1. Chapter 10 Class 6 Mensuration
2. Concept wise
3. Area of Rectangle

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: (a) 100 cm and 144 cm Here, Number of tiles = (๐ด๐๐๐ ๐๐ ๐๐๐๐ก๐๐๐๐ข๐๐๐ ๐๐๐๐๐๐)/(๐ด๐๐๐ ๐๐ ๐๐๐๐ก๐๐๐๐ข๐๐๐ ๐ก๐๐๐) Finding area of rectangular region and rectangular tile Rectangular tile Length of tile = 12 cm Breadth of tile = 5 cm Area of tile = Length ร Breadth = 12 ร 5 = 60 sq. cm = 60 cm2 Rectangular region Length of region = 100 cm Breadth of region = 144cm Area of region = Length ร Breadth = 100 ร 144 = 14400 sq. cm = 14400 cm2 Number of tiles = (๐ด๐๐๐ ๐๐ ๐๐๐๐ก๐๐๐๐ข๐๐๐ ๐๐๐๐๐๐)/(๐ด๐๐๐ ๐๐ ๐๐๐๐ก๐๐๐๐ข๐๐๐ ๐ก๐๐๐๐ ) = 14400/60 = 1440/6 = 240 โด 240 tiles are required to cover the region