Tiling a Panchayat Hall: HCF, LCM and a Diagonal Claim
The panchayat office in a block of Balasore district has approved a rectangular community hall measuring 5.60 m by 3.20 m for floor tiling. The mason must use identical square tiles of the largest possible size, laid without cutting any tile, and the dealer sells only whole tiles. Before releasing payment, the panchayat wants two claims by the supervisor checked. First, he says the hall's diagonal can also be covered exactly by a whole number of these tiles, so no half-tiles will be needed along the diagonal. Second, he wants to reuse the same tile size in an adjoining room measuring 4.40 m by 3.60 m, so that one order covers both rooms. The mason has already bought 30 tiles of the size he worked out for the first hall and wants to know whether any will be left over. All measurements are to be treated as exact, and the panchayat has asked for the working, not just the final answers.
1. Find the side of the largest square tile that can be used in the 5.60 m by 3.20 m hall without cutting, and decide whether the 30 tiles already bought are enough.
Read the worked answer
Convert to centimetres: 560 cm and 320 cm. A tile side must divide both dimensions exactly, so we need the HCF, not the LCM.
560 = 2⁴ × 5 × 7 and 320 = 2⁶ × 5, so HCF = 2⁴ × 5 = 80 cm.
Number of tiles = (560 ÷ 80) × (320 ÷ 80) = 7 × 4 = 28 tiles.
Since 28 < 30, the mason has enough and 2 tiles will be left over (useful as spares for breakage). Using the LCM instead would have given a huge, meaningless number - the LCM answers 'when do two lengths meet again', not 'what size divides both'.
2. The supervisor claims the diagonal can be covered exactly by a whole number of 80 cm tiles. Test this claim.
Read the worked answer
Diagonal = √(560² + 320²) = √(313600 + 102400) = √416000 cm.
√416000 = √(1600 × 260) = 40√260 = 40 × 2√65 = 80√65 cm.
Number of tile-widths along the diagonal = 80√65 ÷ 80 = √65.
Now √65 is irrational: if √65 = p/q in lowest terms, then p² = 65q², so 5 | p and 13 | p, hence 65 | p, say p = 65k. Then 4225k² = 65q², giving q² = 65k², so 65 | q - contradicting 'lowest terms'. Also 8² = 64 and 9² = 81, so √65 is not even an integer; it is about 8.06.
So the diagonal is 80√65 cm ≈ 644.96 cm and the count of tiles along it is not a whole number. The supervisor's claim is false; tiles along the diagonal would have to be cut.
3. Is there a single tile size that works for both rooms (5.60 m × 3.20 m and 4.40 m × 3.60 m)? Give the largest such size and check it.
Read the worked answer
We need a length dividing 560, 320, 440 and 360 cm - that is HCF(560, 320, 440, 360).
560 = 2⁴ × 5 × 7, 320 = 2⁶ × 5, 440 = 2³ × 5 × 11, 360 = 2³ × 3² × 5.
Common part = 2³ × 5 = 40 cm.
Check: 560 ÷ 40 = 14, 320 ÷ 40 = 8, 440 ÷ 40 = 11, 360 ÷ 40 = 9 - all whole numbers, so a 40 cm tile fits both rooms with no cutting.
Tile count for the two rooms = 14 × 8 + 11 × 9 = 112 + 99 = 211 tiles. Note that 80 cm (the best for the first hall alone) fails for the second hall because 440 ÷ 80 is not a whole number.