ABCD is a rhombus whose diagonals intersect at E. Then EA  + EB + EC  +ED   equals to

(a) 0             (b) AD                (c) 2BD             (d) 2AD 

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In Rhombus ABCD , diagonals bisect each other at E Hence, |(𝑬𝑨) βƒ—| = |(𝑬π‘ͺ) βƒ—| |(𝑬𝑩) βƒ—| = |(𝑬𝑫) βƒ—| Since their magnitude is same, but direction is opposite we have (𝐸𝐴) βƒ— = βˆ’ (𝐸𝐢) βƒ— (𝑬𝑨) βƒ— + (𝑬π‘ͺ) βƒ— = 0 And, (𝐸𝐡) βƒ— = βˆ’ (𝐸𝐷) βƒ— (𝑬𝑩) βƒ— + (𝑬𝑫) βƒ— = 0 We need to find (𝑬𝑨) βƒ—+(𝑬𝑩) βƒ—+(𝑬π‘ͺ) βƒ—+(𝑬𝑫) βƒ— = (𝐸𝐴) βƒ—+(𝐸𝐢) βƒ—+(𝐸𝐡) βƒ—+(𝐸𝐷) βƒ— From (1) and (2) = 0 So, the correct answer is (a)

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