and height is removed. Find the remaining volume.
Answer: 72π − 24π = 48π cm3.
Surface Area and Volume Class 10 HOTS Questions with Solutions
How Does Changing the Radius Affect a Sphere?
Question: A sphere’s radius doubles. How do its surface area and volume change?
New surface area = 4π(2r)2
New surface area = 4 × original surface area
New volume = ⁴⁄₃π(2r)3
New volume = 8 × original volume
Surface area depends on the square of the radius, while volume depends on its cube. More generally, multiplying all linear dimensions by k multiplies area by k2 and volume by k3.
Compare a Cone and a Cylinder
Question: A cone and a cylinder have equal radii and heights. What is the ratio of their volumes?
Cone volume : cylinder volume = ⅓πr2h : πr2h
Cone volume : cylinder volume = 1 : 3
NCERT Exercise Practice and Revision
Pair this worksheet with your surface area and volume Class 10 NCERT solutions and textbook exercises. Students may encounter Exercise 12.1 and Exercise 12.2 in one edition, or search for Surface Areas and Volumes Class 10 Exercise 13.1 and Exercise 13.2 from older editions. Match the actual question and diagram with your textbook.
For effective revision, review the formula chart, practise combinations of solids and then attempt suitable NCERT Exemplar questions. When using a surface area and volume Class 10 test paper or PDF with answers, solve it first without referring to the answer key. Check whether each mistake came from an incorrect formula, a hidden face, a measurement or a unit conversion.
Related Notes and Worksheets
- Surface Areas and Volumes Class 10 Notes and Mind Map for formula revision and a chapter overview.
- Areas Related to Circles Class 10 Worksheet with Solutions to revise circular areas used in mensuration.
- Areas Related to Circles Class 10 Notes and Mind Map for circle, sector and segment formulas.
- Triangles Class 10 Worksheet with Solutions for related geometry practice.
Frequently Asked Questions
What is the difference between surface area and volume?
Surface area measures the boundary of a solid and is expressed in square units. Volume measures the space occupied and is expressed in cubic units. For example, painting a tank involves surface area; calculating how much water it holds involves internal volume.
When should I use curved surface area instead of total surface area?
Use curved surface area when only the curved portion is required. Use total surface area when all faces of an individual solid are included. For an open cylinder, include the curved wall and one circular base rather than both bases.
Why are joined faces excluded from the surface area of a combined solid?
Joined faces lie inside the object and are not exposed. If a cone and hemisphere share a circular base, add their curved surface areas. Adding both total surface areas would count the hidden circular faces.
How do I find the cylinder’s height in the gulab jamun question?
Subtract the lengths of the two hemispherical ends from the total length. Each end contributes one radius, so cylindrical height equals total length − 2r. For total length 5 cm and diameter 2.8 cm, the cylindrical height is 5 − 2.8 = 2.2 cm.
How do I calculate the syrup volume in gulab jamuns?
Add the cylinder volume and the volume of two hemispheres to obtain one sweet’s volume. Multiply by the number of sweets, then multiply by the syrup percentage divided by 100. For 30% syrup, multiply the total volume by 0.30.
What is the difference between a cone’s height and slant height?
Height is the perpendicular distance from the vertex to the base plane. Slant height runs along the cone’s surface to the base rim. For a right circular cone, s2 = r2 + h2. Use h for volume and s for curved surface area.
How do I solve questions involving a cavity or hollow space?
Subtract the cavity’s volume from the original solid’s volume. For surface area, identify which cavity walls and outer faces are exposed. A removed portion can reduce volume while creating new surfaces, so the same subtraction rule does not automatically apply to surface area.
How is capacity converted from cubic centimetres to litres?
Divide the volume in cubic centimetres by 1000. For example, 2500 cm3 equals 2.5 litres. Also, 1 cm3 equals 1 millilitre and 1 m3 equals 1000 litres.
Which surface area and volume Class 10 important questions should I practise?
Practise exposed areas of joined solids, container capacity, cone slant height, solids with cavities and percentage-based volume questions. Include questions that require you to calculate a missing dimension before applying the main formula.
Are Class 9 surface area and volume solutions useful for Class 10?
Yes. Class 9 practice helps revise individual-solid formulas. Class 10 problems build on these ideas by combining solids and interpreting more complex objects. Review basic formulas first if you struggle to recognise the component shapes.
How can I avoid common mistakes in this chapter?
Convert diameter to radius, distinguish overall height from the height of each part, exclude hidden joined faces and use matching units. Keep π consistent and round only at the end. Check that your final unit matches the quantity requested.
Where can I find Surface Areas and Volumes Class 10 notes?
Use the Surface Areas and Volumes Class 10 notes and mind map for revision, then return to this worksheet to practise applying the formulas and checking your solutions.
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