Ex 10.3

Chapter 10 Class 6 Mensuration
Serial order wise

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Ex 10.3, 12 How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively: (a) 100 cm and 144 cm Here, Number of tiles = (π΄πππ ππ ππππ‘ππππ’πππ ππππππ)/(π΄πππ ππ ππππ‘ππππ’πππ π‘πππ) Finding area of rectangular region and rectangular tile Rectangular tile Length of tile = 12 cm Breadth of tile = 5 cm Area of tile = Length Γ Breadth = 12 Γ 5 = 60 sq. cm = 60 cm2 Rectangular region Length of region = 100 cm Breadth of region = 144cm Area of region = Length Γ Breadth = 100 Γ 144 = 14400 sq. cm = 14400 cm2 Number of tiles = (π΄πππ ππ ππππ‘ππππ’πππ ππππππ)/(π΄πππ ππ ππππ‘ππππ’πππ π‘ππππ ) = 14400/60 = 1440/6 = 240 β΄ 240 tiles are required to cover the region