The floor of a room has dimensions 5 , m time | vedclass

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(B) Given,the floor of a room is covered with circular tiles.Length of the floor $(l) = 5 \, m$.Breadth of the floor $(b) = 4 \, m$.Area of the floor

Vedclass · Accepted Answer

$4.3$

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(B) Given,the floor of a room is covered with circular tiles.Length of the floor $(l) = 5 \, m$.Breadth of the floor $(b) = 4 \, m$.Area of the floor $= l 	imes b = 5 	imes 4 = 20 \, m^2$.Diameter of each circular tile $= 50 \, cm = 0.5 \, m$.Radius of each circular tile $(r) = \frac{0.5}{2} = 0.25 \, m = \frac{1}{4} \, m$.Area of one circular tile $= \pi r^2 = 3.14 	imes (0.25)^2 = 3.14 	imes 0.0625 = 0.19625 \, m^2$.From the figure,the number of tiles along the length $= \frac{5 \, m}{0.5 \, m} = 10$ tiles.The number of tiles along the breadth $= \frac{4 \, m}{0.5 \, m} = 8$ tiles.Total number of tiles $= 10 	imes 8 = 80$ tiles.Total area covered by $80$ tiles $= 80 	imes 0.19625 = 15.7 \, m^2$.Area of the floor that remains uncovered $=$ Total area of floor $-$ Total area of tiles.Area uncovered $= 20 - 15.7 = 4.3 \, m^2$.

The floor of a room has dimensions $5 \, m \times 4 \, m$ and it is covered with circular tiles of diameter $50 \, cm$ each,as shown in the figure. Find the area of the floor that remains uncovered by the tiles. (Use $\pi = 3.14$) (in $m^2$)

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