A toy made of a cone on top of a hemisphere, a capsule with a cylinder in the middle, a tent shaped like a cylinder topped by a cone: these are the questions that decide marks in the mensuration chapter. The formulas for each single solid are short and well known. The trouble starts when two solids are joined, because a student who knows every formula can still produce the wrong answer by adding the wrong things together.
The pattern behind most errors is one idea: volumes add, surface areas do not. When two solids are joined, the volume of the new object is simply the sum of the two volumes. But some of the surface disappears at the joint, and only the outside of the combined object counts. This article goes through the errors that follow from that idea and from a few other habits, and then works through examples with the arithmetic checked.
For the single-solid formulas, the page on surface areas and volumes and the introduction to surface areas and volumes are handy references to keep open as you read.
The Formulas You Need, in One Place
With radius r, height h and slant height l:
- Cylinder: curved surface area 2 pi r h; total surface area 2 pi r (r + h); volume pi r squared h.
- Cone: curved surface area pi r l; total surface area pi r (l + r); volume one third pi r squared h; and l = square root of (r squared + h squared).
- Sphere: surface area 4 pi r squared; volume four thirds pi r cubed.
- Hemisphere: curved surface area 2 pi r squared; total surface area 3 pi r squared; volume two thirds pi r cubed.
A hemisphere is half a sphere, so its curved surface is half of 4 pi r squared, which is 2 pi r squared. Its total surface adds the flat circular face, pi r squared, giving 3 pi r squared. Keeping that reasoning in mind is better than memorising the numbers, because you will need to decide, in each problem, whether the flat face is exposed or hidden.
Adding Total Surface Areas of the Two Solids
This is the single most common mistake. A cone placed on a hemisphere has a flat circular face on the hemisphere and a flat circular base on the cone. Where they are glued together, both of those faces vanish. If you add the total surface areas, you count both hidden faces and get an answer that is too large.
The correct method is to add only the surfaces you can touch from outside. For a cone on a hemisphere, that is the curved surface of the cone plus the curved surface of the hemisphere. For a cylinder with a hemisphere on each end, it is the curved surface of the cylinder plus two hemispherical curved surfaces. Before writing any formula, ask: which faces are on the outside?
Forgetting That Volume Is Simpler
Students who have struggled with surface area sometimes try to subtract hidden faces from volume too. Volume does not need that. The volume of a combined solid is just the sum of the volumes of its parts. If material is scooped out, the volume of the scooped part is subtracted. There is no adjustment for joints.
Using Height Where Slant Height Is Needed
The curved surface area of a cone uses the slant height l, not the vertical height h. Volume uses h. When a question gives the height, you need to calculate l by Pythagoras first. A typical question gives radius 6 cm and height 8 cm, which leads to l = 10 cm. Students who use 8 in the curved surface formula get a smaller area than the right answer. A useful habit is to write both h and l at the top of the working, with a line saying which formula uses which.
Reading the Total Height Incorrectly
When a hemisphere sits on a cylinder, or when an ice cream cone is topped with a hemisphere, the total height of the object includes the radius of the hemisphere. If a question gives a capsule of total length 14 mm and diameter 5 mm, the cylindrical part is 14 minus 5 = 9 mm long, because each hemispherical end takes up 2.5 mm. Using 14 mm as the cylinder length double counts the ends.
The CBSE class 10 notes for this chapter include several figures of this kind, and it is worth drawing each one with its dimensions marked before substituting.
Diameter Taken as Radius, and the Reverse
A question may give the diameter of a hemispherical bowl, or the radius of the cylinder as a half of something else. The formulas want the radius. Always write r = … as the first line of the solution. Many wrong answers in this chapter come from a diameter that was never halved, and the error is large because r is squared or cubed.
Wrong Treatment of Hollow Shapes
For a hollow pipe or a hollow cylinder, there are two radii: the outer R and the inner r. The volume of the material is pi h (R squared minus r squared), not pi h (R minus r) squared. For a solid with a hemispherical hollow scooped out, the volume of the hollow is subtracted, and the surface area now includes the inside of the hollow, while the flat face that was removed is no longer exposed. Draw it and trace the outside and inside before choosing formulas.
Melting and Recasting, but Using Surface Area
When one solid is melted and recast into another, or into many smaller ones, it is the volume that stays constant, not the surface area. Setting the surface areas equal is a frequent and costly slip. The equation is: volume of the melted solid equals the number of new solids times the volume of each. For the number of pieces, divide the large volume by the small volume.
Mixed Units and Litres
A tank is described in metres but a bucket in centimetres. One cubic metre is 1,000 litres, and 1,000 cubic centimetres is 1 litre. Students who forget this are off by a factor of a thousand or a million. Convert everything to one unit before you start.
Worked Examples With the Arithmetic Checked
Use pi = 22/7 unless stated otherwise.
Example 1: cone on a hemisphere. A toy is a cone of radius 6 cm and height 8 cm placed on a hemisphere of the same radius. Find its total surface area and its volume.
Slant height l = square root of (36 + 64) = 10 cm. Outside surface = curved surface of the cone + curved surface of the hemisphere = pi x 6 x 10 + 2 pi x 36 = 60 pi + 72 pi = 132 pi = 132 x 22/7 = about 414.86 square cm.
Volume = one third pi x 36 x 8 + two thirds pi x 216 = 96 pi + 144 pi = 240 pi = 240 x 22/7 = about 754.29 cubic cm.
Notice that only curved surfaces were added for the area. The joint faces were never counted.
Example 2: a capsule. A medicine capsule is a cylinder with a hemisphere at each end. The total length is 14 mm and the diameter is 5 mm. Find its surface area and volume.
r = 2.5 mm. Cylinder length = 14 minus 5 = 9 mm. Surface area = 2 pi r h + 4 pi r squared = 2 pi x 2.5 x 9 + 4 pi x 6.25 = 45 pi + 25 pi = 70 pi = 220 square mm.
Volume = pi r squared h + four thirds pi r cubed = pi x 6.25 x 9 + four thirds pi x 15.625 = 56.25 pi + 20.83 pi = 77.08 pi, which is about 242.26 cubic mm.
Check: 70 x 22/7 is 220 exactly, which suggests the dimensions were chosen to work with 22/7.
Example 3: melting and recasting. A metal sphere of radius 4.2 cm is melted and recast into a cylinder of radius 6 cm. Find the height of the cylinder.
Volume of sphere = four thirds pi x 4.2 cubed = four thirds pi x 74.088 = 98.784 pi. Volume of cylinder = pi x 36 x h. Equating: 36 h = 98.784, so h = 2.744 cm.
Check: a sphere is compact and the cylinder is wide (radius 6 against 4.2), so the height should be small, and it is.
Example 4: number of pieces. A solid metal cone of radius 6 cm and height 24 cm is melted and recast into spheres of radius 3 cm. How many spheres are formed?
Volume of cone = one third pi x 36 x 24 = 288 pi. Volume of one sphere = four thirds pi x 27 = 36 pi. Number = 288 divided by 36 = 8.
Example 5: a hollow scooped out. A solid cylinder of radius 7 cm and height 10 cm has a hemispherical hollow of the same radius scooped out of one end. Find the total surface area and the volume of the remaining solid.
Exposed surfaces: the curved surface of the cylinder, the flat base at the other end, and the curved surface of the hemispherical hollow. The top face is gone, because the hollow takes up the entire top circle. Surface = 2 pi r h + pi r squared + 2 pi r squared = 2 x 22/7 x 7 x 10 + 3 x 22/7 x 49 = 440 + 462 = 902 square cm.
Volume = pi r squared h minus two thirds pi r cubed = 22/7 x 49 x 10 minus two thirds x 22/7 x 343 = 1540 minus 718.67 = 821.33 cubic cm.
Example 6: units. A cylindrical tank has radius 7 m and height 10 m. How many litres can it hold?
Volume = 22/7 x 49 x 10 = 1,540 cubic metres. Since 1 cubic metre is 1,000 litres, the capacity is 15,40,000 litres.
For more problems like these at board level, see the important questions for Class 10 Maths Chapter 13 and the 2, 3 and 5 mark questions.
A Four-Step Habit That Prevents Most of These Errors
- Sketch the combined solid and mark every dimension, converting diameter to radius.
- Decide which faces are on the outside, and shade them lightly.
- For area, add only the shaded surfaces. For volume, add the volumes, or subtract any hollow.
- Check units and plausibility: does the answer have the right size compared with a rough estimate?
The MCQs on surface areas and volumes are a quick way to test whether this habit is working.
Where This Chapter Sits in Your Syllabus
Surface Areas and Volumes is Chapter 13 in the CBSE Class 10 Maths numbering used on the pages linked here. As far as the recent CBSE syllabus goes, problems are limited to combinations of two solids, and the frustum of a cone is not included. The Karnataka SSLC textbook has traditionally treated the topic differently, and may number the chapter differently, so check whether the frustum and conversion-of-solids problems are in your current syllabus. Please confirm with your own board’s syllabus and sample papers.
The chapter also leans on the circle formulas from areas of sector and segment of a circle, so if the circle area slips, solids will too.
Frequently Asked Questions
Q1. Why can’t I just add the total surface areas of two joined solids?
Because the faces where the solids are joined become hidden inside. Adding both total surface areas counts those hidden faces twice. Add only the outside surfaces.
Q2. Does the volume of a combined solid need any adjustment for the joint?
No. The volume of a combined solid is the sum of the volumes of its parts. If a part is scooped out, subtract its volume.
Q3. When do I use slant height and when do I use height for a cone?
Use slant height for the curved surface area and total surface area. Use height for the volume. Find slant height with l equals the square root of r squared plus h squared.
Q4. What stays the same when a solid is melted and recast?
The volume. The surface area generally changes. The number of new pieces equals the volume of the original divided by the volume of one piece.
Q5. How many litres are in a cubic metre?
1,000 litres. And 1,000 cubic centimetres make 1 litre. Convert all dimensions to one unit before calculating.
Conclusion: Count Only What You Can Touch
Most errors with combined solids disappear when you decide, before using any formula, which surfaces are really on the outside, and remember that volume has no such problem. Draw the solid, halve the diameter, find the slant height, and keep your units consistent.
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