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Ex 10.3, 12 - Chapter 10 Class 6 Mensuration - Part 5

Ex 10.3, 12 - Chapter 10 Class 6 Mensuration - Part 6
Ex 10.3, 12 - Chapter 10 Class 6 Mensuration - Part 7

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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‘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) 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‘šγ€—^2 Now, 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) = 2520/60 = 252/6 = 42 ∴ 42 tiles are required to cover the given region

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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.