Ex 10.3, 12 (b) - Chapter 10 Class 6 Mensuration
Last updated at April 16, 2024 by Teachoo
Ex 10.3
Ex 10.3, 1 (a)
Ex 10.3, 1 (b) Important
Ex 10.3, 1 (c)
Ex 10.3, 1 (d) Important
Ex 10.3, 2 (a)
Ex 10.3, 2 (b) Important
Ex 10.3, 2 (c)
Ex 10.3, 3 Important
Ex 10.3, 4
Ex 10.3, 5 Important
Ex 10.3, 6
Ex 10.3, 7 Important
Ex 10.3, 8
Ex 10.3, 9 Important
Ex 10.3, 10 (a) Important
Ex 10.3, 10 (b)
Ex 10.3, 11 (a)
Ex 10.3, 11 (b) Important
Ex 10.3, 11 (c) Important
Ex 10.3, 12 (a)
Ex 10.3, 12 (b) Important You are here
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 = (𝐴𝑟𝑒𝑎 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑟𝑒𝑔𝑖𝑜𝑛)/(𝐴𝑟𝑒𝑎 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑡𝑖𝑙𝑒) 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 × Breadth = 70 × 36 = 2520 sq. cm = 2520 〖𝑐𝑚〗^2 Now, Number of tiles = (𝐴𝑟𝑒𝑎 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑟𝑒𝑔𝑖𝑜𝑛)/(𝐴𝑟𝑒𝑎 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑡𝑖𝑙𝑒𝑠) = 2520/60 = 252/6 = 42 ∴ 42 tiles are required to cover the given region
