Ex 10.3

Chapter 10 Class 6 Mensuration
Serial order wise

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### Transcript

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: (b) 70 cm and 36 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) From part (a), Area of rectangular tile = 60 cm2 Finding area of rectangular region Rectangular region Length of rectangle = 70 cm Breadth of rectangle = 36 cm Area of rectangle = Length Γ Breadth = 70 Γ 36 = 2520 sq. cm = 2520 γππγ^2 Now, 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 ) = 2520/60 = 252/6 = 42 β΄ 42 tiles are required to cover the given region