Surface area and volume: Questions and Answers

73 surface area and volume questions solved step by step. Open a question for the full working, the marks for each step and exam practice.

• What is the total surface area of a cuboid? You may refer to the net of the cuboid given in the figure.Answer: Total surface area (TSA) of a cuboid = 2(wl + hl + hw), the total area of its six rectangular faces.
• Compare the formula for the volume of a cuboid with the formula for the area of a rectangle (length × width).Answer: Area of a rectangle = l × w (two lengths, square units). Volume of a cuboid = l × w × h = (area of the rectangle) × height (three lengths, cubic units). The cuboid’s formula is the rectangle’s formula with one more length multiplied in.
• Try to work out for yourself why this model explains the formula for the volume of a cuboid, i.e.,
  volume = area of base × height.
  Write the formula for the surface area and the volume of a cube.Answer: Each thin layer has area l × w; piling layers up to height h fills l × w × h unit cubes, so volume = area of base × height. For a cube of side a: surface area = 6a², volume = a³.
• Observe how these two formulas are special cases of the formulas 2(wl + hl + hw) and whl, when w = h = l.Answer: Put w = h = l = a: 2(a² + a² + a²) = 6a² and a × a × a = a³. These are the cube’s surface area and volume.
• Two solid objects are made from the same material: Cube A has side 6 cm and Cuboid B has dimensions 9 cm × 6 cm × 4 cm. Compare their volumes and surface areas and determine which object has a greater surface area. How is this useful in a real-life situation?Answer: Both have volume 216 cm³. Cube A has surface area 216 cm² and Cuboid B has 228 cm², so the cuboid has the greater surface area. For the same capacity a cube needs less covering material; a cuboid exposes more area to its surroundings.
• The volume of a cube is 64 cm³. What is its total surface area?Answer: Side = ∛64 = 4 cm, so total surface area = 6 × 4² = 96 cm².
• How many small cubes with side 20 cm can be packed tight in a cubical box with side 2 m?Answer: 2 m = 200 cm, so 200 ÷ 20 = 10 cubes fit along each edge: 10 × 10 × 10 = 1000 cubes.
• The dimensions of a godown are 40 m × 25 m × 10 m. If it is filled with cuboidal boxes, each of dimensions 2 m × 1.25 m × 1 m, then find the number of boxes.Answer: Godown 10 000 m³ ÷ box 2.5 m³ = 4000 boxes (20 × 20 × 10 fit exactly).
• Two cubes each of volume 125 cm³ are joined end to end. Find the surface area of the resulting cuboid.Answer: Each cube has side 5 cm, so the cuboid is 10 cm × 5 cm × 5 cm. Surface area = 2(50 + 25 + 50) = 250 cm².
• A cube of side 4 cm is cut into cubes of side 1 cm. What is the ratio of the surface areas of the original cube and all the cut-out cubes? (Note that there is no change in volume but a big change in the surface area. This property has major consequences in the biological world.)Answer: Original: 96 cm². Cut-out: 64 cubes × 6 cm² = 384 cm². Ratio = 96 : 384 = 1 : 4.
• The surface areas of the three faces of a cuboid that meet at one of the corners of the cuboid are 6 cm², 15 cm², and 10 cm² respectively. What is the volume of the cuboid?Answer: (lw)(wh)(hl) = (lwh)² = 6 × 15 × 10 = 900, so volume = √900 = 30 cm³ (the cuboid is 2 cm × 3 cm × 5 cm).
• A cube of side 5 cm is painted on all its faces. If it is sliced into 1 cm³ cubes, how many of these 1 cm³ cubes haveAnswer: (i) 8 (ii) 36 (iii) 54 (iv) 27. Check: 8 + 36 + 54 + 27 = 125 = 5³.
• Find a cuboid with edges whose lengths are integers (in cm), given that it has a total surface area of exactly 100 cm².Answer: (i) Yes: for example 1 cm × 2 cm × 16 cm and 2 cm × 4 cm × 7 cm. (ii) These two are the only ones. (iii) With l ≤ w ≤ h, lw + wh + hl = 50 gives (w + l)(h + l) = 50 + l² and l ≤ 4; checking l = 1, 2, 3, 4 gives only these two.
• Suppose a swimming pool is being built in your school. Two choices are available for the dimensions of the pool: (A) 8 ft (depth) × 50 ft × 20 ft and (B) 6 ft (depth) × 100 ft × 15 ft. Which option would be suitable/appropriate for your school? Why? What parameters/aspects did you consider to make the choice? [1 cubic foot ≈ 28.3 litres]Answer: A holds 8000 ft³ ≈ 2,26,400 L; B holds 9000 ft³ ≈ 2,54,700 L. B is the better school pool: it is shallower (6 ft, safer), has 1500 ft² of water surface instead of 1000 ft² (room for more students), and 100 ft ≈ 30 m is long enough for racing. A uses about 28,300 L less water and less tiling. Aspects considered: safety (depth), space per swimmer, water needed, tiling area and cost.
• The following figures show different types of cylinders. What is different in each of these cylinders?Answer: Both have circular ends, but in the right circular cylinder the axis (the line joining the centres of the two circles) is perpendicular to the base, while in the oblique circular cylinder the axis is slanted (inclined) to the base.
• What is the area of the curved surface of a cylinder?
  Let the radius of the circular base of the cylinder be r; let its height be h. Suppose the cylinder is cut as shown in the figure below. We get a rectangle.Answer: Curved surface area (CSA) = circumference × height = 2πrh. (Closed at both ends: TSA = 2πr(h + r); closed at one end: πr(2h + r).)
• Savitri had to make a model of a cylindrical kaleidoscope for her science project. She wanted to use chart paper to make the curved surface of the kaleidoscope (see the figure). What would be the area of chart paper required by her, if she wanted to make a kaleidoscope of length 25 cm with a 3.5 cm radius? You may take π = 227.Answer: Chart paper = curved surface area = 2πrh = 2 × 227 × 3.5 × 25 = 550 cm².
• Two cylinders, A and B, are given. The radius of cylinder B is twice that of cylinder A, and the height of cylinder B is half that of cylinder A. Find the ratio of the curved surface area of A to the curved surface area of B. Also find the ratio of the volume of A to the volume of B.Answer: CSA of A : CSA of B = 1 : 1; volume of A : volume of B = 1 : 2.
• The radii of two cylinders are in the ratio 2:3, and their heights are in the ratio 3:2. FindAnswer: (a) Volumes 2 : 3 (b) Curved surface areas 1 : 1
• The edge of a cube measures r cm. The largest possible right circular cylinder is cut out of the cube. What do you think is the volume of the cylinder (in cm³)?Answer: The cylinder has radius r2 and height r, so volume = π(r2)² × r = πr³4 cm³.
• The radius of the base of a cylinder is increased by 10%. At the same time, the height of the cylinder is decreased by x%. Given that the volume of the cylinder remains unchanged, find the value of x.Answer: (1.1)² × (1 − x100) = 1 ⇒ x = 2100121 = 1743121 ≈ 17.36.
• A solid metallic cube of side 12 cm is melted and recast into solid cylindrical rods, each having radius 2 cm and height 12 cm. Find:Answer: (i) 1728 cm³ (ii) 10567 ≈ 150.86 cm³ (iii) 11 complete rods
• What is the curved surface area of a cone?Answer: Curved surface area (CSA) of a cone = πrl, where r is the base radius and l the slant height (l² = h² + r²). Total surface area = πrl + πr² = πr(l + r).
• What is the volume of a cone?Answer: Volume of a cone = 13πr²h = 13 × area of base × height: exactly one-third of a cylinder with the same base and height.
• Here is another hands-on activity that uses modelling clay (earlier this used to be called ‘plasticine’). Carefully mold the clay into a solid cylinder and measure its radius and height. Then reshape the same portion of clay into identical cones, each having the same radius and the same height as the cylinder. You will find that you are able to make exactly three such cones.Answer: The same clay makes exactly 3 cones of the cylinder’s radius and height, so each cone has 13 of the cylinder’s volume: V = 13πr²h. For example, a cylinder r = 2 cm, h = 6 cm holds 24π ≈ 75.4 cm³, and each cone 8π ≈ 25.1 cm³.
• A right triangle ABC with sides 5 cm, 12 cm and 13 cm is rotated through 360° about the side with length 12 cm. Find the volume of the solid so obtained.Answer: The solid is a cone with height 12 cm and radius 5 cm: volume = 13π × 5² × 12 = 100π cm³ (≈ 314.3 cm³ with π = 227).
• Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.Answer: r = 12 m, l = 21 m: TSA = πr(l + r) = 227 × 12 × 33 = 1244.57 m² (about 1243.44 m² with π = 3.14).
• Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm.Answer: CSA = πrl = 227 × 7 × 10 = 220 cm².
• The height of a cone is 16 cm, and its base radius is 12 cm. Find the curved surface area and the total surface area of the cone.Answer: l = √(16² + 12²) = 20 cm. CSA = π × 12 × 20 = 240π ≈ 753.6 cm²; TSA = π × 12 × 32 = 384π ≈ 1205.76 cm² (π = 3.14).
• A cone has a height of 15 cm. If its volume is 1570 cm³, find the radius of the base.Answer: 13 × 3.14 × r² × 15 = 1570 ⇒ 15.7r² = 1570 ⇒ r² = 100 ⇒ r = 10 cm.
• The curved surface area of a cone is 308 cm² and its slant height is 14 cm. FindAnswer: (i) r = 7 cm (ii) TSA = 462 cm²
• A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.Answer: l = √(24² + 7²) = 25 cm; one cap = πrl = 227 × 7 × 25 = 550 cm²; 10 caps = 5500 cm².
• What length of tarpaulin 3 m wide is required to make a conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that is required for stitching margins and wastage in cutting is 20 cm.Answer: l = 10 m; curved surface = 3.14 × 6 × 10 = 188.4 m²; length = 188.4 ÷ 3 = 62.8 m; add 0.2 m: 63 m (π = 3.14).
• A right triangle with sides 6 cm, 8 cm and 10 cm is rotated through 360° about the side of 8 cm. Find the volume and the curved surface area of the solid so formed.Answer: A cone with h = 8 cm, r = 6 cm, l = 10 cm. Volume = 13π × 36 × 8 = 96π ≈ 301.44 cm³; CSA = π × 6 × 10 = 60π ≈ 188.4 cm² (π = 3.14).
• Suppose you have a cup in the shape of a right circular cone. Fill it with water to half the depth of the cone. What fraction of the volume of the cup is occupied by the water?Answer: The water forms a cone with half the height and half the radius, so its volume is (12)³ = 18 of the cup.
• But you may be able to find your own derivation of the formula by connecting it with the formula for volume of a cone; namely, by writing it as
  V = 13 × (4πr²) × r,
  i.e., Volume of sphere = 13 × surface area of sphere × radius of sphere.
  Try to work out the connecting links on your own.Answer: Split the sphere’s surface into tiny patches and join each to the centre: the sphere becomes many thin cones, all of height r. Their volumes add to r3 × (total base area) = r3 × 4πr² = 43πr³.
• Take a rubber ball and drive a nail into it. Using the nail for support, wind a string tightly around the ball. When you have reached the ‘fullest’ part of the ball, use pins to keep the string in place, and continue to wind the string around the remaining part of the ball, till you have completely covered it; see the figure. Mark the starting and finishing points on the string, and unwind the string from the surface of the ball. Now, measure the diameter of the ball and get its radius. On a sheet of paper, draw four circles with a radius equal to the radius of the ball. Fill the circles with the string you had wound around the ball (as shown in the figure).Answer: The string that covered the ball exactly fills four circles of the ball’s radius. So surface area of a sphere = 4 × πr² = 4πr².
• A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain?Answer: Volume = 23πr³ = 23 × 227 × 3.5 × 3.5 × 3.5 ≈ 89.8 cm³.
• A ball bearing has a radius of 0.7 cm. Find its surface area.Answer: Surface area = 4πr² = 4 × 227 × 0.7 × 0.7 = 6.16 cm².
• Two solid spheres made of the same metal have weights 5920 g and 740 g. Determine the radius of the larger sphere, if the diameter of the smaller one is 5 cm.Answer: Same metal, so weight ∝ volume ∝ r³. 5920740 = 8 = 2³, so the larger radius is 2 × 2.5 = 5 cm.
• The diameter of the moon is approximately one fourth the diameter of the earth. Given that the moon and earth are both roughly spherical, find the ratio of their surface areas.Answer: Radii in ratio 1 : 4, so surface areas in ratio 1² : 4² = 1 : 16 (moon : earth).
• Find (i) the curved surface area and (ii) the total surface area of a hemisphere of radius 21 cm.Answer: (i) CSA = 2πr² = 2772 cm² (ii) TSA = 3πr² = 4158 cm²
• Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm × 8 cm × 8 cm. When 16 spheres are packed, the box is filled with preservative liquid. Find the volume of this liquid. Round your answer to the nearest integer.Answer: Box 1024 cm³ − 16 spheres (16 × 43π × 8 = 512π3 ≈ 535.89 cm³) ≈ 488.11, so the liquid ≈ 488 cm³.
• The radius of a sphere is increased by 10%. Show that the volume increases by approximately 33.1%.Answer: New radius 1.1r ⇒ new volume = (1.1)³ × old = 1.331 × old, an increase of 0.331 = 33.1%.
• The radius of a sphere is increased by x%. The volume of the sphere increases by 72.8%. Find the value of x.Answer: (1 + x100)³ = 1.728 = 1.2³ ⇒ 1 + x100 = 1.2 ⇒ x = 20.
• The hemispherical dome of a building needs to be painted (see the figure). If the circumference of the base of the dome is 35.2 m, find the cost of painting it, given the cost of painting is ₹ 10 per 100 cm².Answer: r = 5.6 m; curved surface = 2πr² = 197.12 m² = 19,71,200 cm²; cost = 19,71,200100 × ₹ 10 = ₹ 1,97,120.
• We learned about different 3D solids. Identify and list objects or structures around you that are cubical, cuboidal, cylindrical, conical, spherical, or pyramidal in shape, or closely resemble these shapes.Answer: Cubical: dice, ice cube. Cuboidal: brick, book, almirah, room. Cylindrical: gas cylinder, pipe, tin can, pillar. Conical: ice-cream cone, funnel, traffic cone, tent. Spherical: ball, globe, orange, marble (hemispherical: bowl, dome). Pyramidal: Egyptian pyramids, some roofs, a triangular pyramid puzzle.
• Activity: Find the approximate volumes of a gas cylinder, a bucket, any cylindrical vessel, a cupboard, a water tank that you can find near you (use 1000 cm³ = 1 litre).Answer: With sample measurements (π = 3.14): gas cylinder ≈ 35 L, bucket ≈ 17 L, cooking vessel ≈ 5.7 L, cupboard ≈ 810 L, cylindrical water tank ≈ 1190 L. Your own numbers will differ with your measurements.
• Lallan’s golgappa (a small hollow ball made of wheat) stall has a cylindrical vessel with pani (spiced water) in it. The vessel is of radius 10 cm and it has pani up to a height of 50 cm. Estimate the number of golgappas he can serve assuming all golgappas are spheres. Before calculating, make a guess.
  Can you guess what number this could be — is it in the hundreds, thousands, or more?Answer: Pani = π × 10² × 50 = 5000π ≈ 15 708 cm³. With a golgappa of radius 1.5 cm filled 60% (≈ 8.5 cm³ of pani), about 1850 golgappas; with radius 1.75 cm filled 80% (≈ 18 cm³), about 870. So the answer is in the high hundreds to about two thousand, depending on the assumptions.
• Anirban saw a few tennis balls lying around. He wondered, “How many tennis balls can fit in an empty classroom?”. The classroom is of dimensions 30 ft × 30 ft × 15 ft. Estimate how many tennis balls of radius 3 cm fit in this classroom. Before calculating, make a guess.Answer: Dividing volumes gives an upper bound: 37,80,00,000 ÷ 113 ≈ 33,45,132, but balls leave gaps. Stacking ball-on-ball (6 cm per ball): 150 × 150 × 75 = 16,87,500, about 17 lakh tennis balls.
• Make an estimate based on the second and third configurations.Answer: Arrangement A fills about 52% of the room with balls (16,87,500 balls). B (a ball resting in the middle of each group of eight) fills about 68%: about 22 lakh balls. C (each layer sitting in the hollows of the layer below, the way oranges are stacked) fills about 74%, the tightest possible: about 24 lakh balls. Both are below the upper bound of about 33 lakh.
• Estimate how many scoops of ice cream can be obtained from a cuboidal container of dimensions 10 cm × 15 cm × 20 cm.Answer: Container = 3000 cm³. A scoop is roughly a ball of diameter 5 cm, about 65 cm³. 3000 ÷ 65 ≈ 46, so about 45 scoops.
• A cube of integer side length a is made of unit cubes. Write an expression giving the number of unit cubes to be added to make a cube of side length a + 1.Answer: (a + 1)³ − a³ = 3a² + 3a + 1 unit cubes.
• Solve the following:Answer: (i) No. A classroom about 8 m × 6 m × 3.5 m holds 168 m³ = 1,68,000 L; a person drinking about 2.5 L a day for 75 years drinks about 68,000 L, only about 40% of the room. (ii) Walls of about 18.5 m³ of brickwork, with each brick (with mortar) about 20 cm × 10 cm × 10 cm = 0.002 m³: about 9000 bricks.
• You are given two options: (A) one chocolate cube of side 50 mm, (B) hundred chocolate cubes of side 10 mm. Which option gives you more chocolate?Answer: A: 50³ = 1,25,000 mm³. B: 100 × 10³ = 1,00,000 mm³. Option A gives more (by 25,000 mm³ = 25 cm³).
• If all the people in the world are crowded together in one location, how much area would it cover?Answer: About 8.2 billion people at about 4 people per m² (0.25 m² each) need about 2.05 × 10⁹ m² ≈ 2000 km²: a square only about 45 km on each side.
• A school provides milk to its students in cylindrical glasses, each with a diameter of 7 cm. If each glass is filled with milk to a height of 12 cm, how many litres of milk are needed to serve 1600 students? Use 1000 cm³ = 1 litre.Answer: One glass = 227 × 3.5² × 12 = 462 cm³; 1600 glasses = 7,39,200 cm³ = 739.2 litres.
• The surface area of a sphere of radius 5 cm is five times the area of the curved surface of a cone of radius 4 cm. Find the height and the volume of the cone.Answer: 4π(5)² = 5 × π(4)l ⇒ l = 5 cm; h = √(25 − 16) = 3 cm; volume = 13π × 16 × 3 = 16π ≈ 50.29 cm³.
• Take Earth to be a perfect sphere with radius 6370 km. Take Jupiter to be a perfect sphere with radius 69,900 km. Take the Sun to be a perfect sphere with radius 6,95,700 km. Compute approximately:
  (Calculator can be used.)Answer: (i) (69 9006370)³ ≈ 10.97³ ≈ 1321 (ii) (6,95,7006370)³ ≈ 109.2³ ≈ 13,03,000 (about 1.3 million)
• Show that the volume of a sphere is equal to 23 of the volume of the smallest cylinder which encloses it.Answer: The smallest enclosing cylinder has radius r and height 2r: volume 2πr³. Sphere: 43πr³ = 23 × 2πr³. Shown.
• Suppose the Earth is perfectly spherical. A string is wrapped tightly around the equator of the Earth. Another string, 1 metre longer, is placed around the Earth so that it forms a larger circle, staying the same distance above the ground everywhere.
  How high above the ground is the second string?
  Repeat this for: (i) the Moon (ii) Jupiter (iii) a volleyball. Are you surprised by the three answers? Why or why not?Answer: 2π(R + x) − 2πR = 1 ⇒ x = 12π m ≈ 0.159 m ≈ 16 cm. The same 16 cm for (i) the Moon, (ii) Jupiter and (iii) a volleyball, because the radius R cancels.
• We have a cylinder with a base radius of r cm and height h cm. A square pyramid is fitted inside it. The square base of the pyramid lies on the base of the cylinder, its corners on the boundary of the cylinder. The apex of the pyramid lies on the top of the cylinder. Find the ratio of the volume of this pyramid to the volume of the cylinder.Answer: Square inscribed in a circle of radius r has diagonal 2r, so area 2r². Pyramid = 13 × 2r² × h = 2r²h3. Ratio = 23π, i.e. 2 : 3π (≈ 0.21).
• What is the change in volume when:Answer: (i) (c) wh cubic units (ii) (e) 2πrh + πh cubic units (iii) the volume decreases by 43π(3r² − 3r + 1) = 4πr² − 4πr + 43π cubic units
• Given a cube with volume V, express its total surface area S in terms of V.Answer: Side a = ∛V, so S = 6V^2/3 = 6∛(V²).
• Given a cube with total surface area S, express its volume V in terms of S.Answer: Side a = √(S6), so V = (S6)^3/2 = S6√(S6).
• A cylindrical glass of height 25 cm and radius 4 cm has water up to a height of 16 cm. A crow wants to drink water from this glass. The water must be at a height of 20 cm for the crow to reach it. There are some marbles lying around. How many marbles of radius 1 cm should the crow drop into the glass to make the water reach the required height?Answer: The water must rise 4 cm: π × 4² × 4 = 64π cm³. One marble = 43π cm³. Number = 64π ÷ 43π = 48 marbles.
• Find the volume of ink in a new ball point pen. Take the necessary measurements and make approximations, as needed.Answer: The ink sits in the refill tube, a thin cylinder. With inside diameter about 2 mm (r ≈ 0.1 cm) and an ink column about 10 cm long: V ≈ 3.14 × 0.1² × 10 ≈ 0.3 cm³ (about 0.3 mL).
• Solve:Answer: (i) Dough = 43π × 6³ ≈ 905 cm³; one chapati ≈ π × 8² × 0.2 ≈ 40 cm³: about 22 chapatis. (ii) For a coconut (shell inner radius ≈ 5.5 cm, flesh 1 cm thick): flesh ≈ 43π(5.5³ − 4.5³) ≈ 315 cm³ (about 158 cm³ in each half). A muskmelon (inner radius 7.5 cm, seed hollow 3.5 cm) gives about 1590 cm³ of fruit.
• Looking at her mathematics book (Part 2), Sheela wonders:Answer: (i) No. A page is about 28 cm × 21 cm ≈ 0.059 m²; even 180 separate pages cover only about 10.6 m², against a classroom floor of about 48 m². (ii) The storeroom is 9000 ft³ ≈ 25,48,53,000 cm³; a 28 cm × 21 cm × 1 cm book takes 588 cm³, so at most about 4,33,000 books by volume, and about 4,24,000 when stacked neatly.
• If the entire human population decided to climb into one giant cube, how long would the side have to be?Answer: About 8.2 billion people, each standing in 0.25 m² on floors 2 m apart (0.5 m³ each): 4.1 × 10⁹ m³, so the side ≈ ∛(4.1 × 10⁹) ≈ 1600 m (about 1.6 km). (Squeezing bodies with no space at all, about 0.065 m³ each, would need only about 810 m.)
• The Earth’s surface has an estimated volume of 1.38 billion km³ of water. Suppose the Earth is a perfect sphere and all this water forms a uniform layer completely covering the Earth’s surface, like a thin water bubble. Estimate the thickness of this water layer. The Earth’s radius is ~6371 km.Answer: (i) t = ∛(R³ + 3V4π) − R, or, since the layer is thin, t ≈ V4πR². (ii) t ≈ 1.38 × 10⁹4π × 6371² ≈ 2.7 km.
• Give the dimension of a cuboid whose volume is halved when its surface area is doubled.Answer: One answer: a cuboid 8 cm × 14 cm × 14 cm (volume 1568 cm³, surface area 840 cm²). Squash it to 1 cm × 28 cm × 28 cm: the surface area doubles to 1680 cm² and the volume halves to 784 cm³.
• Project: Find the volume of your house making necessary approximations. Present how you solved it.Answer: Sample: a flat of six rooms treated as cuboids with total floor area ≈ 62 m² and ceiling height 3 m has an inside volume of about 62 × 3 = 186 m³ (about 1,86,000 litres of space).

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