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Answer :
Given, number of tiles = 3000 and the length of the two diagonals of a tile = 45 cm and 30 cm
We have area of one rhombus shaped tile =\(\frac { 1 }{ 2 }\times d1 \times d2\)
\(=\frac { 1 }{ 2 }\times 45 \times 30\)
\(= 45 \times 15\)
\(= 675 cm^2\)
So, area of one tile=675\(cm^2\)
Therefore, Area covered by 3000 tiles = \(3000 \times 675 cm^2 = 2025000 cm^2 = 202.5 m^2\)
So, the cost of polishing the floor = \(202.5 \times 4 = ₹ 810\quad\)[∵ cost per \(m^2\) is ₹ 4.
]
Hence, the required cost = ₹ 810.