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Surface Areas and Volumes

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Chapter Notes

Overview
This chapter extends the study of surface areas and volumes from Class IX to solids formed by combining two or more basic solids like cuboids, cones, cylinders, spheres, and hemispheres. The big idea is that the total surface area of a combined solid is the sum of the curved surface areas of its visible parts (excluding hidden joining surfaces), while the volume of the combined solid is simply the sum of the volumes of its constituent solids. A student must walk away knowing how to break down a complex object into basic solids, apply the correct formulas, and avoid adding hidden surfaces when calculating surface area.
Key Concepts (12)
Combination of Solids
A solid made up of two or more basic solids (cuboid, cone, cylinder, sphere, hemisphere) joined together.
e.g. A truck container made of a cylinder with two hemispheres at its ends.
Total Surface Area (TSA)
The total area of all the surfaces of a solid that are visible or exposed, excluding any surfaces that are joined or hidden.
e.g. For a toy made of a cone and hemisphere joined at their bases, TSA = CSA of cone + CSA of hemisphere.
Curved Surface Area (CSA)
The area of the curved (lateral) surface of a solid, excluding its base(s).
e.g. The curved surface area of a cylinder is 2πrh.
Volume of a Combination of Solids
The total space occupied by a solid formed by joining two or more basic solids, equal to the sum of the volumes of the individual solids.
e.g. The volume of a shed that is a cuboid surmounted by a half cylinder is the sum of the volume of the cuboid and half the volume of the cylinder.
Cuboid
A solid with six rectangular faces, opposite faces equal, and all angles right angles.
e.g. A brick or a wooden block.
Cone
A solid with a circular base and a vertex, where the lateral surface tapers from the base to the vertex.
e.g. A birthday cap or a conical top of a tent.
Cylinder
A solid with two parallel circular bases of equal radius and a curved lateral surface.
e.g. A pipe or a cylindrical glass.
Sphere
A perfectly round solid where every point on its surface is equidistant from its centre.
e.g. A ball or a spherical glass vessel.
Hemisphere
Half of a sphere, usually with a flat circular base.
e.g. The dome of a bird-bath or the top of a lattu.
Slant Height
The distance along the lateral surface of a cone from the vertex to any point on the circumference of the base.
e.g. For a cone of radius 2.5 cm and height 6 cm, slant height l = √(r² + h²) = 6.5 cm.
Apparent Capacity
The volume of a container calculated assuming it is a perfect basic solid, without accounting for internal depressions or raised portions.
e.g. A cylindrical glass of height 10 cm and diameter 5 cm has apparent capacity πr²h = 196.25 cm³.
Actual Capacity
The actual volume a container can hold, obtained by subtracting the volume of any internal solid (like a hemisphere) from the apparent capacity.
e.g. If the glass has a hemispherical raised bottom of radius 2.5 cm, actual capacity = 196.25 - 32.71 = 163.54 cm³.
Formulas & Equations (19)
TSA of new solid = CSA of one hemisphere + CSA of cylinder + CSA of other hemisphere
Reads as: Total surface area of the new solid equals curved surface area of one hemisphere plus curved surface area of cylinder plus curved surface area of the other hemisphere.
Variables: TSA = Total Surface Area (cm²), CSA = Curved Surface Area (cm²).
Use when: When finding the surface area of a solid made of a cylinder with hemispheres at both ends, like a capsule or a container.
⚠ Students add the areas of the flat circular faces of the hemispheres and cylinder, which are hidden inside the joint.
TSA of toy = CSA of hemisphere + CSA of cone
Reads as: Total surface area of the toy equals curved surface area of hemisphere plus curved surface area of cone.
Variables: TSA = Total Surface Area (cm²), CSA = Curved Surface Area (cm²).
Use when: When finding the surface area of a toy made by joining a cone and a hemisphere at their bases.
⚠ Students add the base areas of the cone and hemisphere, which are not visible after joining.
CSA of hemisphere = 2πr²
Reads as: Curved surface area of a hemisphere equals two times pi times r squared.
Variables: r = radius of hemisphere (cm), π = 22/7 or 3.14.
Use when: When calculating the curved surface area of a hemisphere (half of a sphere).
⚠ Students use 4πr² (surface area of full sphere) instead of 2πr².
CSA of cone = πrl
Reads as: Curved surface area of a cone equals pi times radius times slant height.
Variables: r = radius of base (cm), l = slant height (cm), π = 22/7 or 3.14.
Use when: When calculating the curved surface area of a cone, given radius and slant height.
⚠ Students use πr² (area of base) instead of πrl.
l = √(r² + h²)
Reads as: Slant height equals the square root of r squared plus h squared.
Variables: l = slant height (cm), r = radius of base (cm), h = height of cone (cm).
Use when: When slant height is not given directly but radius and height of cone are known.
⚠ Students forget to take the square root or add r and h directly.
TSA of cube = 6 × (edge)²
Reads as: Total surface area of a cube equals 6 times edge squared.
Variables: edge = side length of cube (cm).
Use when: When calculating the total surface area of a cube.
⚠ Students use 4 × (edge)² (lateral surface area) instead of 6 × (edge)².
Surface area of block = TSA of cube – base area of hemisphere + CSA of hemisphere
Reads as: Surface area of the block equals total surface area of cube minus base area of hemisphere plus curved surface area of hemisphere.
Variables: TSA = Total Surface Area (cm²), CSA = Curved Surface Area (cm²).
Use when: When a hemisphere is mounted on a cube, and the base of the hemisphere is attached to the cube.
⚠ Students forget to subtract the base area of the hemisphere from the cube's TSA.
Area to be painted orange = CSA of cone + base area of cone – base area of cylinder
Reads as: Area to be painted orange equals curved surface area of cone plus base area of cone minus base area of cylinder.
Variables: CSA = Curved Surface Area (cm²), base area = πr² (cm²).
Use when: When a cone is mounted on a cylinder and the cone's base is larger, leaving a ring to be painted.
⚠ Students paint the entire base of the cone without subtracting the cylinder's base area.
Area to be painted yellow = CSA of cylinder + area of one base of the cylinder
Reads as: Area to be painted yellow equals curved surface area of cylinder plus area of one base of the cylinder.
Variables: CSA = 2πrh (cm²), base area = πr² (cm²).
Use when: When the cylindrical portion of a rocket has one base visible (the other is covered by the cone).
⚠ Students add both bases of the cylinder, but one base is hidden under the cone.
TSA of bird-bath = CSA of cylinder + CSA of hemisphere = 2πrh + 2πr² = 2πr(h + r)
Reads as: Total surface area of bird-bath equals curved surface area of cylinder plus curved surface area of hemisphere, which equals 2πr times (h plus r).
Variables: r = common radius of cylinder and hemisphere (cm), h = height of cylinder (cm), π = 22/7 or 3.14.
Use when: When a cylinder has a hemispherical depression at one end (open top).
⚠ Students add the base area of the cylinder, which is not present (open end).
Volume of shed = volume of cuboid + ½ volume of cylinder
Reads as: Volume of shed equals volume of cuboid plus half volume of cylinder.
Variables: Volume of cuboid = l × b × h (m³), volume of cylinder = πr²h (m³).
Use when: When a shed is a cuboid surmounted by a half cylinder (like a tunnel).
⚠ Students use full cylinder volume instead of half.
Apparent capacity of glass = πr²h
Reads as: Apparent capacity of glass equals pi times r squared times h.
Variables: r = inner radius of glass (cm), h = height of glass (cm), π = 3.14 or 22/7.
Use when: When calculating the volume of a cylindrical glass assuming it is a perfect cylinder.
⚠ Students forget to subtract the volume of internal solids like a hemisphere.
Actual capacity of glass = apparent capacity – volume of hemisphere
Reads as: Actual capacity of glass equals apparent capacity minus volume of hemisphere.
Variables: Volume of hemisphere = (2/3)πr³ (cm³).
Use when: When a glass has a hemispherical raised portion at the bottom.
⚠ Students add the volume of the hemisphere instead of subtracting it.
Volume of toy = (2/3)πr³ + (1/3)πr²h
Reads as: Volume of toy equals two-thirds pi r cubed plus one-third pi r squared h.
Variables: r = common radius of hemisphere and cone (cm), h = height of cone (cm), π = 3.14 or 22/7.
Use when: When a toy is a hemisphere surmounted by a cone.
⚠ Students use the wrong formula for hemisphere volume (1/3)πr³ instead of (2/3)πr³.
Volume of cylinder = πr²h
Reads as: Volume of cylinder equals pi times r squared times h.
Variables: r = radius of base (cm), h = height (cm), π = 22/7 or 3.14.
Use when: When calculating the volume of a cylinder.
⚠ Students use 2πrh (CSA formula) instead of πr²h.
Volume of cone = (1/3)πr²h
Reads as: Volume of cone equals one-third pi times r squared times h.
Variables: r = radius of base (cm), h = height (cm), π = 22/7 or 3.14.
Use when: When calculating the volume of a cone.
⚠ Students forget the factor 1/3 and use πr²h.
Volume of sphere = (4/3)πr³
Reads as: Volume of sphere equals four-thirds pi times r cubed.
Variables: r = radius of sphere (cm), π = 22/7 or 3.14.
Use when: When calculating the volume of a sphere.
⚠ Students use (2/3)πr³ (hemisphere volume) instead.
Volume of hemisphere = (2/3)πr³
Reads as: Volume of hemisphere equals two-thirds pi times r cubed.
Variables: r = radius of hemisphere (cm), π = 22/7 or 3.14.
Use when: When calculating the volume of a hemisphere.
⚠ Students use (4/3)πr³ (full sphere) instead.
Volume of cuboid = l × b × h
Reads as: Volume of cuboid equals length times breadth times height.
Variables: l = length (m or cm), b = breadth (m or cm), h = height (m or cm).
Use when: When calculating the volume of a cuboid.
⚠ Students confuse with surface area formula.
How It Works
Finding Surface Area of a Combination of Solids
  1. Step 1: Identify the basic solids that make up the combined object.
  2. Step 2: Determine which surfaces are visible (exposed) and which are hidden (joined).
  3. Step 3: Calculate the curved surface area (CSA) of each visible part using the appropriate formula.
  4. Step 4: Add the CSAs of all visible parts to get the total surface area. Do not add areas of hidden surfaces.
  5. Step 5: If a base of one solid is attached to another, subtract that base area from the total if it was included in the TSA of the first solid.
💡 Never add the areas of surfaces that are joined together — they are not visible. For example, when a cone is mounted on a hemisphere, the base of the cone and the flat face of the hemisphere are hidden.
Finding Volume of a Combination of Solids
  1. Step 1: Identify the basic solids that make up the combined object.
  2. Step 2: Calculate the volume of each constituent solid using the appropriate formula.
  3. Step 3: Add the volumes of all constituent solids to get the total volume of the combined solid.
  4. Step 4: If there is a hollow or depression, subtract the volume of the removed part from the total volume.
💡 Unlike surface area, volume is simply additive — no surfaces are hidden when calculating volume. For example, the volume of a toy with a cone and hemisphere is the sum of their volumes.
Calculating Slant Height of a Cone
  1. Step 1: Identify the radius (r) and height (h) of the cone.
  2. Step 2: Use the Pythagorean theorem: l = √(r² + h²).
  3. Step 3: Compute the square root to get the slant height (l).
💡 Slant height is always greater than the height. Do not forget to take the square root.
Finding Surface Area When a Hemisphere is Mounted on a Cube
  1. Step 1: Calculate the TSA of the cube (6 × edge²).
  2. Step 2: Calculate the base area of the hemisphere (πr²) — this area is hidden because it is attached to the cube.
  3. Step 3: Subtract the base area of the hemisphere from the TSA of the cube.
  4. Step 4: Add the CSA of the hemisphere (2πr²) to the result.
💡 The base of the hemisphere is not visible, so it must be removed from the cube's surface area. The final formula is: TSA of block = 150 + πr² (as in Example 2).
Finding Area to Paint on a Rocket (Cone on Cylinder with Different Base Radii)
  1. Step 1: Identify the radius of the cone (r), radius of the cylinder (r'), height of cone (h), height of cylinder (h'), and slant height of cone (l).
  2. Step 2: For the orange (conical) portion: calculate CSA of cone (πrl) + base area of cone (πr²) – base area of cylinder (πr'²).
  3. Step 3: For the yellow (cylindrical) portion: calculate CSA of cylinder (2πr'h') + area of one base of cylinder (πr'²).
  4. Step 4: Add the two results if total area is required, or keep separate for different colours.
💡 The base of the cone is larger than the base of the cylinder, so a ring-shaped area on the cone's base is visible and must be painted. Subtract the cylinder's base area from the cone's base area.
Finding Apparent and Actual Capacity of a Glass with a Hemispherical Bottom
  1. Step 1: Calculate the apparent capacity assuming the glass is a perfect cylinder: πr²h.
  2. Step 2: Calculate the volume of the hemispherical raised portion at the bottom: (2/3)πr³.
  3. Step 3: Subtract the volume of the hemisphere from the apparent capacity to get the actual capacity.
💡 The hemispherical raised portion reduces the capacity, so it is subtracted, not added.
Finding Volume of a Toy (Hemisphere Surmounted by Cone) and Difference with Circumscribing Cylinder
  1. Step 1: Calculate the volume of the hemisphere: (2/3)πr³.
  2. Step 2: Calculate the volume of the cone: (1/3)πr²h.
  3. Step 3: Add them to get the volume of the toy.
  4. Step 4: For the circumscribing cylinder, its radius equals the radius of the hemisphere, and its height equals the sum of the cone's height and the hemisphere's radius.
  5. Step 5: Calculate the volume of the cylinder: πr²h_cylinder.
  6. Step 6: Subtract the volume of the toy from the volume of the cylinder to get the difference.
💡 The cylinder circumscribes the toy, meaning the toy fits exactly inside the cylinder. The height of the cylinder is the total height of the toy (cone height + hemisphere radius).
Exam Traps — Don't Make These Mistakes
⚠ Adding Hidden Surfaces in Surface Area
✗ TSA of toy = CSA of hemisphere + CSA of cone + area of base of hemisphere + area of base of cone
→ The bases of the hemisphere and cone are joined together and are not visible on the surface of the toy. Adding them counts hidden surfaces.
✓ TSA of toy = CSA of hemisphere + CSA of cone (only the visible curved surfaces).
⚠ Forgetting to Subtract the Base of the Hemisphere from the Cube's TSA
✗ Surface area of block = TSA of cube + CSA of hemisphere = 150 + 2πr²
→ The base of the hemisphere is attached to the cube, so that area is no longer part of the cube's surface. It must be subtracted from the cube's TSA before adding the hemisphere's CSA.
✓ Surface area of block = TSA of cube – base area of hemisphere + CSA of hemisphere = 150 – πr² + 2πr² = 150 + πr².
⚠ Using Full Cylinder Volume Instead of Half for a Shed
✗ Volume of shed = volume of cuboid + volume of cylinder
→ The shed is surmounted by a half cylinder (like a tunnel), not a full cylinder. Using the full cylinder volume overestimates the volume.
✓ Volume of shed = volume of cuboid + ½ × volume of cylinder.
⚠ Adding the Volume of the Hemisphere Instead of Subtracting for Actual Capacity
✗ Actual capacity of glass = apparent capacity + volume of hemisphere
→ The hemispherical raised portion takes up space inside the glass, reducing the capacity. It should be subtracted, not added.
✓ Actual capacity of glass = apparent capacity – volume of hemisphere.
⚠ Painting the Entire Base of the Cone When It Overlaps the Cylinder
✗ Area to be painted orange = CSA of cone + base area of cone
→ The base of the cone is larger than the base of the cylinder, but the part of the cone's base that sits on the cylinder is not visible. Only the ring (base area of cone minus base area of cylinder) is visible.
✓ Area to be painted orange = CSA of cone + (base area of cone – base area of cylinder).
⚠ Using Wrong Formula for Hemisphere Volume or CSA
✗ Volume of hemisphere = (4/3)πr³ or CSA of hemisphere = 4πr²
→ A hemisphere is half of a sphere. The volume of a hemisphere is (2/3)πr³, and the curved surface area is 2πr² (half of 4πr²). Using the full sphere formulas gives double the correct value.
✓ Volume of hemisphere = (2/3)πr³, CSA of hemisphere = 2πr².
Compare & Contrast
Method of calculation
Surface Area of a Combination of Solids
Volume of a Combination of Solids
Surface area is found by adding only the curved surface areas of visible parts, excluding hidden joining surfaces. Volume is found by simply adding the volumes of all constituent solids, with no surfaces hidden.
Definition and calculation
Apparent Capacity
Actual Capacity
Apparent capacity is the volume of the container assuming it is a perfect basic solid (e.g., a cylinder). Actual capacity is the real volume the container can hold, obtained by subtracting the volume of any internal solid (e.g., a hemisphere) from the apparent capacity.
Exam Focus
Marks pattern: This chapter typically appears in CBSE board exams with 2 MCQs (1 mark each), 1 short answer question (3 marks), and 1 long answer question (5 marks), totalling approximately 10 marks.
Most asked: Surface area and volume of solids formed by combining a cone and a hemisphere (e.g., toy, top). · Surface area and volume of solids formed by combining a cylinder and a hemisphere (e.g., capsule, bird-bath, glass). · Finding the area to be painted on objects like rockets or decorative blocks. · Problems involving apparent and actual capacity of containers with internal depressions.
Don't skip: Long answer questions (5 marks) involving a combination of three solids (e.g., cone on cylinder, or cylinder with two hemispheres) where students must calculate both surface area and volume, or find the difference in volumes.
One-Line Revision (read aloud before the exam)
  1. The total surface area of a combined solid is the sum of the curved surface areas of its visible parts only.
  2. The volume of a combined solid is the sum of the volumes of its constituent solids.
  3. When a hemisphere is mounted on a cube, subtract the base area of the hemisphere from the cube's TSA before adding the hemisphere's CSA.
  4. The slant height of a cone is calculated using l = √(r² + h²).
  5. The curved surface area of a hemisphere is 2πr², not 4πr².
  6. The volume of a hemisphere is (2/3)πr³, not (4/3)πr³.
  7. For a cone mounted on a cylinder with different base radii, the visible ring on the cone's base is painted: area = πr² – πr'².
  8. The actual capacity of a glass with a hemispherical bottom is apparent capacity minus the volume of the hemisphere.
  9. A shed in the shape of a cuboid surmounted by a half cylinder has volume = volume of cuboid + ½ volume of cylinder.
  10. When a cylinder circumscribes a toy (cone + hemisphere), the cylinder's height equals the cone's height plus the hemisphere's radius.
  11. Never add the areas of surfaces that are joined together — they are hidden.
  12. Always check whether a base is open or closed when calculating surface area of containers like bird-baths or glasses.
  13. Use π = 22/7 unless stated otherwise in the problem.
  14. In problems with multiple colours, calculate the area for each colour separately and then add if total is required.

MCQs — 10 Questions with Answers

Q1. A container is made by joining a cylinder of radius 7 cm with two hemispheres of the same radius at both ends. When calculating the total surface area, a student claims that the flat circular faces of the hemispheres must be included. Is this claim correct?

  • A. Yes, because hemispheres have two circular faces
  • B. No, because the flat faces are interior to the combined solid and not part of the external surface ✓
  • C. Yes, because we must count all faces of each component
  • D. No, because hemispheres have no flat faces

Answer: B — When two solids are joined, their interface faces become interior and are excluded from the external surface area; only curved surfaces of the hemispheres and cylinder are exposed.

Q2. Assertion (A): When a hemisphere is placed on top of a cube, the total surface area of the combined solid equals the total surface area of the cube plus the curved surface area of the hemisphere. Reason (R): The base of the hemisphere and the top face of the cube overlap when the solids are joined. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A ✓
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false but R is true

Answer: A — Both statements are true: the overlapping circular base of the hemisphere with the cube's top face means neither contributes to external surface area, making the assertion correct and R logically explains why.

Q3. A toy is made by surmounting a cone on top of a hemisphere such that their bases coincide exactly. When finding the painted surface area, should the base area of the cone be included?

  • A. Yes, because the base is part of the cone
  • B. No, because the base is internal to the combined solid and completely covered by the hemisphere ✓
  • C. Yes, because we must paint all surfaces of both solids
  • D. No, because the cone's base is smaller than the hemisphere's base

Answer: B — When the cone and hemisphere are joined with coinciding bases, the circular base becomes an internal interface and is completely hidden, so it cannot be painted or included in surface area.

Q4. A student argues that the total surface area of a cylinder with two hemispheres at its ends can be calculated by adding the curved surface area of the cylinder to the total surface area of both hemispheres (including their flat circular bases). Is this reasoning valid?

  • A. Yes, this method always works for composite solids
  • B. No, because the flat circular bases of the hemispheres become interior surfaces and must be subtracted ✓
  • C. Yes, because hemispheres are complete solids with circular bases
  • D. No, because hemispheres do not have curved surface areas

Answer: B — The flat bases of the hemispheres are glued to the cylinder's ends and become interior surfaces; they must be excluded by subtracting them from the total.

Q5. Assertion (A): In a composite solid made of a cone mounted on a cylinder, if the cone's base diameter is larger than the cylinder's base diameter, then part of the cone's base will be visible and must be included in the painted surface area. Reason (R): The exposed part of the cone's base forms an annular (ring) region that contributes to the external surface. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A ✓
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false but R is true

Answer: A — When the cone's base is larger, the annular region (outer circle of cone minus inner circle of cylinder) is indeed exposed and must be painted, making both A and R true with R correctly explaining A.

Q6. A decorative block consists of a cube with a hemispherical dome on its top face. The hemisphere's diameter is smaller than the cube's edge length. A student calculates the total surface area as: TSA = 6(edge)² − πr² + 2πr², where r is the hemisphere's radius. Is this formula correct?

  • A. Yes, because it accounts for all exposed surfaces of both solids ✓
  • B. No, because the formula subtracts and adds the same term, yielding TSA = 6(edge)² + πr², which is incorrect
  • C. Yes, because the first term represents the cube and the second represents the hemisphere
  • D. No, because it fails to account for the curved surface of the hemisphere

Answer: A — The formula correctly removes the contact area (πr²) from the cube's total and adds the curved surface area of the hemisphere (2πr²), giving TSA = 6(edge)² + πr².

Q7. Assertion (A): For a composite solid made by attaching a cylinder and a hemisphere with the same radius, the total surface area equals the sum of the curved surface areas of both components only. Reason (R): The flat circular base of the hemisphere coincides with one of the bases of the cylinder, so both these faces are interior to the combined solid. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A ✓
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false but R is true

Answer: A — Assertion A is correct because only curved surfaces are external; Reason R correctly explains this by identifying that both flat faces become interior interfaces.

Q8. A cone and a hemisphere have the same radius r. When the cone is placed with its base on the flat face of the hemisphere, does the entire circular base of the cone need to be painted if the combined toy is painted on all external surfaces?

  • A. Yes, because the base of the cone is always external
  • B. No, because the entire base is internal and fully covered by the hemisphere ✓
  • C. Partially, only if the cone's base is larger than the hemisphere's circular face
  • D. Only the curved surface of the cone needs painting, not its base

Answer: B — Since both solids have equal radius and their bases coincide, the cone's base is completely covered and becomes an interior interface, so it is not painted.

Q9. Assertion (A): When calculating the surface area of a cylinder with a hemispherical depression at one end (like a bird-bath), the curved surface area of the hemisphere is added to the lateral surface area of the cylinder and the area of the exposed base. Reason (R): The hemispherical depression is carved inward, so its curved inner surface becomes part of the external surface area of the bird-bath. Choose the correct option:

  • A. Both A and R are true and R is the correct explanation of A ✓
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false but R is true

Answer: A — Both statements correctly describe a bird-bath: the inner curved surface of the hemispherical depression is exposed and must be included, along with the cylinder's lateral surface and one base.

Q10. A solid is formed by joining two cones base-to-base (forming a double cone or bicone). To find its total surface area, should the area of the common base be counted?

  • A. Yes, because each cone has a base area
  • B. No, because the base of each cone is internal and completely hidden at the junction ✓
  • C. Yes, because the double cone has a visible widest section
  • D. No, because double cones have no base area by definition

Answer: B — When two cones are joined base-to-base, their common base becomes an interior surface and is completely concealed, so only the two curved surfaces of the cones are part of the external surface area.

Flashcards

What is the curved surface area of a hemisphere with radius r?

CSA of hemisphere = 2πr²

When combining two solids, what surfaces do we count in total surface area?

Only the exposed curved surfaces; hidden bases where solids touch are NOT counted.

A cone is mounted on a cylinder. What three areas must we calculate?

CSA of cone + CSA of cylinder + ring area (base of cone minus base of cylinder).

Formula for curved surface area of a cone with radius r and slant height l?

CSA of cone = πrl

A hemisphere sits on top of a cube. Which part of the cube's surface is removed?

The circular base of the hemisphere where it touches the cube is subtracted from the cube's top face.

How do you find the slant height of a cone given radius r and height h?

Slant height l = √(r² + h²)

What is the total surface area formula for a cylinder with radius r and height h?

TSA of cylinder = 2πr(r + h) or 2πrh + 2πr²

In a composite solid, how do you find the height of one component if total height is given?

Subtract the known heights of other components from the total height.

A hemispherical depression means the curved surface is inside. Do we count it?

Yes, because it is an exposed internal surface that needs to be painted or treated.

Why is TSA of a composite solid NOT equal to sum of individual TSAs?

Because the base areas where solids are joined together are removed from the total.

Important Board Questions

A hemisphere of radius 3.5 cm is placed on top of a cube of side 5 cm. State the formula for the total surface area of the resulting solid and identify which area from the cube is NOT counted. [2 marks]

TSA = (Total surface of cube) − (Base area of hemisphere) + (Curved surface of hemisphere) = 6(5²) − π(3.5)² + 2π(3.5)². The circular base where hemisphere sits is subtracted because it becomes internal.

A toy is in the form of a cone surmounted on a hemisphere, both having radius 3.5 cm. The total height of the toy is 15.5 cm. Find (i) the height of the cone, and (ii) the slant height of the cone. [3 marks]

Height of cone = 15.5 − 3.5 = 12 cm (subtract radius of hemisphere from total height). Use l = √(r² + h²) = √(3.5² + 12²) to find slant height ≈ 12.5 cm. Show both calculations with substitution.

A wooden rocket consists of a cone mounted on a cylinder. The height of the entire rocket is 26 cm, with the conical part 6 cm high. The cone base diameter is 5 cm (radius 2.5 cm) and the cylinder base diameter is 3 cm (radius 1.5 cm). Find (i) the curved surface area of the cone, (ii) the curved surface area of the cylinder, and (iii) the exposed ring area at the junction. Use π = 3.14. [5 marks]

First calculate slant height of cone: l = √(2.5² + 6²) ≈ 6.5 cm. CSA of cone = πrl = 3.14 × 2.5 × 6.5. Height of cylinder = 26 − 6 = 20 cm. CSA of cylinder = 2πrh = 2 × 3.14 × 1.5 × 20. Ring area = π(R² − r²) = 3.14(2.5² − 1.5²). Show all three calculations separately and identify which surfaces are painted orange and which yellow based on the problem context.

Assertion-Reason Questions for Surface Areas and Volumes

Q1. Assertion (A):

When a cone is surmounted on a hemisphere of the same radius, the total surface area equals the sum of the curved surface area of the cone and curved surface area of the hemisphere.

Reason (R):

The base of the cone coincides with the base of the hemisphere, so both bases are internal surfaces and not counted in the total surface area.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true. When two solids join, the contact surfaces become internal and are excluded from TSA calculation. The TSA indeed equals CSA of cone + CSA of hemisphere only, making the reason the correct explanation for the assertion.

Q2. Assertion (A):

A decorative block consists of a cube with edge 6 cm and a hemisphere of radius 3 cm placed on its top face. The TSA of the block is 6(6)² + 2π(3)² cm².

Reason (R):

The hemisphere covers the entire top face of the cube completely, so we subtract the base area of the hemisphere from the cube's TSA.

Answer: C — A is true but R is false — The assertion is correct: TSA = TSA of cube - base area of hemisphere + CSA of hemisphere = 216 - 9π + 18π = 216 + 9π = 6(6)² + 2π(3)². However, the reason is incorrectly stated. The hemisphere covers only part of the top face (since radius 3 cm < half of edge 6 cm). The subtraction accounts for the covered portion, but the reasoning about 'covering completely' is false.

Q3. Assertion (A):

In a cylinder with a hemispherical depression at one end, the total interior surface area is given by 2πr(h + r), where h is the height of the cylinder and r is the common radius.

Reason (R):

This formula represents the curved surface area of the cylinder plus the curved surface area of the hemisphere.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — Both assertion and reason are true. 2πr(h + r) = 2πrh + 2πr², which is exactly the sum of CSA of cylinder (2πrh) and CSA of hemisphere (2πr²). However, the reason does not uniquely explain the assertion because the formula would be the same regardless of whether the hemisphere is on top or forming a depression. The reason is correct but not the best explanation of this particular configuration.

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Case-Based Questions (CBQ) for Surface Areas and Volumes

Case Study:

A truck manufacturer designs a water storage tank by combining two basic shapes. The main body is a cylinder with radius 1.5 m and height 2 m. On top of the cylinder, a hemispherical tank (radius 1.5 m) is mounted to increase storage capacity. The entire structure is to be painted for waterproofing. The painter needs to know the total surface area to estimate the amount of paint required. (Use π = 22/7)

Q1. What is the curved surface area of the cylindrical portion in m²?

  • A. 18.86 m²
  • B. 18 m² ✓
  • C. 13.5 m²
  • D. 28.3 m²

Answer: B

Q2. What is the curved surface area of the hemispherical top in m²?

  • A. 14.14 m²
  • B. 10.71 m²
  • C. 13.5 m² ✓
  • D. 9 m²

Answer: C

Q3. The bottom of the cylindrical tank has an area that must be painted. What is this area?

  • A. 7.07 m² ✓
  • B. 7.5 m²
  • C. 10.5 m²
  • D. 12 m²

Answer: A

Q4. What is the total outer surface area that needs to be painted?

  • A. 32.36 m²
  • B. 39.86 m²
  • C. 42 m²
  • D. 36.21 m² ✓

Answer: D

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Previous Year Questions (PYQ) for Surface Areas and Volumes

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 31/4/1 · 1 mark

Two wooden solid hemispheres of same radii r, are joined at a point, as shown in the figure. The total surface area of the object is:

  • 4πr²
  • 6πr²
  • 3πr²
  • 5πr²

CBSE 2026 · Set 31/4/1 · 1 mark

A cone with slant height 10 cm and radius 6 cm is surmounted on a hemisphere of same radius. The height of the toy is:

  • 12 cm
  • 16 cm
  • 14 cm
  • 10 cm

CBSE 2026 · Set 31/4/1 · 1 mark

'Gilli Danda' is a very popular traditional game of India which is played with two wooden sticks – the larger one is called 'Danda' and smaller one 'Gilli'. 'Danda' – It is cylindrical in shape with diameter 4 cm and length 42 cm. Gilli – It is cylindrical in middle with identical conical ends of same radius 1.5 cm and length 2.8 cm. The length of cylindrical part is 7 cm.

Find the volume of wood used in making both the conical parts of Gilli.

CBSE 2026 · Set 31/5/1 · 1 mark

A cylinder of radius r is surmounted on a hemisphere of same radius. If total height of the object is 13 cm, then its inner surface area is

  • 2πr(r + 13)
  • 13πr
  • 2π(13 + r)²
  • 26πr

CBSE 2026 · Set 31/5/1 · 1 mark

The total surface area of a solid cone of radius 7 cm and slant height 25 cm, is

  • 724 cm²
  • 704 cm²
  • 550 cm²
  • 616 cm²

CBSE 2026 · Set 31/5/1 · 1 mark

There are many varieties of mushrooms available in the world. One such mushroom 'Amanita muscaria' has a upper part which is like red cap (hemispherical) and lower part is like white stem (cylindrical). The hemispherical cap's radius = 3 cm and cylindrical stem is 2 cm high with diameter 1.4 cm. Considering mushroom a solid object.

What is the total height of a mushroom?

CBSE 2026 · Set 31/5/2 · 4 marks

There are many varieties of mushrooms available in the world. One such mushroom 'Amanita muscaria' has a upper part which is like red cap (hemispherical) and lower part is like white stem (cylindrical). The hemispherical cap's radius = 3 cm and cylindrical stem is 2 cm high with diameter 1.4 cm.

(i) What is the total height of a mushroom?

CBSE 2026 · Set 31/1164/30 · 1 mark

The number of conical bottles of radius 2 cm and height 1.2 cm that can be filled from a cylindrical bottle of radius 6 cm and height 8 cm, full of liquid, is

  • 60
  • 130
  • 180
  • 18

CBSE 2026 · Set 31/1164/30 · 5 marks

From a solid cylinder whose height is 2.8 cm and radius 2.1 cm, a conical cavity of the same height and same radius is hollowed out. Find the volume and the total surface area of the remaining solid.

CBSE 2026 · Set 31/1167/430 · 1 mark

Volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is:

  • 4 : 3
  • 3 : 4
  • 16 : 9
  • 9 : 16

CBSE 2026 · Set 31/1167/430 · 4 marks

In a coffee shop, coffee is served in two types of cups. One is cylindrical shape with each of diameter 8 cm and height 7 cm and the other hemispherical with each of diameter 14 cm.

What is the outer curved surface area of the cylindrical cup?

CBSE 2026 · Set 31/5/1 · 1 mark

A conical cavity of maximum volume is carved out from a wooden solid hemisphere of radius 10 cm. Curved surface area of the cavity carved out is (use π = 3.14)

  • 314√2 cm²
  • 314 cm²
  • 3140/3 cm²
  • 3140√2 cm²

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Class 10 Maths Formula Sheet

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Class 10 Mathematics Previous Year Questions, Chapter-wise

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