Mensuration | ICSE Class 10 Maths Notes

Mensuration covers curved and total surface areas, volumes of cylinders, cones, spheres and hemispheres, hollow solids, combinations of solids, capacity, costs, and melting and recasting.

What do the measurements and symbols in mensuration mean?

Mensuration is the measurement of lengths, areas and volumes of geometrical figures. A solid occupies space. Its surface area measures its boundary, while its volume measures the space it occupies. These quantities answer different questions and have different units.

A radius, written as r, joins the centre of a circle or sphere to its boundary. A diameter, written as d, passes through the centre and joins opposite boundary points. Thus d = 2r. Convert a given diameter to a radius before substituting into a radius formula.

The symbol π, read as pi, denotes the ratio of a circle's circumference to its diameter. Circumference means the distance around a circle. Use the approximation to π specified in a numerical question; 22/7 and 3.14 are approximations, not exact values.

How are area and volume distinguished?

Curved surface area, abbreviated CSA, measures the curved part of a boundary. Total surface area, abbreviated TSA, includes all the surfaces bounding the solid. A flat circular face contributes its area when it is part of the required surface.

The symbol h denotes perpendicular height, measured at right angles to a base. For a cone, l denotes slant height, measured along its sloping surface from the vertex, or pointed end, to the base circumference. Height and slant height are different measurements.

Capacity is the internal space available for a substance in a container. Use internal dimensions for capacity. Keep all lengths in the same unit before multiplication, and attach square or cubic units to the final answer as appropriate.

How are a right circular cylinder's area and volume calculated?

A right circular cylinder has equal, parallel circular bases. Its axis, the line through their centres, is perpendicular to the bases. Its height is the perpendicular separation of the bases. The radius in its formulae is the radius of either circular base.

Result: Cylinder formulae

CSA = 2πrh, TSA = 2πrh + 2πr² and V = πr²h, where V denotes volume. The two area terms in TSA represent the curved wall and the two circular ends. A cylinder's curved area does not include either end.

The curved wall can be opened into a rectangle whose length is the base circumference and whose breadth is the cylinder's height. Multiplying these measurements gives its area. Multiplying the circular base area by perpendicular height gives the cylinder's volume.

A closed cylinder includes both circular ends. A thin container open at the top has a curved wall and one base. If both ends are open, neither circular opening is itself a material face. Identify the surfaces actually present before choosing an area expression.

How is colouring a cylindrical surface costed?

Worked example 1. A factory makes 120000 cylindrical pencils daily. Each is 25 cm long with base circumference 1.5 cm. Find the cost of colouring their curved surfaces at Rs 0.05 per square decimetre. A decimetre (dm) is 10 cm, so 1 dm² = 100 cm².

Answer: One curved area = circumference × length = 1.5 × 25 = 37.5 cm². The daily area = 120000 × 37.5 = 4500000 cm² = 45000 dm². Cost = 45000 × 0.05 = Rs 2250.

The symbol Rs denotes rupees. The quoted rate is per square decimetre, so the area must be expressed in that unit before multiplying. The circular ends are excluded because the question specifies colouring the curved surfaces.

How do a cone's height and slant height enter its formulae?

A right circular cone has a circular base and a vertex directly above its centre. The perpendicular height runs from the vertex to the base centre. Slant height runs from the vertex to a point on the circumference.

Result: Cone formulae

l² = r² + h², so l = √(r² + h²). The square-root sign √ means the positive number whose square equals the expression inside it. Radius, perpendicular height and slant height form a right-angled triangle in a section through the cone's axis.

This is an application of Pythagoras' theorem: in a right-angled triangle, the square of the longest side equals the sum of the squares of the other two sides. The longest side here is the slant height, not the perpendicular height.

CSA = πrl, TSA = πrl + πr² and V = πr²h/3. Use l for curved area and h for volume. The cone's volume is one-third of the volume of a cylinder with the same base radius and perpendicular height.

Which surfaces remain when cones are joined?

Worked example 2. Two cones, each of radius 8 cm and height 15 cm, are joined along their entire bases. Find the surface area of the resulting solid. Give an exact answer in terms of π.

Answer: Each slant height is √(8² + 15²) = √289 = 17 cm. The circular bases are hidden at the join. Required area = π × 8 × 17 + π × 8 × 17 = 272π cm².

The phrase in terms of π means leaving π in the final expression instead of replacing it by a decimal or fraction. The joined solid has two exposed curved surfaces. Adding the total areas of the separate cones would incorrectly count both hidden bases.

Note: Perpendicular height determines a cone's volume; slant height determines its curved surface area. Calculate the missing height from the right triangle before applying the relevant formula.

How are spheres and hemispheres measured?

A sphere has a curved surface whose points are at the same distance from its centre. That distance is its radius. A hemisphere is half a sphere, obtained by a plane through the centre. It has a curved surface and a flat circular base.

Result: Sphere and hemisphere formulae

For a sphere, surface area = 4πr² and V = 4πr³/3. There is no separate flat base to add. For a hemisphere, CSA = 2πr², TSA = 3πr² and V = 2πr³/3.

A hemisphere's curved area is half the surface area of its sphere. Its total area also includes the circular base, of area πr². Its volume is half the sphere's volume. The height of a hemisphere, measured from its flat base to its farthest point, equals its radius.

How do volume and area ratios differ?

A ratio compares quantities by division. Quantities are proportional when their quotient is constant. Sphere volumes are proportional to the cubes of their radii, while their surface areas are proportional to the squares of their radii.

Worked example 3. The volumes of two spheres are in the ratio 64:27. Find the ratio of their surface areas.

Answer: Since 64 = 4³ and 27 = 3³, the radii are in the ratio 4:3. Surface areas are proportional to squared radii, giving 4²:3² = 16:9.

The common factors in the sphere formulae cancel when ratios are taken. Cubic dependence for volume and square dependence for area mean that the surface-area ratio is the volume ratio raised to the power 2/3. For the same pair of spheres, these ratios differ unless both are 1:1.

How do inner and outer dimensions determine capacity and material volume?

A hollow solid contains a cavity, or empty region. Its outer dimensions describe its external boundary; its inner dimensions describe the cavity. The volume of material is found by subtracting the cavity's volume from the volume enclosed by the outer boundary.

For a spherical shell, let R be its outer radius and r its inner radius. Then material volume = 4π(R³ − r³)/3. The empty internal volume is 4πr³/3. It must not be counted as material available for melting.

How is a hollow cylinder treated?

For a pipe with a cylindrical hole running through its length, let R and r again denote outer and inner radii, and let h denote its length. Subtracting the inner cylinder from the outer cylinder gives material volume = π(R² − r²)h.

The wall thickness is R − r when the two circular boundaries have the same centre. Their diameter difference is twice this thickness. For a container with a bottom, distinguish its internal height from its external height; the base occupies material too.

Surface area requires a separate decision. A hole removes a volume but exposes an internal surface. Thus subtraction of volumes does not imply that every surface-area calculation is also a subtraction. Identify whether the question asks for the inside, outside, or complete material boundary.

How can a raised base reduce capacity?

Worked example 4. A cylindrical glass has inner diameter 5 cm and internal height 10 cm. A hemispherical raised portion of radius 2.5 cm occupies its bottom. Find its apparent cylindrical capacity and actual capacity, using π = 3.14.

Answer: Apparent capacity = 3.14 × 2.5² × 10 = 196.25 cm³. The raised hemisphere occupies (2/3) × 3.14 × 2.5³, approximately 32.71 cm³. Actual capacity = 196.25 − 32.71 = 163.54 cm³, approximately.

Apparent capacity here means the volume suggested by the full cylindrical interior before allowing for the raised portion. Actual capacity is the remaining space. A hemispherical depression extending into a solid and a hemisphere protruding into a vessel affect different quantities, so interpret the shape carefully.

How is the exposed surface area of combined solids found?

A combination of solids is an object assembled from recognisable basic shapes. Its exposed surface consists of the parts that remain on its boundary after joining. The circular faces used to join equal-radius pieces lie inside the object and are not exposed.

How are shared faces removed from the calculation?

A cylinder with a hemisphere on each end has the cylinder's curved area and both hemispherical curved areas on its outside. Its two circular end faces are covered. The hemispheres' flat faces are also hidden at the joins.

Two separate hemispheres appear on either side of a horizontal cylinder. Arrows show them being brought together and then joined. Shading marks the circular joining faces in the separated and intermediate views.

See Fig. 12.4 in your NCERT textbook

Worked example 5. A toy consists of a cone mounted on a hemisphere of the same radius, 3.5 cm. Its total height is 15.5 cm. Find its total surface area using π = 22/7.

Answer: The hemisphere's height is 3.5 cm, so the cone's height is 15.5 − 3.5 = 12 cm. Slant height = √(12² + 3.5²) = 12.5 cm. Exposed area = π × 3.5 × 12.5 + 2π × 3.5² = 214.5 cm².

What changes when the joined radii are unequal?

If a wider cone rests on a narrower cylinder, part of the cone's base remains exposed. This region is an annulus, a circular ring between circles with a common centre. Its area is the larger circle's area minus the smaller circle's area.

The rocket drawing labels overall height 26 cm and cone height 6 cm. An adjacent circular view labels the cone's base diameter 5 cm and the cylinder's base diameter 3 cm, showing the surrounding ring.

See Fig. 12.8 in your NCERT textbook

Worked example 6. A wooden rocket has total height 26 cm and cone height 6 cm. The cone's base diameter is 5 cm; the cylinder's is 3 cm. Find the orange area of the cone and its exposed base ring, and the yellow area of the cylinder including its lower base. Use π = 3.14.

Answer: Cone radius = 2.5 cm, cylinder radius = 1.5 cm, cylinder height = 20 cm. Cone slant height = √(2.5² + 6²) = 6.5 cm. Orange area = 3.14(2.5 × 6.5 + 2.5² − 1.5²) = 63.585 cm². Yellow area = 3.14(2 × 1.5 × 20 + 1.5²) = 195.465 cm².

Track each exposed face separately: cone curve, exposed ring, cylinder curve and lower circular base. The whole cone base is not hidden, and the cylinder's lower base is not covered by the cone. This surface inventory explains both expressions.

How are volumes of combined solids added or subtracted?

When basic solids are joined without overlapping interiors, their volumes add. Joining faces do not represent extra volume. This differs from surface area, where formerly exposed faces can become hidden. Decide whether parts are being added or material is being removed.

If a cavity is cut from a solid, remaining volume = original volume − removed volume. If an object occupies space inside a container, subtract its volume from the container's capacity when calculating the unoccupied space. Include each part once.

How is a cone on a hemisphere compared with its enclosing cylinder?

Worked example 7. A solid toy has a cone of height 2 cm and base diameter 4 cm standing on a hemisphere of the same radius. Find its volume and the difference between this and the volume of the smallest right circular cylinder enclosing it. The toy and cylinder share an axis. Use π = 3.14.

Answer: Common radius = 2 cm. Toy volume = (2/3) × 3.14 × 2³ + (1/3) × 3.14 × 2² × 2 = 25.12 cm³. The enclosing cylinder has radius 2 cm and height 2 + 2 = 4 cm. Its volume = 3.14 × 2² × 4 = 50.24 cm³. Difference = 25.12 cm³.

An enclosing cylinder surrounds the toy. In this close-fitting arrangement, the cylinder's radius equals the toy's greatest radius. Its height includes both the cone's perpendicular height and the hemisphere's height. It does not use the cone's slant height.

What should be checked before adding volumes?

1. Identify every basic solid and whether it contributes material or empty space.
2. Determine each part's radius and perpendicular height from the full dimensions.
3. Write a volume formula for each part, keeping all measurements in a common unit.
4. Add occupied parts or subtract cavities, then state the result in cubic units.

Keep fractions during the calculation when separate parts contain thirds. Adding exact expressions before rounding avoids small differences caused by rounding each part independently. An approximate final value should be labelled as approximate.

How does melting and recasting connect different solids?

Recasting means melting material and forming it into a new shape. In these problems, the volume of material before melting equals the volume after recasting, assuming no material is lost and the material's volume is unchanged. Surface area need not remain equal.

If n denotes the number of identical new solids, then original material volume = n × volume of one new solid. When several old solids are melted, add their material volumes first. For a hollow object, first subtract its cavity.

How many cones can be made from a sphere?

Worked example 8. A solid metallic sphere of radius 10.5 cm is melted into identical cones, each of radius 3.5 cm and height 3 cm. Assuming no loss of material or change in its volume, find the number of cones.

Answer: Let n be the number of cones. Equating volumes gives (4/3)π × 10.5³ = n × (1/3)π × 3.5² × 3. Cancelling the common factors gives n = (4 × 10.5³)/(3.5² × 3) = 126 cones.

The factors π and one-third cancel because they occur on both sides of the equation. No approximation to π is needed. The calculation uses the sphere's cubic radius and the cone's squared radius multiplied by perpendicular height.

How is a shell recast into a cone?

Worked example 9. A metallic spherical shell has internal diameter 4 cm and external diameter 8 cm. It is melted into a solid cone with base diameter 8 cm. Assuming no loss of material or change in its volume, find the cone's height.

Answer: Inner radius = 2 cm, outer radius = 4 cm and cone radius = 4 cm. Shell volume = (4/3)π(4³ − 2³). Equating this to π × 4² × h/3 gives 4(64 − 8) = 16h. Hence h = 14 cm.

The cavity supplies no metal. Using the outer sphere alone would overestimate the available material. After finding a new solid's missing dimension from volume, use that dimension in its area formula if the question also asks for its surface area.

How are material costs calculated for a combination of solids?

A cost calculation combines a geometrical measurement with a rate, the charge per unit of that measurement. Canvas and painting problems use area when the rate is per square unit. The required area is the surface actually covered, including or excluding bases as stated.

Cost = required area × cost per unit area. Match the area unit to the quoted rate before multiplication. The shape description determines the geometrical calculation; the rate determines the final monetary calculation. Do these as separate steps.

How much canvas does a cylindrical tent with a conical roof require?

Worked example 10. A tent has a cylindrical portion of height 2.1 m and diameter 4 m, topped by a cone of the same base radius and slant height 2.8 m. The floor has no canvas. Find the canvas area and its cost at Rs 500 per m², using π = 22/7.

Answer: Radius = 2 m. Canvas area = cylinder CSA + cone CSA = 2 × (22/7) × 2 × 2.1 + (22/7) × 2 × 2.8 = 26.4 + 17.6 = 44 m². Cost = 44 × 500 = Rs 22000.

The cylindrical wall contributes a curved rectangular surface when opened out. The roof contributes the cone's curved surface. The circular joining face is internal, and the circular floor is explicitly uncovered. Consequently, neither contributes canvas area in this calculation.

Why is the roof's perpendicular height unnecessary here?

The cone's slant height is already supplied, which is the measurement needed for its curved area. Calculating its perpendicular height would be an extra step with no role in the required answer. A volume question about the tent would need that perpendicular height instead.

The cost follows directly from the calculated 44 m² and the given rate. Do not add an unstated allowance for extra cloth: a quantity for seams, wastage or overlaps would need to be supplied before it could enter a numerical answer.

How can a mensuration solution be organised and checked?

Start with the requested quantity: surface area, material volume, capacity, a missing dimension, or cost. The same object can give different answers to these questions. Next identify the solids involved and record which lengths belong to which component.

What sequence keeps the geometry clear?

1. List the supplied dimensions and convert diameters to radii where necessary.
2. Distinguish overall height, each component's height and any slant height.
3. For area, list exposed surfaces; for volume, list occupied parts and cavities.
4. Write the formula or conservation equation before substituting the numbers.
5. Calculate with consistent units and use the stated approximation to π.
6. Check the result against the required quantity and label its unit.

Read combined heights carefully. A hemisphere contributes one radius to the overall height of a cone-and-hemisphere toy. Two hemispherical ends contribute a full diameter to the length of a capsule-shaped object. The cylindrical part occupies the remaining length.

Which checks reveal a wrong expression?

An area expression should contain products of two lengths; a volume expression should contain products of three. A cone's slant height exceeds either its radius or its perpendicular height. A cavity's removal reduces material volume, and a raised base reduces available capacity.

Check the geometry before the arithmetic. Correct multiplication cannot repair a hidden face counted as exposed, a diameter used as a radius, or an overall height substituted for a component's height. Recasting begins with equal material volumes, while a coating calculation begins with the exposed area.

Glossary

• Mensuration — Measurement of lengths, surface areas and volumes of geometrical figures and solids.
• Radius — Distance from the centre of a circle or sphere to its boundary.
• Diameter — Length through the centre joining opposite boundary points, equal to twice the radius.
• Perpendicular height — Distance measured at right angles from a base to the opposite level or vertex.
• Slant height — Distance along a right circular cone's surface from its vertex to its base circumference.
• Curved surface area — Area of the curved boundary, excluding any flat circular faces of the solid.
• Total surface area — Sum of the areas of all surfaces forming the boundary of a solid.
• Volume — Measure of the space occupied by a solid, expressed in cubic units.
• Capacity — Internal space available in a container for holding a substance.
• Hemisphere — Half a sphere, with a curved surface and a flat circular base.
• Spherical shell — Material between an outer spherical boundary and an inner spherical cavity.
• Annulus — Circular ring between two circles that have a common centre.
• Recasting — Forming melted material into a new solid, using unchanged material volume when no loss occurs.

Common errors and misconceptions

• Misconception: A supplied diameter can be substituted for r. Correct: The symbol r denotes radius, so halve the diameter first.
• Misconception: A cone's slant height is used in its volume formula. Correct: Volume uses perpendicular height; curved area uses slant height.
• Misconception: A hemisphere's total area is half a sphere's area. Correct: Its curved area is half; total area also includes its circular base.
• Misconception: Add the total areas of all separate solids after joining. Correct: Count the exposed boundary and exclude hidden joining faces.
• Misconception: A hollow object supplies its entire outer volume for recasting. Correct: Subtract the cavity to obtain the material volume.
• Misconception: Surface area is conserved when a solid is melted. Correct: Equate material volumes under the no-loss assumption; areas can differ.
• Misconception: Overall height equals the height of every component. Correct: Separate the contributions of cones, cylinders and hemispheres before substitution.
• Misconception: A coating cost can use area in any unit. Correct: Convert the area to the square unit specified by the rate.

Exam-style questions with model answers

Q1. State the curved and total surface areas of a solid hemisphere of radius r, explaining their difference. [2 marks]

1. The curved surface area is 2πr², where π is the circle circumference-to-diameter ratio and r is the hemisphere's radius.
2. The total surface area is 3πr² because the flat circular base contributes an additional area πr².

Q2. Two cones, each of radius 8 cm and perpendicular height 15 cm, are joined along their entire bases. Find the exposed surface area in terms of π. [3 marks]

1. Let l denote either cone's slant height. Using the right triangle containing its radius and height, l = √(8² + 15²) = 17 cm.
2. The circular bases are hidden where the cones join. The exposed boundary therefore consists of the two curved surfaces, giving area = 2πrl, with radius r = 8 cm.
3. Substitution gives 2 × π × 8 × 17 = 272π cm². This is the area of the joined solid's exposed surface.

Q3. A cylindrical glass has inner diameter 5 cm and internal height 10 cm. Its bottom contains a raised hemisphere of radius 2.5 cm. Find the actual capacity. Use π = 3.14 and give the answer to two decimal places. [3 marks]

1. The cylindrical interior has radius 5/2 = 2.5 cm. Its apparent capacity, before allowing for the raised base, is 3.14 × 2.5² × 10 = 196.25 cm³.
2. The raised hemisphere occupies (2/3) × 3.14 × 2.5³, approximately 32.71 cm³. This occupied region cannot hold liquid and must be subtracted.
3. The actual capacity is therefore approximately 196.25 − 32.71 = 163.54 cm³. It is smaller than the full cylindrical volume because of the raised base.

Q4. A solid sphere of radius 10.5 cm is melted into identical cones of radius 3.5 cm and perpendicular height 3 cm. Assuming no loss of material or change in its volume, calculate how many cones are formed. [4 marks]

1. The original sphere's material volume is (4/3)π × 10.5³ cm³. All of this material is available because the sphere is solid.
2. Each new cone has volume (1/3)π × 3.5² × 3 cm³, using its perpendicular height.
3. Let n be the number of cones. Conservation of material volume gives (4/3)π × 10.5³ = n × (1/3)π × 3.5² × 3.
4. Cancel π/3 and divide by one cone's remaining volume factor: n = 4 × 10.5³/(3.5² × 3) = 126 cones.

Q5. A tent consists of a cylinder 2.1 m high and 4 m in diameter, topped by a cone of the same base radius and slant height 2.8 m. The floor is uncovered. Find the canvas area and its cost at Rs 500 per m². Use π = 22/7. [5 marks]

1. The common radius is half the diameter: 4/2 = 2 m. The cone's supplied 2.8 m measurement is its slant height, suitable for its curved-area formula.
2. The cylindrical wall requires area 2πrh = 2 × (22/7) × 2 × 2.1 = 26.4 m², where h is the cylindrical height.
3. The conical roof requires curved area πrl = (22/7) × 2 × 2.8 = 17.6 m², where l is the roof's slant height.
4. The floor is uncovered and the joining circular faces are internal. Required canvas area is therefore 26.4 + 17.6 = 44 m².
5. Multiply by the stated rate in the matching area unit. Total canvas cost = 44 × 500 = Rs 22000.

Q6. A toy consists of a cone of height 2 cm and base diameter 4 cm joined to a hemisphere of the same radius. Find the toy's volume and the unoccupied volume inside the smallest right circular cylinder enclosing it, with the same axis. Use π = 3.14. [5 marks]

1. The cone and hemisphere have common radius 4/2 = 2 cm. The hemisphere contributes 2 cm to the total height, so the toy is 4 cm high.
2. Add the two component volumes: hemisphere volume = (2/3) × 3.14 × 2³ and cone volume = (1/3) × 3.14 × 2² × 2, both in cm³.
3. Their sum is 25.12 cm³. The parts meet at their flat faces, so their interiors do not overlap and their volumes can be added.
4. The enclosing cylinder has radius 2 cm and height 4 cm. Its volume is 3.14 × 2² × 4 = 50.24 cm³.
5. Subtract the toy's occupied space from the cylinder's volume. The unoccupied volume is 50.24 − 25.12 = 25.12 cm³.

Q7. A spherical metal shell has inner diameter 4 cm and outer diameter 8 cm. It is melted into a solid cone of base diameter 8 cm. Assuming no loss of material or change in its volume, find the cone's height. [4 marks]

1. Halve each diameter. The shell's inner radius is 2 cm and outer radius is 4 cm; the new cone's radius is also 4 cm.
2. Subtract the cavity from the outer sphere. Material volume = (4/3)π(4³ − 2³) = (4/3)π × 56 cm³.
3. Let h be the cone's perpendicular height in centimetres. Equating volumes gives (1/3)π × 4² × h = (4/3)π × 56.
4. Cancel π/3 to obtain 16h = 224. Division by 16 gives the required height, h = 14 cm.

Key takeaways

• Convert diameters to radii and express every length in a common unit before substituting into area or volume formulae.
• A cylinder's volume equals its circular base area multiplied by perpendicular height; its curved area excludes both circular ends.
• A cone uses slant height for curved area and perpendicular height for volume, linked by Pythagoras' theorem.
• A hemisphere's total surface area includes its circular base, while its curved area is half the sphere's surface area.
• For combined solids, add exposed surface areas and exclude joining faces; unequal radii may leave an exposed circular ring.
• Add volumes of non-overlapping components, subtract cavities, and use internal dimensions when calculating a container's capacity.
• Melting and recasting equate material volumes under the no-loss assumption; the resulting surface area may change.
• For canvas or painting costs, multiply the required exposed area by the rate after matching their area units.

Test yourself

Why must a diameter be halved before using a radius formula?

A diameter equals twice the radius. Substituting it directly would use the wrong length in the formula.

What is the relation between a right cone's radius r, perpendicular height h and slant height l?

They satisfy l² = r² + h² because these lengths form a right-angled triangle, with slant height as the longest side.

Why does a solid hemisphere have total area 3πr² rather than 2πr², where r is its radius?

Its curved area is 2πr², and the flat circular base contributes another πr².

Which surfaces are exposed on a cylinder with equal-radius hemispheres attached at both ends?

The cylinder's curved wall and the two hemispherical curved surfaces are exposed; the joining circular faces are hidden.

How is material volume calculated for a hollow solid?

Subtract the volume of the internal cavity from the volume enclosed by the outer boundary.

What quantity is equated when a solid is recast without loss or change in material volume?

The original material volume equals the combined volume of the new solids; their surface areas need not be equal.

Why does a raised hemispherical bottom reduce a cylindrical glass's capacity?

The hemisphere occupies space within the cylindrical interior, leaving less empty space available to hold liquid.

What must be checked before multiplying an area by a coating rate?

The area must describe the surfaces being coated and use the same square unit as the quoted rate.