Surface Areas and Volumes takes the formulas for the cuboid, cube, cylinder, cone, sphere and hemisphere from Class IX and applies them to objects made by joining two of these solids: a capsule, a toy top, a tent, a tanker or a block with a hollow scooped out. The chapter has two ideas. For surface area, add only the surfaces still visible after joining, since glued faces disappear. For volume, add the volumes of the parts and subtract anything removed. Board questions are practical word problems where the setup earns as many marks as the arithmetic.
Key Concepts
1. The Basic Solids and Their Formulas
Every object in this chapter is built from the basic solids you studied in Class IX. Before you can handle a toy, a tent or a capsule, you need each of these formulas ready without thinking. Two short forms are used throughout the chapter: TSA means Total Surface Area and CSA means Curved Surface Area.
| Solid | CSA or lateral surface | TSA | Volume |
|---|---|---|---|
| Cuboid, length l, breadth b, height h | 2h(l + b) | 2(lb + bh + hl) | lbh |
| Cube, edge a | 4a2 | 6a2 | a3 |
| Right circular cylinder, radius r, height h | 2πrh | 2πr(h + r) | πr2h |
| Right circular cone, radius r, height h, slant height l | πrl | πr(l + r) | (1/3)πr2h |
| Sphere, radius r | 4πr2 | 4πr2 | (4/3)πr3 |
| Hemisphere, radius r | 2πr2 | 3πr2 | (2/3)πr3 |
Slant height of a cone. The height h, the radius r and the slant height l of a cone form a right triangle with l as the hypotenuse. So l = √(r2 + h2). For r = 5 cm and h = 12 cm, l = √(25 + 144) = √169 = 13 cm. Many questions give h and expect you to find l before the CSA of the cone.
Where the hemisphere formulas come from. A hemisphere is half a sphere. Its curved surface is half of 4πr2, that is 2πr2. A solid hemisphere also has a flat circular face of area πr2, so its TSA is 2πr2 + πr2 = 3πr2. Its volume is half of (4/3)πr3, that is (2/3)πr3.
2. Breaking a Combined Solid into Basic Solids
The NCERT chapter starts with a truck carrying oil or water. Its container is a cylinder with a hemisphere at each end. A test tube is a cylinder with a hemisphere at the bottom. Neither is one of the basic solids, so no single formula fits. The method the chapter uses is to break the new problem into smaller problems you have already solved.
Step 1: Name the parts. Read the question and write down the basic solids, for example “cone surmounted on a hemisphere” or “cylinder with two hemispherical ends”. The word surmounted means placed on top of.
Step 2: Match the shared measurements. Where two parts meet on a flat circular face, they usually share the same radius. In the NCERT toy, the base radius of the cone is taken equal to the radius of the hemisphere so that the toy has a smooth surface.
Step 3: Split the heights and lengths. The total height of an object is the sum of the heights of its parts. The height of a hemisphere is its radius. So:
- A top 5 cm tall, made of a cone on a hemisphere of radius 1.75 cm, has a cone of height 5 − 1.75 = 3.25 cm.
- A capsule 14 mm long with diameter 5 mm has two hemispheres of radius 2.5 mm, so its cylindrical part is 14 − 2 × 2.5 = 9 mm long.
- A vessel of total height 13 cm made of a hemisphere of diameter 14 cm and a cylinder has a cylinder of height 13 − 7 = 6 cm.
Step 4: Decide what is asked. Paint, colour, canvas, cloth, polish or sheet metal means surface area. Water, air, juice, syrup, wood, iron or capacity means volume. Once you know this, choose which surfaces or volumes to add or subtract.
3. Surface Area of Solids Joined Together
When two solids are joined, the faces that are stuck together are hidden. They are no longer on the outside, so they are not part of the surface area of the new solid. The rule is simple: add only the surfaces you can still see.
The tanker (cylinder with two hemispherical ends). After joining, you see only the curved surfaces of the two hemispheres and the curved surface of the cylinder. The flat faces of the hemispheres and the two circular ends of the cylinder are all hidden. So
TSA of new solid = CSA of one hemisphere + CSA of cylinder + CSA of other hemisphere
With common radius r and cylinder length h, this is 2πr2 + 2πrh + 2πr2 = 2πr(h + 2r).
The toy (cone on a hemisphere). The flat base of the cone and the flat face of the hemisphere are pressed together and vanish. The visible surface is
TSA of the toy = CSA of hemisphere + CSA of cone = 2πr2 + πrl
The NCERT text points out that the total surface area of the top is not the sum of the total surface areas of the cone and the hemisphere. Adding the two TSAs would count two hidden circles.
A hemisphere on a cube. This case is different because the cube face is larger than the circle it touches. The part of the cube top covered by the hemisphere is hidden, the rest of the top is still visible, and the curved surface of the hemisphere is added. So
Surface area = TSA of cube − base area of hemisphere + CSA of hemisphere = 6a2 − πr2 + 2πr2 = 6a2 + πr2
For a cube of edge 5 cm with a hemisphere of diameter 4.2 cm (NCERT Example 2), this gives 150 + (22/7) × 2.1 × 2.1 = 150 + 13.86 = 163.86 cm2. The greatest hemisphere that fits on a cube of edge a has diameter a.
A cone resting on a narrower cylinder (the rocket). When the base of the cone is wider than the cylinder below it, a ring of the cone’s base stays visible. Its area is πr2 − πr′2, where r is the cone radius and r′ the cylinder radius. The visible parts of the cone are then its CSA plus this ring.
The tent (cone on a cylinder). Canvas covers the curved wall of the cylinder and the curved top of the cone. The base is the ground, so it has no canvas. Canvas area = 2πrh + πrl = πr(2h + l).
| Object | Visible surfaces | Surface area |
|---|---|---|
| Capsule or tanker | 2 hemisphere CSAs + cylinder CSA | 2πr(h + 2r) |
| Cone on hemisphere (toy, top) | Cone CSA + hemisphere CSA | πrl + 2πr2 |
| Hemisphere on cube | Cube TSA − circle + hemisphere CSA | 6a2 + πr2 |
| Tent (open at the ground) | Cylinder CSA + cone CSA | πr(2h + l) |
| Two cubes of edge a joined | A cuboid 2a by a by a | 10a2 |
4. Surface Area After Scooping or Hollowing
Sometimes a shape is cut out of a solid instead of being stuck on. A hemispherical depression in a block, a conical cavity drilled into a cylinder, and a bird-bath are all examples. Two things happen when you scoop a cavity out of a flat face:
- The flat area of the opening (a circle) is removed from the outer face.
- The inner curved surface of the cavity is now exposed and must be added.
So the surface area of the remaining solid usually increases, even though the volume decreases. This surprises many students.
Hemisphere scooped from a cube face. If the diameter of the hemisphere equals the edge l of the cube, r = l/2. Surface area = 6l2 − πr2 + 2πr2 = 6l2 + πr2 = 6l2 + πl2/4 = (l2/4)(π + 24).
Conical cavity in a cylinder, same height and same diameter. The top circle of the cylinder is removed completely because the cone base covers it. What remains is the curved surface of the cylinder, its bottom circle, and the inside of the cone. Surface area = 2πrh + πr2 + πrl.
Hemisphere scooped from each end of a cylinder. Both circular ends vanish, and two hemispherical hollows appear. Surface area = 2πrh + 2 × 2πr2 = 2πr(h + 2r).
The bird-bath. NCERT Example 4 describes a cylinder with a hemispherical depression at one end. Its total surface area is taken as CSA of cylinder + CSA of hemisphere = 2πrh + 2πr2 = 2πr(h + r). The depression has the same radius as the cylinder, so no flat ring is left at the top: the curved hollow takes the place of the top circle.
Inner surface area. A hollow vessel has an inside and an outside. When a question asks for the inner surface area, use the inner dimensions and count only the inside surfaces. For a vessel shaped as a hollow hemisphere with a hollow cylinder above it, the inner surface is 2πr2 + 2πrh. The top is open and has no surface.
5. Volume of Solids Joined Together
The NCERT text makes a clear contrast. For surface area, you cannot simply add the areas of the parts, because some surface disappears when they are joined. For volume, nothing disappears: the space taken by each part is still inside the new solid. So
Volume of the combined solid = sum of the volumes of its parts
| Object | Volume |
|---|---|
| Cone on hemisphere, common radius r, cone height h | (1/3)πr2h + (2/3)πr3 = (1/3)πr2(h + 2r) |
| Cylinder with two hemispherical ends, cylinder length h | πr2h + (4/3)πr3 |
| Cylinder with a cone at each end, cone height h′ | πr2h + 2 × (1/3)πr2h′ |
| Cuboid with a half cylinder on top (shed) | lbh + (1/2)πr2l |
| Cylinder on a larger cylinder (iron pole) | πR2H + πr2h |
Two hemispheres make a sphere. In a capsule or a gulab jamun, the two hemispherical ends together have volume (4/3)πr3. Writing it this way saves one step.
The half cylinder in a shed. In NCERT Example 5 the roof is half a cylinder lying on its side. Its diameter is the 7 m width of the shed and its length (the “height” of the cylinder) is the 15 m length of the shed. Its volume is (1/2)πr2h with r = 3.5 m and h = 15 m.
Illustration. A solid is a cone standing on a hemisphere. Both radii are 1 cm and the height of the cone equals its radius. Volume = (1/3)π(1)2(1) + (2/3)π(1)3 = π/3 + 2π/3 = π cm3.
6. Volume After Removal, Capacity and Displacement
When part of a solid is removed, or part of a container is already occupied, you subtract. The same idea appears in several forms.
Remaining material. A wooden pen stand is a cuboid with four conical depressions. Wood left = volume of cuboid − 4 × volume of one cone.
Actual capacity. A glass looks like a cylinder but has a raised hemisphere at the bottom. Its apparent capacity is the cylinder volume πr2h. Its actual capacity is less by the volume of the raised hemisphere, (2/3)πr3.
Space left over. The shed in Example 5 holds 1128.75 m3 of air when empty. Machinery and workers take up space, so air left = total volume − volume of machinery − volume of workers.
Water left in a container. A solid is lowered into a cylinder full of water. The water that stays equals the cylinder volume minus the volume of the solid, because the solid pushes out (displaces) exactly its own volume of water.
Counting objects. Lead shots dropped into a cone full of water push out water. If one-fourth of the water flows out, the total volume of the shots equals one-fourth of the cone’s volume. Number of shots = volume that flowed out ÷ volume of one shot.
Difference of volumes. In Example 7 a cylinder circumscribes a toy (the toy fits exactly inside it, touching it). The difference of volumes is the empty space: cylinder volume − toy volume.
Percentages and mass. If syrup fills 30% of a gulab jamun, syrup = 0.3 × volume. If 1 cm3 of iron has a mass of about 8 g, mass = volume in cm3 × 8 g.
7. Value of π, Units and Approximation
The exercises say “Unless stated otherwise, take π = 22/7”. Some questions ask for π = 3.14 or for the answer “in terms of π”. Use exactly what the question asks, because the final number changes with the value of π.
- With π = 22/7, radii like 7, 3.5, 1.4 or 0.7 cancel the 7 cleanly. Keep 22/7 as a fraction until the end.
- With π = 3.14, take π out as a common factor first and multiply once. NCERT Example 3 does this: 3.14 × 20.25 = 63.585 cm2.
- For an answer in terms of π, leave π as a symbol, for example 8π cm3.
Units. Convert every length to one unit before you start. The bird-bath in Example 4 has height 1.45 m and radius 30 cm, so use 145 cm and 30 cm. The useful conversions:
| Quantity | Conversion |
|---|---|
| Length | 1 m = 100 cm, 1 cm = 10 mm |
| Area | 1 m2 = 10,000 cm2 |
| Volume | 1 m3 = 1,000,000 cm3 |
| Capacity | 1 litre = 1000 cm3, 1 m3 = 1000 litres |
| Mass | 1 kg = 1000 g |
Approximation. When a slant height is not a whole number, NCERT rounds it (√13.625 ≈ 3.7 cm in Example 1) and writes “approx.” with the answer. When a question says “to the nearest cm2”, round only the final answer.
Formula and Theorem Sheet
| Situation | Formula |
|---|---|
| Slant height of a cone | l = √(r2 + h2) |
| Cylinder with two hemispherical ends: surface area | 2πrh + 4πr2 = 2πr(h + 2r) |
| Cylinder with two hemispherical ends: volume | πr2h + (4/3)πr3 |
| Cone on hemisphere: surface area | πrl + 2πr2 |
| Cone on hemisphere: volume | (1/3)πr2h + (2/3)πr3 |
| Hemisphere on a cube of edge a: surface area | 6a2 + πr2 |
| Hemisphere scooped from a cube face: surface area | 6a2 + πr2 |
| Tent (cone on cylinder, no base): canvas | 2πrh + πrl |
| Cylinder with hemispherical depression (bird-bath) | 2πrh + 2πr2 = 2πr(h + r) |
| Cylinder with conical cavity of same height and radius | 2πrh + πr2 + πrl |
| Hemisphere scooped from each end of a cylinder | 2πrh + 4πr2 |
| Actual capacity of a glass with raised hemispherical bottom | πr2h − (2/3)πr3 |
| Volume of any combined solid | Sum of the volumes of the parts |
| Number of small objects | Volume they must fill ÷ volume of one object |
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Important Definitions
- Combination of solids: a solid formed by joining two or more basic solids, such as a cylinder with two hemispheres at its ends.
- Curved Surface Area (CSA): the area of only the curved part of a solid, leaving out its flat faces.
- Total Surface Area (TSA): the area of the whole outer surface of a solid, curved and flat parts together.
- Lateral surface area: for a cuboid or cube, the area of the four side faces, leaving out the top and bottom.
- Volume: the amount of space a solid occupies, measured in cubic units such as cm3 or m3.
- Capacity: the amount a hollow container can hold, found from its inner dimensions.
- Apparent capacity: the capacity a container seems to have from its outer shape, before allowing for any part that takes up space inside it.
- Hemisphere: one half of a sphere, cut by a plane through its centre. Its height equals its radius.
- Slant height of a cone: the distance from the vertex of the cone to any point on the circle of its base, l = √(r2 + h2).
- Surmounted: placed on top of. “A cylinder surmounted by a cone” has the cone above the cylinder.
- Depression or cavity: a hollow scooped out of a solid. It removes volume and exposes an inner surface.
- Circumscribe: a solid circumscribes another when the second fits exactly inside it, touching it. A cylinder circumscribing a cone-on-hemisphere toy has the toy’s radius and total height.
Solved Examples (NCERT-Based)
Example 1: The playing top (NCERT Example 1)
Question: Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere. The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area he has to colour. (Take π = 22/7)
Solution: Radius r = 3.5/2 = 1.75 cm. TSA of the toy = CSA of hemisphere + CSA of cone.
CSA of hemisphere = 2πr2 = 2 × (22/7) × 1.75 × 1.75 = 19.25 cm2.
Height of cone = height of top − radius of hemisphere = 5 − 1.75 = 3.25 cm.
Slant height l = √(1.752 + 3.252) = √(3.0625 + 10.5625) = √13.625 ≈ 3.7 cm.
CSA of cone = πrl = (22/7) × 1.75 × 3.7 = 5.5 × 3.7 = 20.35 cm2.
Area to colour = 19.25 + 20.35 = 39.6 cm2 (approx.)
Example 2: The decorative block (NCERT Example 2)
Question: A decorative block is made of a cube of edge 5 cm with a hemisphere of diameter 4.2 cm fixed on top. Find the total surface area of the block. (Take π = 22/7)
Solution: TSA of cube = 6 × 5 × 5 = 150 cm2. The circle where the hemisphere sits is hidden, and the hemisphere’s curved surface is added. r = 2.1 cm.
Surface area = 150 − πr2 + 2πr2 = 150 + πr2 = 150 + (22/7) × 2.1 × 2.1 = 150 + 13.86 = 163.86 cm2
Example 3: The toy rocket (NCERT Example 3)
Question: A wooden toy rocket is a cone mounted on a cylinder. The entire rocket is 26 cm high and the conical part is 6 cm high. The base of the cone has diameter 5 cm and the base of the cylinder has diameter 3 cm. The cone is painted orange and the cylinder yellow. Find the area painted with each colour. (Take π = 3.14)
Solution: Cone: r = 2.5 cm, h = 6 cm, l = √(2.52 + 62) = √(6.25 + 36) = √42.25 = 6.5 cm. Cylinder: r′ = 1.5 cm, h′ = 26 − 6 = 20 cm.
The cone’s base is wider than the cylinder, so a ring of the cone’s base shows. Orange area = CSA of cone + base of cone − base of cylinder = πrl + πr2 − πr′2 = π[(2.5 × 6.5) + 2.52 − 1.52] = π[16.25 + 6.25 − 2.25] = 3.14 × 20.25 = 63.585 cm2.
Yellow area = CSA of cylinder + one base of the cylinder = 2πr′h′ + πr′2 = πr′(2h′ + r′) = 3.14 × 1.5 × (40 + 1.5) = 4.71 × 41.5 = 195.465 cm2.
Example 4: The bird-bath (NCERT Example 4)
Question: Mayank made a bird-bath in the shape of a cylinder with a hemispherical depression at one end. The height of the cylinder is 1.45 m and its radius is 30 cm. Find the total surface area of the bird-bath. (Take π = 22/7)
Solution: h = 145 cm, r = 30 cm. TSA = CSA of cylinder + CSA of hemisphere = 2πrh + 2πr2 = 2πr(h + r).
= 2 × (22/7) × 30 × (145 + 30) = 2 × (22/7) × 30 × 175 = 44 × 30 × 25 = 33000 cm2 = 3.3 m2
Example 5: The hollow vessel (Exercise 12.1, Q2)
Question: A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution: r = 7 cm. Height of the cylinder = 13 − 7 = 6 cm. Inner surface = CSA of hemisphere + CSA of cylinder = 2πr2 + 2πrh = 2πr(r + h).
= 2 × (22/7) × 7 × (7 + 6) = 44 × 13 = 572 cm2
Example 6: Cone on a hemisphere (Exercise 12.1, Q3)
Question: A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution: Height of cone = 15.5 − 3.5 = 12 cm. l = √(3.52 + 122) = √(12.25 + 144) = √156.25 = 12.5 cm.
TSA = πrl + 2πr2 = πr(l + 2r) = (22/7) × 3.5 × (12.5 + 7) = 11 × 19.5 = 214.5 cm2
Example 7: Hemispherical depression in a cube (Exercise 12.1, Q5)
Question: A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Solution: Edge = l, radius of hemisphere = l/2.
Surface area = TSA of cube − area of circular opening + CSA of hemisphere = 6l2 − π(l/2)2 + 2π(l/2)2 = 6l2 + πl2/4.
∴ Surface area = (l2/4)(π + 24) square units
Example 8: Air in the shed (NCERT Example 5)
Question: Shanta’s shed is a cuboid surmounted by a half cylinder. The base of the shed is 7 m × 15 m and the cuboidal part is 8 m high. Find the volume of air the shed can hold. The machinery occupies 300 m3 and there are 20 workers, each occupying about 0.08 m3. How much air is then in the shed? (Take π = 22/7)
Solution: Cuboid: 15 × 7 × 8 = 840 m3. Half cylinder: diameter 7 m (r = 3.5 m), length 15 m. Volume = (1/2) × (22/7) × 3.5 × 3.5 × 15 = (1/2) × 38.5 × 15 = 288.75 m3.
Volume of air in the empty shed = 840 + 288.75 = 1128.75 m3.
Space taken = 300 + 20 × 0.08 = 300 + 1.6 = 301.6 m3. Air left = 1128.75 − 301.6 = 827.15 m3.
Example 9: The juice glass (NCERT Example 6)
Question: A cylindrical glass has inner diameter 5 cm and height 10 cm, but its bottom has a raised hemispherical portion. Find the apparent capacity and the actual capacity. (Use π = 3.14)
Solution: r = 2.5 cm. Apparent capacity = πr2h = 3.14 × 2.5 × 2.5 × 10 = 196.25 cm3.
Volume of hemisphere = (2/3)πr3 = (2/3) × 3.14 × 2.5 × 2.5 × 2.5 = 32.71 cm3 (approx.).
Actual capacity = 196.25 − 32.71 = 163.54 cm3.
Example 10: Toy inside a cylinder (NCERT Example 7)
Question: A solid toy is a hemisphere surmounted by a right circular cone. The cone is 2 cm high and the diameter of the base is 4 cm. Find the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. (Take π = 3.14)
Solution: r = 2 cm, cone height h = 2 cm.
Volume of toy = (2/3)πr3 + (1/3)πr2h = (2/3) × 3.14 × 8 + (1/3) × 3.14 × 4 × 2 = 3.14 × (16/3 + 8/3) = 3.14 × 8 = 25.12 cm3.
The cylinder has radius 2 cm and height = cone height + hemisphere radius = 2 + 2 = 4 cm. Its volume = 3.14 × 22 × 4 = 50.24 cm3.
Difference = 50.24 − 25.12 = 25.12 cm3.
Example 11: Rachel’s model (Exercise 12.2, Q2)
Question: Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Solution: r = 1.5 cm. Cylinder length = 12 − 2 − 2 = 8 cm.
Volume = πr2(8) + 2 × (1/3)πr2(2) = πr2(8 + 4/3) = π × 2.25 × 28/3 = 21π.
= 21 × 22/7 = 66 cm3
Example 12: Checking the glass vessel (Exercise 12.2, Q8)
Question: A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
Solution: Sphere: r = 4.25 cm. Volume = (4/3) × 3.14 × 4.253 = (4/3) × 3.14 × 76.765625 ≈ 321.39 cm3.
Neck: r = 1 cm, h = 8 cm. Volume = 3.14 × 1 × 8 = 25.12 cm3.
Total = 321.39 + 25.12 = 346.51 cm3. She is not correct: the vessel holds about 346.51 cm3, not 345 cm3.
Competency-Based Questions (with answers)
For each assertion-reason item, choose from these four options:
- Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
- Both A and R are true, but R is not the correct explanation of A.
- A is true, but R is false.
- A is false, but R is true.
1. Case-based: The ice-cream cone
An ice-cream seller fills a wafer cone of radius 3 cm and height 4 cm with ice cream and adds a hemisphere of ice cream of the same radius on top. Take π = 3.14.
(a) Find the slant height of the cone. l = √(9 + 16) = 5 cm.
(b) Find the area of wafer used for the cone. The wafer is the curved surface: πrl = 3.14 × 3 × 5 = 47.1 cm2.
(c) Find the total volume of ice cream. Cone: (1/3)π(9)(4) = 12π. Hemisphere: (2/3)π(27) = 18π. Total = 30π = 94.2 cm3.
2. Case-based: The water tanker
A tanker’s container is a cylinder 6 m long with a hemisphere of radius 1 m at each end. Take π = 3.14.
(a) Find how much water it can hold. Cylinder: π(1)2(6) = 6π. Two hemispheres make one sphere: (4/3)π(1)3. Total = 6π + 4π/3 = 22π/3 = 22 × 3.14/3 ≈ 23.03 m3, which is about 23,027 litres.
(b) Find the outer surface area to be painted. 2πr(h + 2r) = 2π(1)(6 + 2) = 16π = 50.24 m2.
(c) Find the cost of painting at ₹ 40 per m2. 50.24 × 40 = ₹ 2009.60.
3. Case-based: The circus tent
A circus tent is a cylinder of radius 7 m and height 3 m, surmounted by a cone of the same radius and height 24 m. Take π = 22/7.
(a) Find the slant height of the conical top. l = √(49 + 576) = √625 = 25 m.
(b) Find the canvas needed (no canvas on the ground). πr(2h + l) = (22/7) × 7 × (6 + 25) = 22 × 31 = 682 m2. At ₹ 120 per m2, the cost is ₹ 81,840.
(c) Find the volume of air inside. πr2h + (1/3)πr2H = 154 × 3 + (1/3) × 154 × 24 = 462 + 1232 = 1694 m3.
4. Case-based: Marbles in a beaker
A cylindrical beaker of radius 4 cm has water in it. Glass balls of radius 1 cm are dropped in until the water level rises by 3 cm. No water spills.
(a) Find the volume of water pushed up. π(4)2(3) = 48π cm3.
(b) Find the volume of one ball. (4/3)π(1)3 = 4π/3 cm3.
(c) Find the number of balls. 48π ÷ (4π/3) = 48 × 3/4 = 36 balls.
5. Assertion-Reason
Assertion (A): The total surface area of a toy made by joining a cone and a hemisphere of equal radius is less than the sum of the total surface areas of the cone and the hemisphere.
Reason (R): When the two solids are joined, the base of the cone and the flat face of the hemisphere are hidden.
Answer: (a). Both statements are true. The toy’s surface is πrl + 2πr2, which is 2πr2 less than (πrl + πr2) + 3πr2, because the two hidden circles are left out. R explains A.
6. Assertion-Reason
Assertion (A): The volume of a solid formed by joining a cylinder and a cone is the sum of the volumes of the cylinder and the cone.
Reason (R): When two solids are joined, part of their surface area disappears.
Answer: (b). Both are true. A is true because no space is lost when solids are joined. R is a fact about surface area and says nothing about volume, so it does not explain A.
7. Assertion-Reason
Assertion (A): When a hemisphere whose diameter equals the edge of a cube is scooped out of one face, the surface area of the remaining solid is less than that of the cube.
Reason (R): Scooping removes a circle of area πr2 from the face and exposes a curved surface of area 2πr2.
Answer: (d). R is true. By R, the surface changes by −πr2 + 2πr2 = +πr2, so the surface area increases. A is false.
8. Assertion-Reason
Assertion (A): The slant height of a cone of radius 5 cm and height 12 cm is 13 cm.
Reason (R): The slant height of a cone is given by l = r + h.
Answer: (c). A is true: l = √(25 + 144) = 13 cm. R is false: the correct relation is l = √(r2 + h2), and r + h would give 17 cm.
9. Assertion-Reason
Assertion (A): The total surface area of a solid hemisphere of radius r is 3πr2.
Reason (R): The curved surface area of a hemisphere is 2πr2 and its flat face is a circle of area πr2.
Answer: (a). Both are true, and adding the two parts in R gives exactly A.
Important Questions for Board Exams
1-Mark Questions
Q1. Find the total surface area of a solid hemisphere of radius 7 cm. (Take π = 22/7)
Answer: 3πr2 = 3 × (22/7) × 49 = 462 cm2.
Q2. A cone and a cylinder have the same base radius and the same height. Find the ratio of their volumes.
Answer: (1/3)πr2h : πr2h = 1 : 3.
Q3. Two cubes, each of edge 5 cm, are joined end to end. Find the surface area of the resulting cuboid.
Answer: The cuboid is 10 cm × 5 cm × 5 cm. Surface area = 2(50 + 25 + 50) = 250 cm2.
2-Mark Questions
Q4. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid. (Exercise 12.1, Q1)
Answer: Edge = cube root of 64 = 4 cm, since 4 × 4 × 4 = 64. Cuboid: 8 cm × 4 cm × 4 cm. Surface area = 2(8 × 4 + 4 × 4 + 4 × 8) = 2(32 + 16 + 32) = 160 cm2.
Q5. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid. (Exercise 12.1, Q4)
Answer: The hemisphere must sit on a 7 cm face, so the greatest diameter is 7 cm (r = 3.5 cm). Surface area = 6a2 + πr2 = 294 + (22/7) × 12.25 = 294 + 38.5 = 332.5 cm2.
Q6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area. (Exercise 12.1, Q6)
Answer: r = 2.5 mm, cylinder length = 14 − 5 = 9 mm. Surface area = 2πr(h + 2r) = 2 × (22/7) × 2.5 × 14 = 220 mm2.
Q7. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π. (Exercise 12.2, Q1)
Answer: (1/3)π(1)2(1) + (2/3)π(1)3 = π/3 + 2π/3 = π cm3.
3-Mark Questions
Q8. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹ 500 per m2. (Note that the base of the tent will not be covered with canvas.) (Exercise 12.1, Q7)
Answer: r = 2 m. Canvas = 2πrh + πrl = πr(2h + l) = (22/7) × 2 × (4.2 + 2.8) = (22/7) × 2 × 7 = 44 m2. Cost = 44 × 500 = ₹ 22,000.
Q9. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2. (Exercise 12.1, Q8)
Answer: r = 0.7 cm, h = 2.4 cm, l = √(0.49 + 5.76) = √6.25 = 2.5 cm. Surface area = 2πrh + πr2 + πrl = πr(2h + r + l) = (22/7) × 0.7 × (4.8 + 0.7 + 2.5) = 2.2 × 8 = 17.6 cm2 ≈ 18 cm2.
Q10. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article. (Exercise 12.1, Q9)
Answer: Surface area = 2πrh + 2 × 2πr2 = 2πr(h + 2r) = 2 × (22/7) × 3.5 × (10 + 7) = 22 × 17 = 374 cm2.
Q11. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand. (Exercise 12.2, Q4)
Answer: Cuboid = 15 × 10 × 3.5 = 525 cm3. One cone = (1/3) × (22/7) × 0.25 × 1.4 = 0.3667 cm3 (approx.). Four cones = 1.4667 cm3. Wood = 525 − 1.47 = 523.53 cm3 (approx.).
5-Mark Questions
Q12. A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm. (Exercise 12.2, Q3)
Answer: r = 1.4 cm, cylinder length = 5 − 2.8 = 2.2 cm. Cylinder = (22/7) × 1.96 × 2.2 = 13.552 cm3. Two hemispheres = (4/3) × (22/7) × 2.744 ≈ 11.499 cm3. One gulab jamun ≈ 25.051 cm3. 45 of them ≈ 1127.28 cm3. Syrup = 30% of 1127.28 ≈ 338 cm3 (approx.).
Q13. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel. (Exercise 12.2, Q5)
Answer: Volume of water = (1/3)π(25)(8) = 200π/3 cm3. Water out = (1/4)(200π/3) = 50π/3 cm3, which equals the volume of the shots. One shot = (4/3)π(0.5)3 = π/6 cm3. Number = (50π/3) ÷ (π/6) = 100 lead shots.
Q14. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8g mass. (Use π = 3.14) (Exercise 12.2, Q6)
Answer: Lower cylinder = 3.14 × 144 × 220 = 99475.2 cm3. Upper cylinder = 3.14 × 64 × 60 = 12057.6 cm3. Total = 111532.8 cm3. Mass = 111532.8 × 8 = 892262.4 g = 892.26 kg (approx.).
Q15. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm. (Exercise 12.2, Q7)
Answer: Cylinder = π(3600)(180) = 648000π cm3. Cone = (1/3)π(3600)(120) = 144000π. Hemisphere = (2/3)π(216000) = 144000π. Solid = 288000π. Water left = 360000π = 360000 × 22/7 ≈ 1131428.57 cm3 ≈ 1.131 m3.
Common Mistakes and Examiner Tips
- Adding the TSAs of the parts. For a cone on a hemisphere, 3πr2 + πr(l + r) counts two hidden circles. Fix: list the visible surfaces first and add only those.
- Using the full height for the cone. In a top 5 cm tall, the cone is 5 − r tall. Fix: subtract the hemisphere’s radius (or other part heights) from the total height before anything else.
- Using h in place of l. The CSA of a cone is πrl, with the slant height. Fix: if the question gives h, find l = √(r2 + h2) first.
- Taking diameter as radius. Questions usually give diameters (4.2 cm, 3.5 cm, 14 mm). Fix: halve every diameter and write r = … on its own line.
- Forgetting the exposed circle on a cube. With a hemisphere on a cube, the rest of the cube top is still visible. Fix: 6a2 − πr2 + 2πr2, not 5a2 + 2πr2.
- Thinking a cavity lowers surface area. A scooped hemisphere adds 2πr2 and removes only πr2. Fix: remove the flat opening, add the inner curved surface.
- Covering the base of a tent. A tent stands on the ground. Fix: canvas = CSA of cylinder + CSA of cone only, as the NCERT question notes.
- Mixing units. The bird-bath has 1.45 m and 30 cm. Fix: convert to one unit before substituting, then convert the final answer if asked (33000 cm2 = 3.3 m2).
- Wrong conversion of area or volume units. 1 m2 = 10,000 cm2 and 1 m3 = 1,000,000 cm3, not 100. Fix: square or cube the length factor.
- Using the wrong value of π. Fix: read the instruction; exercises default to 22/7, but NCERT Examples 3, 6 and 7 and some exercise questions use 3.14.
- Rounding too early. Rounding the volume of each gulab jamun to 25 cm3 before multiplying by 45 drifts the answer. Fix: keep two or three decimals until the last step.
- Missing units or the final sentence. Fix: write cm2 for area and cm3 for volume, and answer the question asked (“She is not correct”, “100 lead shots”).
Quick Revision Points
- Every object in this chapter is a combination of two basic solids: cuboid, cube, cylinder, cone, sphere or hemisphere.
- TSA means Total Surface Area; CSA means Curved Surface Area.
- Slant height of a cone: l = √(r2 + h2).
- Hemisphere: CSA 2πr2, TSA 3πr2, volume (2/3)πr3.
- The height of a hemisphere equals its radius.
- Surface area of a combined solid: add only the surfaces that remain visible.
- The TSA of a combined solid is not the sum of the TSAs of its parts.
- Tanker or capsule: 2πr(h + 2r). Cone on hemisphere: πrl + 2πr2.
- Hemisphere on a cube: 6a2 + πr2. The greatest hemisphere on a cube of edge a has diameter a.
- A wider cone on a narrower cylinder leaves a visible ring of area πr2 − πr′2.
- Tent canvas: πr(2h + l); no canvas on the base.
- A cavity removes a flat opening and adds an inner curved surface.
- Volume of a combined solid = sum of the volumes of the parts.
- Removed or occupied parts are subtracted: remaining wood, actual capacity, air left, water left.
- Actual capacity of a glass with a raised hemisphere = πr2h − (2/3)πr3.
- A submerged solid pushes out its own volume of water.
- Number of objects = volume to be filled ÷ volume of one object.
- A cylinder circumscribing a cone-on-hemisphere toy has height equal to the toy’s total height.
- Two hemispherical ends together make one sphere: (4/3)πr3.
- Default π = 22/7 unless the question says 3.14 or “in terms of π”.
- 1 litre = 1000 cm3; 1 m3 = 1000 litres.
- Paint, canvas or sheet means area; water, air or wood means volume.
Weightage in Board Exams
Surface Areas and Volumes is Chapter 12 of Class 10 Maths. Check the current CBSE course structure for the marks it carries. The chapter is assessed in a predictable way.
| Question type | What is usually asked |
|---|---|
| MCQ and Assertion-Reason | Hemisphere CSA and TSA, slant height, ratio of cone and cylinder volumes, surface area of two joined cubes, why TSAs cannot be added |
| Short answers | Capsule or toy surface area, hemisphere on a cube, cone-on-hemisphere volume in terms of π |
| Long answers | Tent canvas and cost, cavity problems, water left in a cylinder, counting lead shots, mass of a pole |
| Case-based | An everyday object (tanker, ice-cream cone, tent, glass) split into parts: find a dimension, a surface area and a volume |
The marks go to the setup as much as the arithmetic. Write the parts, the shared radius, the split heights and the formula before substituting numbers. Practise every NCERT example and exercise question first.
Class 10 Maths Β· Chapter 12 β swipe through all 10 cards to understand the whole chapter.
The Basic Solids
Every object in this chapter is built from a cuboid, cube, cylinder, cone, sphere or hemisphere.
TSA means Total Surface Area; CSA means Curved Surface Area.
- Cylinder: CSA 2Οrh, TSA 2Οr(h + r).
- Cone: CSA Οrl, TSA Οr(l + r).
- Sphere: surface area 4Οr2.
Hemisphere Formulas
A hemisphere has a curved surface and one flat circular face; its height equals its radius.
Two hemispherical ends together make one full sphere.
- Radius 7 cm: CSA = 308 cm2, TSA = 462 cm2.
- Subtract r from a total height to get the other part’s height.
- In a capsule, the two ends give (4/3)Οr3 of volume.
Slant Height of a Cone
The radius, height and slant height of a cone form a right triangle.
The CSA of a cone uses l, never h.
- r = 5, h = 12 gives l = 13.
- r = 2.5, h = 6 gives l = 6.5 (toy rocket).
- r = 0.7, h = 2.4 gives l = 2.5.
Surface Area After Joining
Add only the surfaces that stay visible; glued faces disappear.
The TSA of a combined solid is not the sum of the TSAs of its parts.
- Capsule or tanker: 2Οr(h + 2r).
- Tent: Οr(2h + l), no canvas on the ground.
- Top of 5 cm, diameter 3.5 cm: about 39.6 cm2.
Hemisphere on a Cube
The hemisphere hides a circle on the cube top but adds its curved surface.
The greatest hemisphere on a cube of edge a has diameter a.
- Edge 5 cm, diameter 4.2 cm: 163.86 cm2.
- Edge 7 cm, greatest hemisphere: 332.5 cm2.
- Two cubes of edge 4 cm joined: 160 cm2.
Surface Area After Scooping
A cavity removes the flat opening and exposes an inner curved surface.
Scooping a hemisphere out raises surface area by Οr2.
- Bird-bath: 2Οr(h + r) = 3.3 m2 for h = 1.45 m, r = 30 cm.
- Conical cavity in cylinder: 2Οrh + Οr2 + Οrl.
- Hemisphere from each end: 2Οr(h + 2r).
Wider Cone on a Narrow Cylinder
When the cone’s base is wider than the cylinder, a ring of the base stays visible.
Toy rocket: orange 63.585 cm2, yellow 195.465 cm2 with Ο = 3.14.
- Orange part = Οrl + Οr2 β Οrβ²2.
- Yellow part = 2Οrβ²hβ² + Οrβ²2.
- Cylinder height = 26 β 6 = 20 cm.
Volume of Joined Solids
No space is lost when solids are joined, so their volumes add.
Cone on hemisphere: (1/3)Οr2h + (2/3)Οr3.
- Radii 1 cm, cone height 1 cm: V = Ο cm3.
- Rachel’s model (cylinder + 2 cones): 66 cm3.
- Shed = cuboid + half cylinder = 1128.75 m3.
Capacity and Removal
Subtract anything scooped out, raised inside or already occupying the space.
A submerged solid pushes out exactly its own volume of water.
- Juice glass: 196.25 β 32.71 = 163.54 cm3.
- Toy in cylinder: 50.24 β 25.12 = 25.12 cm3.
- Lead shots: 1/4 of the cone’s water Γ· one shot = 100.
Units, Ο and Rounding
Convert every length to one unit and use the value of Ο the question asks for.
Default Ο = 22/7 unless the question says 3.14 or in terms of Ο.
- 1.45 m becomes 145 cm before mixing with 30 cm.
- 1 m2 = 10,000 cm2, so 33000 cm2 = 3.3 m2.
- Round only the final answer.
π Practice Surface Areas and Volumes - 10 board questions
CBSE previous-year and competency-based Β· with answers & explanations
Start βClose β
Why A: Step 1 (radius is half the diameter): r = 2d/2 = d. Step 2 (TSA of a solid hemisphere = 2ΟrΒ² + ΟrΒ² = 3ΟrΒ²): TSA = 3ΟdΒ².
Why not B: 2ΟdΒ² is only the curved surface 2ΟrΒ², with the flat circular base left out.
Why not C: (1/2)ΟdΒ² treats d as the diameter, then also drops the base.
Why not D: (3/4)ΟdΒ² is the answer when the diameter is d, not 2d; here the radius is d itself.
Remember: Diameter 2d means radius d, so TSA = 3ΟdΒ².
Why A: Step 1 (largest cone in a hemisphere): its base is the flat face, so r = 10 cm, and its apex touches the top, so h = 10 cm. Step 2 (slant height, l = β(rΒ² + hΒ²)): l = β(100 + 100) = 10β2 cm. Step 3 (cone CSA = Οrl): 3.14 Γ 10 Γ 10β2 = 314β2 cmΒ².
Why not B: 314 cmΒ² is ΟrΒ², the area of the circular base, not the curved surface.
Why not C: 3140/3 is (1/3)ΟrΒ²h, the volume of the cone, not an area.
Why not D: 3140β2 is ten times too big, a place-value slip in 3.14 Γ 100.
Remember: The biggest cone in a hemisphere has r = h = radius, so l = rβ2.
Why A: Step 1 (the balls are stacked and just fit): the jar’s radius is r and its height is 3 Γ 2r = 6r. Step 2 (cylinder volume = ΟrΒ²h): ΟrΒ² Γ 6r = 6ΟrΒ³. Step 3 (three spheres, each (4/3)ΟrΒ³): 3 Γ (4/3)ΟrΒ³ = 4ΟrΒ³. Step 4 (air = jar minus balls): 6ΟrΒ³ – 4ΟrΒ³ = 2ΟrΒ³.
Why not B: 3ΟrΒ³ uses a jar height of 3r, the sum of the radii instead of the diameters.
Why not C: 5ΟrΒ³ would leave only ΟrΒ³ for the balls, less than even one ball (4/3)ΟrΒ³, so the balls are badly undercounted.
Why not D: 4ΟrΒ³ is the volume of the three balls, not the air around them.
Remember: Air = cylinder (height 6r for three stacked balls) minus balls.
Why C: Step 1 (CSA of a hemisphere = 2ΟrΒ², the tent has no floor): 2 Γ (22/7) Γ 1.4 Γ 1.4 = 2 Γ 22 Γ 0.2 Γ 1.4 = 12.32 mΒ². Step 2 (the door is not canvas, so subtract it): 12.32 – 0.50 = 11.82 mΒ².
Why not A: 11.78 mΒ² is a subtraction slip; 12.32 – 0.50 is 11.82.
Why not B: 12.32 mΒ² is the full curved surface, with the door opening not removed.
Why not D: 12.86 mΒ² does not come from 2ΟrΒ² with or without the door; it overshoots the full curved surface itself.
Remember: A tent has no floor, so use 2ΟrΒ² and take away any opening.
Why B: Step 1 (sphere volume = (4/3)ΟRΒ³): one scoop = (4/3)Ο(r/2)Β³ = (1/6)ΟrΒ³, so two scoops = (1/3)ΟrΒ³. Step 2 (cone volume = (1/3)ΟrΒ²h, set equal): (1/3)ΟrΒ²h = (1/3)ΟrΒ³, so h = r. Step 3 (form the ratio): h : 2r = r : 2r = 1 : 2.
Why not A: 1 : 8 is (r/2)Β³ : rΒ³, a ratio of volumes, not the ratio of lengths h : 2r.
Why not C: 1 : 1 is h : r, not the h : 2r asked for.
Why not D: 2 : 1 writes the ratio upside down, as 2r : h.
Remember: Filled completely means volume of cone = total volume of the scoops.
Why C: Two hemispheres of equal radius placed base to base close up into a full sphere, so A is true. A sphere’s surface area is 4ΟrΒ²; 3ΟrΒ² is the TSA of a solid hemisphere, so R is false.
Why not A: A needs R to be true, but a sphere’s surface area is 4ΟrΒ², not 3ΟrΒ².
Why not B: B also needs R to be true, which it is not.
Why not D: D says A is false, but two equal hemispheres joined along their bases do form a sphere.
Remember: Sphere 4ΟrΒ², solid hemisphere 3ΟrΒ², and two hemispheres make one sphere.
Why B: Step 1 (cube volume = aΒ³): aβΒ³ : aβΒ³ = 8 : 125, so aβ : aβ = 2 : 5 (cube roots). Step 2 (cube surface area = 6aΒ²): 6aβΒ² : 6aβΒ² = 2Β² : 5Β² = 4 : 25.
Why not A: 8 : 125 repeats the volume ratio, as if area scaled like volume.
Why not C: 2 : 5 is the ratio of the edges, one step short of squaring.
Why not D: 16 : 25 squares 4 : 5, but the cube root of 8 is 2, not 4.
Remember: Edges in ratio k give areas kΒ² and volumes kΒ³.
Why D: Step 1 (the new cuboid is 20 cm Γ 10 cm Γ 10 cm, TSA = 2(lb + bh + hl)): 2(200 + 100 + 200) = 1000 cmΒ², so A (1200 cmΒ²) is false. Step 2 (area of a square face = sideΒ²): 10 Γ 10 = 100 cmΒ², so R is true.
Why not A: A is false, so R cannot explain it; 1200 cmΒ² counts all 12 faces of the two cubes.
Why not B: B also needs A to be true, but the two faces that touch are hidden and must be removed.
Why not C: C says R is false, but each face really is 10 Γ 10 = 100 cmΒ².
Remember: Joining two cubes hides two faces: 12 – 2 = 10 faces of 100 cmΒ² each.
Why C: A solid hemisphere has a curved surface 2ΟrΒ² and a flat circular base ΟrΒ², so TSA = 3ΟrΒ². Dividing by rΒ² gives 3Ο, so the ratio is 3Ο : 1.
Why not A: 2Ο : 1 uses only the curved surface 2ΟrΒ² and leaves out the flat base.
Why not B: 4Ο : 1 uses 4ΟrΒ², the surface area of a full sphere.
Why not D: 1 : 4Ο is the sphere ratio written the wrong way round.
Remember: Sphere 4ΟrΒ², hemisphere curved 2ΟrΒ², solid hemisphere 3ΟrΒ².
Why C: Step 1 (distance = speed Γ time): 2 km/h = 2000 m in 60 minutes, so in 2 minutes the water moves 2000 Γ 2/60 = 200/3 m. Step 2 (cuboid volume = length Γ breadth Γ height): 200/3 Γ 40 Γ 3 = 8000 mΒ³.
Why not A: 800 mΒ³ is a factor of 10 too small, a slip in converting km to m or hours to minutes.
Why not B: 4000 mΒ³ is the flow in 1 minute, half the required time.
Why not D: 2000 mΒ³ does not match 3 Γ 40 Γ 200/3; it drops a factor of 4.
Remember: Flowing water in time t forms a cuboid of length speed Γ t over the cross-section.
Chapter Navigation
Previous: Areas Related to Circles Class 10 Notes
Next: Statistics Class 10 Notes
Related Chapters in Class 10 Maths
- Real Numbers Class 10 Notes
- Polynomials Class 10 Notes
- Pair of Linear Equations in Two Variables Class 10 Notes
- Quadratic Equations Class 10 Notes
- Arithmetic Progressions Class 10 Notes
- Triangles Class 10 Notes
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Frequently Asked Questions
When two solids are joined, the faces that touch are hidden inside the new solid. For a cone on a hemisphere, the base of the cone and the flat face of the hemisphere both disappear. Adding the two TSAs would count these hidden circles. So the surface area of the toy is only CSA of cone + CSA of hemisphere, which is Οrl + 2ΟrΒ².
Yes. Joining two solids hides some surface, but no space is lost. The volume of a solid formed by joining two basic solids is the sum of their volumes. For a cone of height h standing on a hemisphere of the same radius r, the volume is (1/3)ΟrΒ²h + (2/3)ΟrΒ³. When a part is scooped out, subtract its volume instead.
Start with the TSA of the cube, 6aΒ². The hemisphere covers a circle of area ΟrΒ² on the top face, so subtract it. Then add the curved surface of the hemisphere, 2ΟrΒ². The result is 6aΒ² + ΟrΒ². For a cube of edge 5 cm and a hemisphere of diameter 4.2 cm, this is 150 + 13.86 = 163.86 cmΒ².
It increases the surface area. The circular opening of area ΟrΒ² is removed from the flat face, but the inside of the hollow, a curved surface of area 2ΟrΒ², is now exposed. The net change is +ΟrΒ². The volume goes down, but the surface area goes up. For a cube of edge l with diameter l, the answer is (lΒ²/4)(Ο + 24).
The height of a hemisphere equals its radius. Subtract the radius from the total height of the toy. In NCERT Example 1, the top is 5 cm tall with radius 1.75 cm, so the cone is 5 β 1.75 = 3.25 cm tall. Then find the slant height with l = β(rΒ² + hΒ²) before using Οrl.
Apparent capacity is what a container seems to hold from its outer shape. Actual capacity allows for any part that takes up space inside. In NCERT Example 6, a glass of radius 2.5 cm and height 10 cm looks like it holds 196.25 cmΒ³, but a raised hemisphere at the bottom takes 32.71 cmΒ³, so it actually holds 163.54 cmΒ³.
The NCERT exercises say to take Ο = 22/7 unless stated otherwise. Some examples and questions ask for Ο = 3.14, and some ask for the answer in terms of Ο. Always follow the instruction in the question, because the final number changes. With 22/7, keep the fraction until the end so that radii like 7, 3.5 or 1.4 cancel cleanly.
