# Ex 6.5, 7 - Chapter 6 Class 7 Triangle and its Properties

Last updated at Sept. 10, 2018 by Teachoo

Last updated at Sept. 10, 2018 by Teachoo

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

Pythagoras Theorem - Statement Questions

Chapter 6 Class 7 Triangle and its Properties

Concept wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.