CBSE Class 12 Sample Paper for 2024 Boards

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Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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 August 14, 2026 by

ABCD is a rhombus whose diagonals intersect at E. Then EAΒ  + EB + ECΒ  +EDΒ  Β equals to

(a) 0Β  Β  Β  Β  Β  Β  Β (b) ADΒ  Β  Β  Β  Β  Β  Β  Β  (c) 2BDΒ  Β  Β  Β  Β  Β  Β (d) 2ADΒ 

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[Class 12] ABCD is a rhombus whose diagonals intersect at E. Then EA + - CBSE Class 12 Sample Paper for 2024 Boards

part 2 - Question 8 - CBSE Class 12 Sample Paper for 2024 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12

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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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