Ex 2.1, 5 - Find perimeters of (i) ABE (ii) rectangle BCDE - Teachoo

Ex 2.1, 5 - Chapter 2 Class 7 Fractions and Decimals - Part 2
Ex 2.1, 5 - Chapter 2 Class 7 Fractions and Decimals - Part 3 Ex 2.1, 5 - Chapter 2 Class 7 Fractions and Decimals - Part 4 Ex 2.1, 5 - Chapter 2 Class 7 Fractions and Decimals - Part 5 Ex 2.1, 5 - Chapter 2 Class 7 Fractions and Decimals - Part 6 Ex 2.1, 5 - Chapter 2 Class 7 Fractions and Decimals - Part 7

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Question 5 Find the perimeters of (i) ABE Perimeter = AB + AE + BE = 5/2 + 3 3/5 + 2 3/4 = 5/2 + (3+3/5)+ (2+3/4) = (3+2) + (5/2+3/5+3/4) = 5 + (5/2+(3 4 + 3 5 )/(5 4)) = 5 + (5/2+(12 + 15)/20) = 5 + (5/2+27/20) = 5 + (5 20 + 27 2)/(2 20) = 5 + (100 + 54)/40 = 5 + 154/40 = 5 + 77/20 = (5 20 + 77)/20 = (100 + 77)/20 = 177/20 = 8 17/20 cm Perimeter of ABE = 177/20 = 8 / cm Question 5 Find the perimeters of (ii) the rectangle BCDE in this figure. Whose perimeter is greater?BCDE As it is a rectangle, Opposite sides are equal Now, Perimeter = BC + CD + DE + BE = 7/6 + 2 3/4 + 7/6 + 2 3/4 = 7/6 2 + 2 3/4 2 = 7/3 + (2+3/4) 2 = 7/3 + 2 2 + 3/2 2 = 7/3 + 4 + 3/2 = 4 + 7/3 + 3/2 = 4 + (7 2 + 3 3)/(3 2) = 4 + (14 + 9)/6 = 4 + 23/6 = (4 6 + 23)/6 = 47/6 Since Numerator > denominator Converting to mixed fraction = 7 / cm Perimeter of BCDE = 47/6 = 7 5/6 cm Also, we have to find which perimeter is greater Perimeter ABE = 177/20 = 8 17/20 cm. Perimeter BCDE = 47/6 = 7 5/6 cm. So, we need to compare 177/20 & 47/6 To find which fraction is greater, we make its denominator equal. Common denominator = LCM of 20 & 6 = 2 2 3 5 = 4 3 5 = 12 5 = 60 So, 177/20 = 47/6 Therefore, 531/60 > 470/60 i.e. 177/20 > 47/60 Perimeter ABE > Perimeter BCDE Thus, Perimeter of ABE is greater.

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.