# The floor of a room is of dimension 5m x 4m and it is coverd with circular tiles of diameter 50 cm. Find the area of the floor that remains uncovered. pi=3.14

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The floor of the room has dimensions 5 m by 4 m. This can be split into segments of dimension 0.5 m by 0.5 cm. One circular tile can fit into each segment. The number of such segments that are formed is 10*8 = 80.

The area of each segment is 0.5*0.5 = 0.25 m^2 and the area of the tile is 3.14*0.25^2 = 0.19625 m^2. The area left uncovered is 0.25 - 0.19625 = 0.05375 m^2.

The total area left uncovered is 0.05375*80 = 4.3 m^2

**When the floor is covered 4.3 m^2 is left uncovered.**

So we have 20 square meters, with 100x100 square centimeters in a square meter, so 200000 cm^2.

We can fit 10x8 = 80 circular tiles on that floor (if we assume they're lined up as though inside square tiles of side length 50cm, which is NOT the most efficient way to fill the plane, but probably what the teacher means).

So the covered area is 80 * 3.14 * 25^2 = 157000 cm^2

Therefore the uncovered area is 200000-157000 = 43000 cm^2.