Volume and surface area in real life · 4 marks

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.

Step-by-step solution

Idea: Work out what each pool needs (water = volume, tiles = floor + four walls) and what it offers (depth, swimming area, length), then decide which matters most for a school.

Water needed. A: 8 × 50 × 20 = 8000 ft³ ≈ 8000 × 28.3 = 2,26,400 L. B: 6 × 100 × 15 = 9000 ft³ ≈ 9000 × 28.3 = 2,54,700 L. B needs about 28,300 L more.1 mark
Swimming area (water surface). A: 50 × 20 = 1000 ft². B: 100 × 15 = 1500 ft², half as much again, so more students can swim at once, and the 100 ft (about 30 m) length suits races.1 mark
Tiling (floor + 4 walls, open top). A: 1000 + 2(8 × 50 + 8 × 20) = 1000 + 1120 = 2120 ft². B: 1500 + 2(6 × 100 + 6 × 15) = 1500 + 1380 = 2880 ft². B costs more to tile.1 mark
Choice: B. Depth matters most for students’ safety: 6 ft (about 1.8 m) is safer than 8 ft (about 2.4 m), and B gives more room and a racing length. A school with little land or water could choose A, but A’s 8 ft depth needs closer supervision. (Either way, a shallow end for beginners would be wise.)1 mark

Volumes: A = 8000 ft³ ≈ 2,26,400 L, B = 9000 ft³ ≈ 2,54,700 L. Choose B: shallower (safer), larger swimming area (1500 ft² vs 1000 ft²) and racing length, at the cost of more water and tiling. Aspects: safety/depth, space, water, tiling cost, land available.

Check: 9000 − 8000 = 1000 ft³ more water for B, and 1000 × 28.3 = 28,300 L ✓ (2,54,700 − 2,26,400 = 28,300).

Answer to write in the exam

A: V = 8 × 50 × 20 = 8000 ft³ ≈ 8000 × 28.3 = 2,26,400 L

B: V = 6 × 100 × 15 = 9000 ft³ ≈ 9000 × 28.3 = 2,54,700 L

Water surface: A = 1000 ft², B = 1500 ft²

Area to tile: A = 1000 + 2(400 + 160) = 2120 ft²; B = 1500 + 2(600 + 90) = 2880 ft²

∴ Option B: shallower (6 ft, safer), more swimming space, racing length; needs more water and tiles.

Aspects considered: depth/safety, space per swimmer, water volume, tiling cost, land.

Common mistakes that cost marks

Comparing only volumes and choosing the pool with less water, without thinking about safety or space.
Forgetting to convert cubic feet to litres, or multiplying by 28.3 twice.
Counting the top as a surface to tile. A pool is open at the top.

How this can come in the exam

Case-based (4 marks)

A housing society plans a children’s paddling pool 1.5 ft deep, 30 ft long and 20 ft wide.
(i) Find its volume in cubic feet. (ii) How many litres of water fill it? (1 ft³ ≈ 28.3 L) (iii) Find the area to be tiled (floor and four walls). (iv) If tiling costs ₹ 60 per ft², find the cost.

Show answer

(i) 1.5 × 30 × 20 = 900 ft³ (1 mark). (ii) 900 × 28.3 = 25,470 L (1 mark). (iii) 600 + 2(1.5 × 30 + 1.5 × 20) = 600 + 150 = 750 ft² (1 mark). (iv) 750 × 60 = ₹ 45,000 (1 mark).

Try one yourself

Which holds more water: a tank 4 ft × 5 ft × 6 ft or one 3 ft × 8 ft × 5 ft? By how many litres? (1 ft³ ≈ 28.3 L)