Ex 6.5, 7 - Find the perimeter of the rectangle whose length is 40 cm

Ex 6.5, 7 - Chapter 6 Class 7 Triangle and its Properties - Part 2
Ex 6.5, 7 - Chapter 6 Class 7 Triangle and its Properties - Part 3


Transcript

Ex 6.5, 7 Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm. Let the rectangle be ABCD where Length = AB = 40 cm Breadth = BC and diagonal BD = 41 cm Perimeter of ABCD = 2(Length + Breadth) Let’s find breadth BC In rectangle, all angles are 80° ∴ ∠A = ∠B = ∠C = ∠D 80° Therefore, In right triangle Δ ABC By Pythagoras Theorem, 〖𝐴𝐶〗^2 = 〖𝐵𝐶〗^2 + 〖𝐴𝐵〗^2 〖(41)〗^2 = 〖𝐵𝐶〗^2 + 〖(40)〗^2 1681 = 〖𝐵𝐶〗^2 + 1600 1681 − 1600 = 〖𝐵𝐶〗^2 1681 − 1600 = 〖𝐵𝐶〗^2 81 = 〖𝐵𝐶〗^2 〖(9)〗^2 = 〖𝐵𝐶〗^2 Cancelling square 〖(9)〗^2 = 〖𝐵𝐶〗^2 ⇒ BC = 9 cm ∴ Breadth = BC = 9 cm Perimeter of ABCD = 2(Length + Breadth) = 2 (40 + 9) = 2 (49) = 98 cm

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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.