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Question 8 - CBSE Class 12 Sample Paper for 2024 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at April 16, 2024 by

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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Transcript

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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