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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CBSE Class 12 Sample Paper for 2024 Boards
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CBSE Class 12 Sample Paper for 2024 Boards
Last updated at April 16, 2024 by Teachoo
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In Rhombus ABCD , diagonals bisect each other at E Hence, |(π¬π¨) β| = |(π¬πͺ) β| |(π¬π©) β| = |(π¬π«) β| Since their magnitude is same, but direction is opposite we have (πΈπ΄) β = β (πΈπΆ) β (π¬π¨) β + (π¬πͺ) β = 0 And, (πΈπ΅) β = β (πΈπ·) β (π¬π©) β + (π¬π«) β = 0 We need to find (π¬π¨) β+(π¬π©) β+(π¬πͺ) β+(π¬π«) β = (πΈπ΄) β+(πΈπΆ) β+(πΈπ΅) β+(πΈπ·) β From (1) and (2) = 0 So, the correct answer is (a)