Floor of a room is of dimenstions $5 m \times 4 m$ and it is covered with circular tiles of diameters 50 cm each as shown in figure. Find the area of floor that remains uncovered with tiles $(T a k e π = 3.14)$

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Solution

No of circular titles = $\frac{5}{0.5} \times \frac{4}{0.5}$
$= \frac{20}{0.25} = \frac{2000}{25} = 80 t i l e s$
Area of each tile = $π r^{2}$
$= \frac{22}{7} \times {(\frac{0.5}{2})}^{2}$
$= \frac{22}{7} \times \frac{5}{20} \times \frac{5}{20} = \frac{11}{56}$
Area of 80 tiles = $80 \times \frac{11}{56} = \frac{110}{7} s q . m$
Area of the flooring= $5 \times 4 = 20 s q . m$
Area of uncovered region = $20 - \frac{110}{7}$
= 20 - 15.71
= 4.285 sq.m

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Floor of a room is of dimenstions 5m×4m and it is covered with circular tiles of diameters 50 cm each as shown in figure. Find the area of floor that remains uncovered with tiles (Take π=3.14)

No of circular titles =50.5×40.5 =200.25=200025=80 tiles Area of each tile =πr2 =227×(0.52)2 =227×520×520=1156 Area of 80 tiles = 80×1156=1107sq.m Area of the flooring= 5×4=20 sq.m Area of uncovered region = 20−1107 = 20 - 15.71 = 4.285 sq.m

Floor of a room is of dimenstions $5 m \times 4 m$ and it is covered with circular tiles of diameters 50 cm each as shown in figure. Find the area of floor that remains uncovered with tiles $(T a k e π = 3.14)$