SOLIDS [Surface Area and Volume of 3-D Solids] - Questions & Answers

EXERCISE 20

1. Multiple Choice Type : Choose the correct answer from the options given below.
(a) The area of a square on the diagonal of a cube is 48 cm². Each edge of the cube is :
(i) 2 cm (ii) 4 cm (iii) 6 cm (iv) 8 cm
Step 1: Let the edge of the cube be 'a' cm.
Step 2: The length of the diagonal of a cube is a√3.
Step 3: The area of the square on this diagonal is (a√3)² = 3a².
Step 4: We are given 3a² = 48.
Step 5: a² = 16, so a = 4 cm.
Step 6: The correct option is (ii) 4 cm.

(b) A cuboid has length = 4 cm, breadth = 3 cm and diagonal = 13 cm. The volume of the cuboid is :
(i) 156 cm³ (ii) 288 cm³ (iii) 144 cm³ (iv) 78 cm³
Step 1: The formula for the diagonal of a cuboid is √(l² + b² + h²).
Step 2: Given diagonal = 13, so √(4² + 3² + h²) = 13.
Step 3: Squaring both sides: 16 + 9 + h² = 169.
Step 4: 25 + h² = 169, which gives h² = 144, so height h = 12 cm.
Step 5: Volume = length × breadth × height = 4 × 3 × 12 = 144 cm³.
Step 6: The correct option is (iii) 144 cm³.

(c) A cuboid with dimensions 12 cm × 9 cm × 2 cm is made of a metal. It is melted and recast into a solid cube. The edge of the cube is :
(i) 6 cm (ii) 12 cm (iii) 15 cm (iv) 8 cm
Step 1: Volume of the cuboid = 12 × 9 × 2 = 216 cm³.
Step 2: When melted and recast, the volume remains the same.
Step 3: Volume of the new cube = a³ = 216.
Step 4: Taking the cube root, a = 6 cm.
Step 5: The correct option is (i) 6 cm.

(d) The dimensional ratio of the sides of a cuboid is 3 : 2 : 1. If its volume is 1296 cm³; the actual dimensions of the cuboid are :
(i) 12 cm, 12 cm and 8 cm (ii) 8 cm, 8 cm and 12 cm (iii) 12 cm, 12 cm and 12 cm (iv) 18 cm, 12 cm and 6 cm
Step 1: Let the dimensions be 3x, 2x, and x.
Step 2: Volume = 3x × 2x × x = 6x³.
Step 3: We are given 6x³ = 1296.
Step 4: x³ = 216, which gives x = 6.
Step 5: The dimensions are 3(6) = 18 cm, 2(6) = 12 cm, and 1(6) = 6 cm.
Step 6: The correct option is (iv) 18 cm, 12 cm and 6 cm.

(e) A tank with dimensions 12 m, 10 m and 8 m is dug and the soil taken out of it is spread uniformly on a field of length = 40 m and breadth = 32 m. The rise in level of the field is :
(i) 7·5 m (ii) 7·5 cm (iii) 75 cm (iv) 75 m
Step 1: Volume of soil taken out = Volume of tank = 12 × 10 × 8 = 960 m³.
Step 2: Area of the field = length × breadth = 40 × 32 = 1280 m².
Step 3: Rise in level = Volume of soil / Area of field.
Step 4: Rise in level = 960 / 1280 = 0.75 m.
Step 5: Convert meters to centimeters: 0.75 × 100 = 75 cm.
Step 6: The correct option is (iii) 75 cm.

(f) Through the pipe of uniform cross-section (12 cm²) water flows with the speed of 20 cm/s. The volume of water that flows in 1 minute is :
(i) 0·0144 m³ (ii) 0·144 m³ (iii) 144 cm³ (iv) 240 cm³
Step 1: Volume of water flowing in 1 second = Area of cross-section × speed = 12 × 20 = 240 cm³.
Step 2: Volume of water flowing in 1 minute (60 seconds) = 240 × 60 = 14400 cm³.
Step 3: Convert cm³ to m³ by dividing by 1,000,000 (since 1 m³ = 10⁶ cm³).
Step 4: Volume = 14400 / 1000000 = 0.0144 m³.
Step 5: The correct option is (i) 0·0144 m³.

(g) In the given figure, all the dimensions are given in cm. The volume of the solid is :
(i) 7 cm × 13 cm × 20 cm
(ii) (7 + 13) × 20 × 3 cm³
(iii) 1/4 (7 + 13) × 3 × 20 cm³
(iv) 1/2 (7 + 13) × 3 × 20 cm³
Step 1: The given solid is a prism with a uniform cross-section which is a trapezium.
Step 2: The parallel sides of the trapezium are 7 cm and (7 + 13) cm = 20 cm. (Assuming bottom edge is 20).
Step 3: The perpendicular distance (height) between the parallel sides is 3 cm.
Step 4: Area of the cross-section = 1/2 × (Sum of parallel sides) × height = 1/2 × (7 + 13) × 3 cm².
Step 5: Volume of the solid = Area of cross-section × length = 1/2 × (7 + 13) × 3 × 20 cm³.
Step 6: The correct option is (iv).

2. The length, breadth and height of a rectangular solid are in the ratio 5 : 4 : 2. If the total surface area is 1216 cm², find the length, the breadth and the height of the solid.
Step 1: Let the length, breadth, and height be 5x, 4x, and 2x respectively.
Step 2: Total surface area formula = 2(lb + bh + hl).
Step 3: Substitute the values: 2(5x·4x + 4x·2x + 5x·2x) = 1216.
Step 4: 2(20x² + 8x² + 10x²) = 1216.
Step 5: 2(38x²) = 1216, which means 76x² = 1216.
Step 6: x² = 1216 / 76 = 16, so x = 4.
Step 7: Length = 5(4) = 20 cm.
Step 8: Breadth = 4(4) = 16 cm.
Step 9: Height = 2(4) = 8 cm.

3. The volume of a cube is 729 cm³. Find its total surface area.
Step 1: Volume of a cube = a³ = 729 cm³.
Step 2: Taking the cube root, the edge a = 9 cm.
Step 3: Total surface area of a cube = 6a².
Step 4: Surface Area = 6 × (9)² = 6 × 81 = 486 cm².

4. The dimensions of a Cinema Hall are 100 m, 60 m and 15 m. How many persons can sit in the hall, if each requires 150 m³ of air for healthy breathing ?
Step 1: Find the total volume of the cinema hall = 100 × 60 × 15 = 90,000 m³.
Step 2: Volume of air required by one person = 150 m³.
Step 3: Number of persons = Total volume / Volume per person.
Step 4: Number of persons = 90,000 / 150 = 600 persons.

5. 75 persons can sleep in a room 25 m by 9·6 m. If each person requires 16 m³ of air; find the height of the room.
Step 1: Total volume of air required for 75 persons = 75 × 16 = 1200 m³.
Step 2: Area of the floor = 25 × 9.6 = 240 m².
Step 3: Volume of the room = Base Area × Height.
Step 4: 1200 = 240 × Height.
Step 5: Height = 1200 / 240 = 5 m.

6. The edges of three cubes of metal are 3 cm, 4 cm and 5 cm. They are melted and formed into a single cube. Find the edge of the new cube.
Step 1: Find the sum of volumes of the three cubes = 3³ + 4³ + 5³.
Step 2: Total volume = 27 + 64 + 125 = 216 cm³.
Step 3: Let the edge of the new cube be 'a'. So, a³ = 216.
Step 4: Taking the cube root, a = 6 cm.

7. Three cubes, whose edges are x cm, 8 cm and 10 cm respectively, are melted and recast into a single cube of edge 12 cm. Find 'x'.
Step 1: The total volume of the three original cubes equals the volume of the new single cube.
Step 2: x³ + 8³ + 10³ = 12³.
Step 3: x³ + 512 + 1000 = 1728.
Step 4: x³ + 1512 = 1728.
Step 5: x³ = 1728 - 1512 = 216.
Step 6: x = ³√216 = 6 cm.

8. Three equal cubes are placed adjacently in a row. Find the ratio of the total surface area of the resulting cuboid to that of the sum of the total surface areas of the three cubes. (HOTS)
Step 1: Let the edge of each cube be 'a'.
Step 2: When placed in a row, length of the cuboid = 3a, breadth = a, height = a.
Step 3: Total surface area of the cuboid = 2(3a·a + a·a + a·3a) = 2(3a² + a² + 3a²) = 2(7a²) = 14a².
Step 4: Sum of total surface areas of three separate cubes = 3 × 6a² = 18a².
Step 5: Required ratio = 14a² / 18a² = 7 / 9, or 7:9.

9. The cost of papering the four walls of a room at 75 paise per square metre is ₹ 240. The height of the room is 5 metres. Find the length and the breadth of the room, if they are in the ratio 5 : 3.
Step 1: Total cost = ₹ 240. Rate = 75 paise = ₹ 0.75 per m².
Step 2: Area of the four walls = Total Cost / Rate = 240 / 0.75 = 320 m².
Step 3: Let length = 5x and breadth = 3x. Height = 5 m.
Step 4: Area of four walls = 2(l + b) × h = 2(5x + 3x) × 5.
Step 5: 320 = 10(8x) = 80x.
Step 6: x = 320 / 80 = 4.
Step 7: Length = 5 × 4 = 20 m, Breadth = 3 × 4 = 12 m.

10. The area of a playground is 3650 m². Find the cost of covering it with gravel 1·2 cm deep, if the gravel costs ₹ 6·40 per cubic metre.
Step 1: Depth of gravel = 1.2 cm = 0.012 m.
Step 2: Volume of gravel required = Area × depth = 3650 × 0.012 = 43.8 m³.
Step 3: Cost = Volume × Rate = 43.8 × 6.40.
Step 4: Total Cost = ₹ 280.32.

11. A square plate of side 'x' cm is 8 mm thick. If its volume is 2880 cm³; find the value of x.
Step 1: Thickness = 8 mm = 0.8 cm.
Step 2: Volume of the square plate = Base Area × Thickness = x² × 0.8.
Step 3: We are given x² × 0.8 = 2880.
Step 4: x² = 2880 / 0.8 = 3600.
Step 5: x = √3600 = 60.

12. The external dimensions of a closed wooden box are 27 cm, 19 cm and 11 cm. If the thickness of the wood in the box is 1·5 cm; find :
(i) volume of the wood in the box;
(ii) the cost of the box, if wood costs ₹ 1.20 per cm³;
(iii) number of 4 cm cubes that could be placed into the box.
Step 1: External Volume = 27 × 19 × 11 = 5643 cm³.
Step 2: Internal dimensions = (27 - 2×1.5), (19 - 2×1.5), (11 - 2×1.5) = 24 cm, 16 cm, 8 cm.
Step 3: Internal Volume = 24 × 16 × 8 = 3072 cm³.
Step 4: (i) Volume of wood = External Volume - Internal Volume = 5643 - 3072 = 2571 cm³.
Step 5: (ii) Cost of the box = 2571 × 1.20 = ₹ 3085.20.
Step 6: (iii) Cubes along length = 24 / 4 = 6. Cubes along breadth = 16 / 4 = 4. Cubes along height = 8 / 4 = 2.
Step 7: Total number of cubes = 6 × 4 × 2 = 48.

13. A tank 20 m long, 12 m wide and 8 m deep is to be made of iron sheet. It is open at the top. Determine the cost of iron-sheet, at the rate of ₹ 12·50 per metre, if the sheet is 2·5 m wide.
Step 1: Since it is open at the top, Surface Area = Area of 4 walls + Area of base.
Step 2: Area = 2(20 + 12) × 8 + (20 × 12) = 2(32) × 8 + 240 = 512 + 240 = 752 m².
Step 3: Length of the sheet required = Total Area / width of sheet = 752 / 2.5 = 300.8 m.
Step 4: Total Cost = Length of sheet × Rate = 300.8 × 12.50 = ₹ 3760.

14. A closed rectangular box is made of wood of 1·5 cm thickness. The exterior length and breadth are respectively 78 cm and 19 cm, and the capacity of the box is 15 cubic decimetres. Calculate the exterior height of the box.
Step 1: Capacity = Internal volume = 15 dm³ = 15 × 1000 = 15000 cm³.
Step 2: Internal length = 78 - (2 × 1.5) = 75 cm.
Step 3: Internal breadth = 19 - (2 × 1.5) = 16 cm.
Step 4: Let internal height be 'h'. Then 75 × 16 × h = 15000.
Step 5: 1200 × h = 15000, so h = 15000 / 1200 = 12.5 cm.
Step 6: Exterior height = Internal height + (2 × 1.5) = 12.5 + 3 = 15.5 cm.

15. The square on the diagonal of a cube has an area of 1875 sq. cm. Calculate : (HOTS)
(i) the side of the cube.
(ii) the total surface area of the cube.
Step 1: Let the side of the cube be 'a'. The diagonal of the cube is a√3.
Step 2: Area of the square on the diagonal = (a√3)² = 3a².
Step 3: We are given 3a² = 1875.
Step 4: a² = 625, so a = √625 = 25 cm. (i) Side of the cube = 25 cm.
Step 5: (ii) Total surface area = 6a² = 6 × 625 = 3750 cm².

16. The following figure shows a solid of uniform cross-section. Find the volume of the solid. All measurements are in centimetres. Assume that all angles in the figure are right angles.
Step 1: Break the uniform cross-section (front face) into three vertical rectangles from left to right.
Step 2: Rectangle 1 (left): width = 3 cm, total height = 9 cm. Area = 3 × 9 = 27 cm².
Step 3: Rectangle 2 (middle): width = 4 cm, height = 9 - 3 = 6 cm. Area = 4 × 6 = 24 cm².
Step 4: Rectangle 3 (right): width = 9 - 3 - 4 = 2 cm, height = 3 cm. Area = 2 × 3 = 6 cm².
Step 5: Total Area of cross-section = 27 + 24 + 6 = 57 cm².
Step 6: The depth of the solid is given as 4 cm (from the top receding edge).
Step 7: Volume = Area × depth = 57 × 4 = 228 cm³.

17. A swimming pool is 40 m long and 15 m wide. Its shallow and deep ends are 1·5 m and 3 m deep respectively. If the bottom of the pool slopes uniformly, find the amount of water in litres required to fill the pool. (LS)
Step 1: The longitudinal cross-section of the pool is a trapezium.
Step 2: Parallel sides are the depths = 1.5 m and 3 m. The height of trapezium (length) = 40 m.
Step 3: Area of cross-section = 1/2 × (1.5 + 3) × 40 = 1/2 × 4.5 × 40 = 90 m².
Step 4: Volume of pool = Area of cross-section × width = 90 × 15 = 1350 m³.
Step 5: 1 m³ = 1000 litres. Amount of water = 1350 × 1000 = 1,350,000 litres.

18. The cross-section of a tunnel perpendicular to its length is a trapezium ABCD as shown in the following figure; also given that :
AM = BN; AB = 7 m; CD = 5 m. The height of the tunnel is 2·4 m. The tunnel is 40 m long. Calculate :
(i) the cost of painting the internal surface of the tunnel (excluding the floor) at the rate of ₹ 5 per m² (sq. metre).
(ii) the cost of paving the floor at the rate of ₹ 18 per m².
Step 1: Draw perpendiculars CM and DN to the base AB. MN = CD = 5 m.
Step 2: AM + BN = AB - MN = 7 - 5 = 2 m. Since AM = BN, AM = 1 m and BN = 1 m.
Step 3: In right triangle ADM, AD² = AM² + DM² = 1² + 2.4² = 1 + 5.76 = 6.76. AD = √6.76 = 2.6 m. Similarly, BC = 2.6 m.
Step 4: Perimeter of internal walls and roof = AD + DC + CB = 2.6 + 5 + 2.6 = 10.2 m.
Step 5: (i) Area to be painted = Perimeter × length = 10.2 × 40 = 408 m².
Step 6: Cost of painting = 408 × 5 = ₹ 2040.
Step 7: (ii) Area of floor = AB × length = 7 × 40 = 280 m².
Step 8: Cost of paving floor = 280 × 18 = ₹ 5040.

19. Water is discharged from a pipe of cross-section area 3·2 cm² at the speed of 5m/s. Calculate the volume of water discharged :
(i) in cm³ per sec.
(ii) in litres per minute.
Step 1: Speed = 5 m/s = 500 cm/s.
Step 2: (i) Volume per second = Area × Speed = 3.2 × 500 = 1600 cm³ per sec.
Step 3: (ii) Volume per minute = 1600 × 60 = 96,000 cm³.
Step 4: Convert to litres: 96,000 / 1000 = 96 litres per minute.

20. A hose-pipe of cross-section area 2 cm² delivers 1500 litres of water in 5 minutes. What is the speed of water in m/s through the pipe ?
Step 1: Volume of water in 5 minutes = 1500 litres = 1,500,000 cm³.
Step 2: Volume of water in 1 minute = 1,500,000 / 5 = 300,000 cm³.
Step 3: Volume of water per second (discharge rate) = 300,000 / 60 = 5000 cm³.
Step 4: Speed = Volume per sec / Area = 5000 / 2 = 2500 cm/s.
Step 5: Convert to m/s: 2500 / 100 = 25 m/s.

21. The cross-section of a piece of metal 4 m in length is shown below. Calculate :
(i) the area of the cross-section;
(ii) the volume of the piece of metal in cubic centimetres.
If 1 cubic centimetre of the metal weighs 6·6 g, calculate the weight of the piece of metal to the nearest kg.
Step 1: Split the cross-section vertically into a rectangle and a trapezium.
Step 2: The left part is a rectangle: width = 10 cm, height = 12 cm. Area = 10 × 12 = 120 cm².
Step 3: The right part is a trapezium from x=10 to x=16 (width = 6 cm). Its left vertical height is 12 cm. The rightmost edge is given a drop of 7.5 cm from the top, so its length is 7.5 cm.
Step 4: Area of right trapezium = 1/2 × (12 + 7.5) × 6 = 1/2 × 19.5 × 6 = 58.5 cm².
Step 5: (i) Total Area = 120 + 58.5 = 178.5 cm².
Step 6: (ii) Length = 4 m = 400 cm. Volume = Area × length = 178.5 × 400 = 71,400 cm³.
Step 7: Total weight = Volume × 6.6 g = 71,400 × 6.6 = 471,240 g.
Step 8: Weight in kg = 471,240 / 1000 = 471.24 kg ≈ 471 kg.

22. A rectangular water-tank measuring 80 cm × 60 cm × 60 cm is filled from a pipe of cross-sectional area 1·5 cm², the water emerging at 3·2 m/s. How long does it take to fill the tank?
Step 1: Volume of tank = 80 × 60 × 60 = 288,000 cm³.
Step 2: Speed of water = 3.2 m/s = 320 cm/s.
Step 3: Volume of water flowing per second = 1.5 × 320 = 480 cm³.
Step 4: Time taken = Total Volume / Volume per second = 288,000 / 480 = 600 seconds.
Step 5: Convert to minutes: 600 / 60 = 10 minutes.

TEST YOURSELF

1. Multiple Choice Type : Choose the correct answer from the options given below.
(a) The radius of a sphere is 2r, then its volume is equal to :
(i) 4/3 πr³ (ii) 4πr³ (iii) 8πr³/3 (iv) 32πr³/3
Step 1: Volume of a sphere = 4/3 π(Radius)³.
Step 2: Volume = 4/3 π(2r)³ = 4/3 π(8r³) = 32πr³ / 3.
Step 3: The correct option is (iv) 32πr³/3.

(b) The total surface area of a cube is 96 cm². The volume of the cube is :
(i) 8 cm³ (ii) 512 cm³ (iii) 64 cm³ (iv) 27 cm³
Step 1: TSA = 6a² = 96, so a² = 16, resulting in a = 4 cm.
Step 2: Volume = a³ = 4³ = 64 cm³.
Step 3: The correct option is (iii) 64 cm³.

(c) If a solid cube of side 16 cm is cut into eight identical cubes. The side of each cube is :
(i) 2 cm (ii) 4 cm (iii) 6 cm (iv) 8 cm
Step 1: Volume of large cube = 16³ = 4096 cm³.
Step 2: Volume of one small cube = 4096 / 8 = 512 cm³.
Step 3: Side of small cube = ³√512 = 8 cm.
Step 4: The correct option is (iv) 8 cm.

(d) The diameters of two solid spheres are in the ratio 5 : 7. The ratio between areas of their curved surfaces is :
(i) 5 : 7 (ii) 7 : 5 (iii) 49 : 25 (iv) 25 : 49
Step 1: Ratio of radii is the same as diameters, 5 : 7.
Step 2: Ratio of surface areas = (r₁/r₂)² = (5/7)² = 25 : 49.
Step 3: The correct option is (iv) 25 : 49.

(e) The radius of a cylinder is doubled and its curved surface area is kept as same, the height of the cylinder is :
(i) same (ii) doubled (iii) halved (iv) none of these
Step 1: CSA of cylinder = 2πrh. If radius becomes 2r, new CSA = 2π(2r)H = 4πrH.
Step 2: Equating old and new: 2πrh = 4πrH, which means H = h / 2.
Step 3: The correct option is (iii) halved.

(f) Statement (1) : Each side of a cuboid is doubled, its total surface area is also doubled.
Statement (2) : The surface area of the resulting cuboid is 2 x 2 x 2 times the original area.
(i) Both the statements are true.
(ii) Both the statements are false.
(iii) Statement 1 is true, and statement 2 is false.
(iv) Statement 1 is false, and statement 2 is true.
Step 1: If each side is doubled (2l, 2b, 2h), the new surface area is 2(2l·2b + 2b·2h + 2h·2l) = 4 × 2(lb + bh + hl) = 4 times the original. Statement 1 is false.
Step 2: Statement 2 says it is 2 × 2 × 2 = 8 times, which is also false.
Step 3: The correct option is (ii) Both the statements are false.

(g) Assertion (A) : The radius of a hemisphere increases from r cm to 2r cm. The ratio between the surface areas of the original hemisphere and the resulting hemisphere is 1 : 4.
Reason (R) : Surface area in the first case = πr² + 2πr²
Step 1: TSA of a solid hemisphere = 3πr². For radius 2r, TSA = 3π(2r)² = 12πr².
Step 2: Ratio = 3πr² : 12πr² = 1 : 4. So, Assertion (A) is true.
Step 3: Reason (R) states TSA = πr² + 2πr² = 3πr², which is the correct formula and logic.
Step 4: The correct option is (iii) Both A and R are true and R is the correct reason for A.

(h) Assertion (A) : A sphere is inscribed in a cylinder, the ratio of the volume of the cylinder to the volume of the sphere is 1 : 4.
Reason (R) : Required ratio = πr² × 2r : 4/3 πr³
Step 1: If a sphere is inscribed, cylinder radius = r, and cylinder height = 2r.
Step 2: Volume of cylinder = πr²(2r) = 2πr³. Volume of sphere = 4/3 πr³.
Step 3: Ratio = 2πr³ : 4/3 πr³ = 6 : 4 = 3 : 2. Assertion (A) stating 1 : 4 is false.
Step 4: Reason (R) correctly gives the formula setup for the required ratio. So, R is true.
Step 5: The correct option is (ii) A is false, R is true.

2. A hollow square-shaped tube open at both ends is made of iron. The internal square is of 5 cm side and the length of the tube is 8 cm. There are 192 cm³ of iron in this tube. Find its thickness.
Step 1: Let the thickness be 'x' cm. The external side = (5 + 2x) cm.
Step 2: Volume of iron = External Volume - Internal Volume.
Step 3: 192 = (5 + 2x)² × 8 - (5)² × 8.
Step 4: Divide by 8: 24 = (5 + 2x)² - 25.
Step 5: (5 + 2x)² = 49, which gives 5 + 2x = 7.
Step 6: 2x = 2, so x = 1 cm. Thickness is 1 cm.

3. Four identical cubes are joined end to end to form a cuboid. If the total surface area of the resulting cuboid is 648 cm²; find the length of edge of each cube.
Also, find the ratio between the surface area of the resulting cuboid and the surface area of a cube.
Step 1: Let the edge of each cube be 'a'. The cuboid dimensions will be l = 4a, b = a, h = a.
Step 2: TSA of cuboid = 2(4a·a + a·a + a·4a) = 2(4a² + a² + 4a²) = 18a².
Step 3: We are given 18a² = 648, so a² = 36, meaning a = 6 cm.
Step 4: TSA of one cube = 6a².
Step 5: Ratio = 18a² : 6a² = 3 : 1.

4. A rectangular card-board sheet has length 32 cm and breadth 26 cm. Squares each of side 3 cm, are cut from the corners of the sheet and the sides are folded to make a rectangular container. Find the capacity of the container formed.
Step 1: Length of container = 32 - 2(3) = 26 cm.
Step 2: Breadth of container = 26 - 2(3) = 20 cm.
Step 3: Height of the container = 3 cm.
Step 4: Capacity (Volume) = 26 × 20 × 3 = 1560 cm³.

5. A swimming pool is 18 m long and 8 m wide. Its deep and shallow ends are 2 m and 1·2 m respectively. Find the capacity of the pool, assuming that the bottom of the pool slopes uniformly.
Step 1: The cross-section along the length is a trapezium with parallel sides 2 m and 1.2 m, and height 18 m.
Step 2: Area of cross-section = 1/2 × (2 + 1.2) × 18 = 1/2 × 3.2 × 18 = 28.8 m².
Step 3: Capacity = Area of cross-section × width = 28.8 × 8 = 230.4 m³.

6. The following figure shows a closed victory-stand whose dimensions are given in cm. Find the volume and the surface area of the victory stand.
Step 1: Split the stand into 3 vertical blocks. Left (3) has width=40, depth=30, height=20. Volume = 24,000 cm³.
Step 2: Middle (1) has width=30, depth=30. Its height is 30 cm more than block 3, so height = 20+30 = 50. Volume = 45,000 cm³.
Step 3: Right (2) has width=40, depth=30. Its vertical right face is 30. Volume = 40×30×30 = 36,000 cm³.
Step 4: Total Volume = 24,000 + 45,000 + 36,000 = 105,000 cm³.
Step 5: To find surface area, add areas of all exposed faces.
Step 6: Front faces = (40×20) + (30×50) + (40×30) = 3500 cm². Back faces = 3500 cm².
Step 7: Top faces = (40×30) + (30×30) + (40×30) = 3300 cm². Bottom face = (40+30+40)×30 = 3300 cm².
Step 8: Left face = 20×30 = 600 cm². Right face = 30×30 = 900 cm².
Step 9: Inside vertical steps = 30×30 + (50-30)×30 = 900 + 600 = 1500 cm².
Step 10: Total Surface Area = 3500 + 3500 + 3300 + 3300 + 600 + 900 + 1500 = 16,600 cm².

7. Each face of a cube has perimeter equal to 32 cm. Find its surface area and its volume.
Step 1: The face of a cube is a square. Perimeter = 4a = 32 cm, so edge a = 8 cm.
Step 2: Surface Area = 6a² = 6 × 8² = 6 × 64 = 384 cm².
Step 3: Volume = a³ = 8³ = 512 cm³.

8. A school auditorium is 40m long, 30 m broad and 12 m high. If each student requires 1·2 m² of the floor area; find the maximum number of students that can be accommodated in this auditorium. Also, find the volume of air available in the auditorium, for each student. (LS)
Step 1: Area of the floor = 40 × 30 = 1200 m².
Step 2: Max students = Floor Area / Area per student = 1200 / 1.2 = 1000 students.
Step 3: Total volume of auditorium = 40 × 30 × 12 = 14400 m³.
Step 4: Air available per student = Total Volume / Number of students = 14400 / 1000 = 14.4 m³.

9. The internal dimensions of a rectangular box are 12 cm × x cm × 9 cm. If the length of the longest rod that can be placed in this box is 17cm; find x.
Step 1: Length of longest rod = diagonal of the cuboid = √(l² + b² + h²).
Step 2: √(12² + x² + 9²) = 17.
Step 3: Squaring both sides: 144 + x² + 81 = 289.
Step 4: x² + 225 = 289, which means x² = 64.
Step 5: x = 8 cm.

10. The internal length, breadth and height of a box are 30 cm, 24 cm and 15 cm. Find the largest number of cubes which can be placed inside this box if the edge of each cube is (i) 3 cm (ii) 4 cm (iii) 5 cm
Step 1: (i) For 3 cm cube: (30/3) × (24/3) × (15/3) = 10 × 8 × 5 = 400 cubes.
Step 2: (ii) For 4 cm cube: Take integer parts of (30/4), (24/4), (15/4) = 7 × 6 × 3 = 126 cubes.
Step 3: (iii) For 5 cm cube: Take integer parts of (30/5), (24/5), (15/5) = 6 × 4 × 3 = 72 cubes.

11. A rectangular field is 112 m long and 62 m broad. A cubical tank of edge 6 m is dug at each of the four corners of the field and the earth so removed is evenly spread on the remaining field. Find the rise in level.
Step 1: Volume of earth dug = 4 × (6³) = 4 × 216 = 864 m³.
Step 2: Total area of the field = 112 × 62 = 6944 m².
Step 3: Area of the 4 tanks = 4 × (6 × 6) = 144 m².
Step 4: Area of the remaining field = 6944 - 144 = 6800 m².
Step 5: Rise in level = Volume / Remaining Area = 864 / 6800 m = 54 / 425 m.
Step 6: Convert to cm: (54 / 425) × 100 = 12.7 cm (approx).

12. When length of each side of a cube is increased by 3 cm, its volume is increased by 2457 cm³. Find its side. How much will its volume decrease, if length of each side of it is reduced by 20% ?
Step 1: Let the original side be 'a'. We have (a + 3)³ - a³ = 2457.
Step 2: a³ + 9a² + 27a + 27 - a³ = 2457.
Step 3: 9a² + 27a - 2430 = 0, dividing by 9 gives a² + 3a - 270 = 0.
Step 4: (a + 18)(a - 15) = 0, so original side a = 15 cm.
Step 5: If reduced by 20%, new side = 15 - 20%(15) = 15 - 3 = 12 cm.
Step 6: Decrease in volume = Original Volume - New Volume = 15³ - 12³ = 3375 - 1728 = 1647 cm³.

13. A rectangular tank 30 cm × 20 cm × 12 cm contains water to a depth of 6 cm. A metal cube of side 10 cm is placed in the tank with its one face resting on the bottom of the tank. Find the volume of water, in litres, that must be poured in the tank so that the metal cube is just submerged in the water.
Step 1: To just submerge the cube, the water level needs to reach a height of 10 cm.
Step 2: Initial volume of water = 30 × 20 × 6 = 3600 cm³.
Step 3: When water level is 10 cm, the total volume up to 10cm height is 30 × 20 × 10 = 6000 cm³.
Step 4: This total volume consists of the water and the submerged cube. Volume of submerged cube = 10³ = 1000 cm³.
Step 5: Total water required = 6000 - 1000 = 5000 cm³.
Step 6: Water to be poured = 5000 - 3600 = 1400 cm³ = 1.4 litres.

14. The dimensions of a solid metallic cuboid are 72 cm × 30 cm × 75 cm. It is melted and recast into identical solid metal cubes with each of edge 6 cm. Find the number of cubes formed. (HOTS)
Also, find the cost of polishing the surfaces of all the cubes formed at the rate ₹ 150 per sq. m.
Step 1: Volume of cuboid = 72 × 30 × 75 = 162,000 cm³.
Step 2: Volume of one cube = 6³ = 216 cm³.
Step 3: Number of cubes = 162,000 / 216 = 750 cubes.
Step 4: Total surface area of one cube = 6 × 6² = 216 cm².
Step 5: Total surface area of 750 cubes = 750 × 216 = 162,000 cm².
Step 6: Convert to square meters: 162,000 / 10,000 = 16.2 m².
Step 7: Cost of polishing = 16.2 × 150 = ₹ 2430.

15. The dimensions of a car petrol tank are 50 cm × 32 cm × 24 cm, which is full of petrol. If a car's average consumption is 15 km per litre, find the maximum distance that can be covered by the car.
Step 1: Capacity of the tank = 50 × 32 × 24 = 38400 cm³.
Step 2: Convert to litres: 38400 / 1000 = 38.4 litres.
Step 3: Maximum distance = Total litres × average consumption = 38.4 × 15 = 576 km.

16. The dimensions of a rectangular box are in the ratio 4 : 2 : 3. The difference between cost of covering it with paper at ₹ 12 per m² and with paper at the rate of 13.50 per m² is ₹ 1,248. Find the dimensions of the box.
Step 1: Difference in rate = 13.50 - 12.00 = ₹ 1.50 per m².
Step 2: Total surface area of box = Total difference in cost / difference in rate = 1248 / 1.50 = 832 m².
Step 3: Let dimensions be 4x, 2x, 3x.
Step 4: TSA = 2(4x·2x + 2x·3x + 3x·4x) = 2(8x² + 6x² + 12x²) = 52x².
Step 5: 52x² = 832, which means x² = 16, so x = 4.
Step 6: Dimensions are 4(4) = 16 m, 2(4) = 8 m, 3(4) = 12 m.

17. The length of the diagonal of a cuboid is 13√2 cm and its volume and total surface area is respectively 780 cm³ and 562 cm². Find the dimension of the cuboid.
Step 1: Diagonal = √(l² + b² + h²) = 13√2, so l² + b² + h² = 338.
Step 2: TSA = 2(lb + bh + hl) = 562.
Step 3: We know (l + b + h)² = (l² + b² + h²) + 2(lb + bh + hl) = 338 + 562 = 900.
Step 4: Therefore, l + b + h = √900 = 30.
Step 5: We have l+b+h = 30, lb+bh+hl = 281, and lbh = 780. Testing dimensions that sum to 30 and multiply to 780, we find 13, 12, 5 (13+12+5=30, 13×12×5=780).
Step 6: Dimensions are 13 cm, 12 cm, and 5 cm.

Case-Study Based Question
1. A warehouse is a large building which, in general, is used to store goods. The dimensions of a warehouse vary with respect to the goods to be stored.
The dimensions of a particular warehouse are 88 m × 66 m × 5·5 m. Its owner wants to fill it completely with identical cubical cartons of maximum volume.
(i) What is the largest cube size that can fit perfectly ?
(ii) What is the volume of the warehouse ?
(iii) What is the volume of the cubical carton ?
(iv) What is the maximum number of cartons that can be stored in the warehouse ?
Step 1: (i) Find the HCF of 88 m, 66 m, and 5.5 m. To avoid decimals, use decimeters: 880, 660, 55. Their HCF is 55 dm = 5.5 m. So, the largest cube size is 5.5 m.
Step 2: (ii) Volume of warehouse = 88 × 66 × 5.5 = 31,944 m³.
Step 3: (iii) Volume of cubical carton = (5.5)³ = 166.375 m³.
Step 4: (iv) Maximum number of cartons = (88 / 5.5) × (66 / 5.5) × (5.5 / 5.5) = 16 × 12 × 1 = 192 cartons.

2. Study the following picture of a study table made by Rohan for his SUPW project with match boxes of each dimension 6 cm by 4 cm by 1.5 cm.
Answer the following:
(i) How many match boxes are used to make the table?
(ii) What is the floor area covered by the table?
(iii) What is the surface area of the top of the table?
(iv) What is the total volume of the table?
Step 1: The dimensions of the entire table and exact box placements are not provided in the diagram text.
Step 2: Therefore, exact mathematical calculations cannot be accurately performed with the given image alone as sufficient count data is missing.