Answer:
Given, length of the floor = 8 m 96 cm \[=8\times 100\text{ }cm+96\text{ }cm\] [\[\therefore \]\[1\,m=100\text{ }cm\]] \[=800+96\text{ }cm\] \[=896\text{ }cm\] and breadth of the floor \[=6m\text{ }72cm\] \[=6\times 100\text{ }cm+72\text{ }cm\] [\[\therefore \] \[1\,m=100\text{ }cm\]] \[=672\text{ }cm\] Now, size of the square tile = HCF of 896 and 672 Prime factorization of 896 and 672
2 896 2 448 2 224 2 112 2 56 2 28 2 14 7 7 1
\[896=2\times 2\times 2\times 2\times 2\times 2\times 2\times 7\] \[672=2\times 2\times 2\times 2\times 2\times 3\times 7\] Common factors of 896 and 672 \[=2\times 2\times 2\times 2\times 2\times 7\] \[=224\] \[\therefore \] Minimum no. of square tiles \[\text{=}\frac{\text{Area}\,\,\text{of}\,\,\text{floor}}{\text{Area}\,\,\text{of}\,\,\text{square}\,\,\text{tile}}\] \[=\frac{896\times 672}{224\times 224}\] [\[\therefore \] area of square\[={{(side)}^{2}}\]] \[=\frac{896\times 3}{224}\] \[=4\times 3=12\] 2 672 2 336 2 168 2 84 2 42 3 21 7 7 1
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