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Ex 9.2, 10 From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1cm are removed. (as shown in the adjoining figure). Find the area of the remaining sheet. (๐‘‡๐‘Ž๐‘˜๐‘’ ๐œ‹=22/7) Remaining Area = Area of larger circle โˆ’ 2 ร— Area of smaller circle โˆ’ Area of rectangle Area of larger circle Radius of larger circle = R = 14 cm Area of larger circle = ๐œ‹R2 = ๐Ÿ๐Ÿ/๐Ÿ• ร— (14)2 = 22/7 ร— 14 ร— 14 = 22/1 ร— 2 ร— 14 = 44 ร— 14 = 616 cm2 Area of smaller circle Radius of smaller circle = r = 3.5 cm Area of smaller circle = ๐œ‹r2 = ๐Ÿ๐Ÿ/๐Ÿ• ร— (3.5)2 = 22/7 ร— (35/10)^2 = 22/1 ร— (7/2)^2 = ๐Ÿ๐Ÿ/๐Ÿ• ร— ๐Ÿ•/๐Ÿ ร— ๐Ÿ•/๐Ÿ = 11/7 ร— 7/1 ร— 7/2 = 22/7 ร— 1 ร— 7/2 Area of rectangle Length of rectangle = l = 3 cm Breadth of rectangle = b = 1 cm Area of rectangle = l ร— b = 3 ร— 1 = 3 cm2 Therefore, โˆด Remaining Area = Area of larger circle โˆ’ 2 ร— Area of smaller circle โˆ’ Area of rectangle = 616 โˆ’ 2 ร— (๐Ÿ•๐Ÿ•/๐Ÿ) โˆ’ 3 = 616 โˆ’ 77 โˆ’ 3 = 616 โˆ’ 80 = 536 cm2 โˆด Required area is 536 cm2

  1. Chapter 9 Class 7 Perimeter and Area
  2. Serial order wise

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo