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Ex 10.3, 12 - How many tiles whose length and breadth are 12 cm, 5 cm

Ex 10.3, 12 - Chapter 10 Class 6 Mensuration - Part 2
Ex 10.3, 12 - Chapter 10 Class 6 Mensuration - Part 3
Ex 10.3, 12 - Chapter 10 Class 6 Mensuration - Part 4

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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: (a) 100 cm and 144 cm Here, Number of tiles = (π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘Ÿπ‘’π‘π‘‘π‘Žπ‘›π‘”π‘’π‘™π‘Žπ‘Ÿ π‘Ÿπ‘’π‘”π‘–π‘œπ‘›)/(π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘Ÿπ‘’π‘π‘‘π‘Žπ‘›π‘”π‘’π‘™π‘Žπ‘Ÿ 𝑑𝑖𝑙𝑒) 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‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘Ÿπ‘’π‘π‘‘π‘Žπ‘›π‘”π‘’π‘™π‘Žπ‘Ÿ π‘Ÿπ‘’π‘”π‘–π‘œπ‘›)/(π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘Ÿπ‘’π‘π‘‘π‘Žπ‘›π‘”π‘’π‘™π‘Žπ‘Ÿ 𝑑𝑖𝑙𝑒𝑠) = 14400/60 = 1440/6 = 240 ∴ 240 tiles are required to cover the region

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.