Sep 2, 2015

Diagonals of rhombus ABCD intersect each other at point O. Prove that OA2+ OC2=2AD2-BD2/2

Question: Diagonals of rhombus ABCD intersect each other at point O. Prove that OA2 + OC2= 2AD2 + BD2/2

Solution: In ∆ AOD
OA2 = AD2 - OD2 -------------Eqn (i)
In ∆ OBC
OC2 = BC2 - OB2 -------------Eqn (ii)
Adding Eqn (i) & Eqn (ii) We get,
OA2 + OC2 = AD2 - OD2 + BC2 - OB2
        = AD2 - OD2 + AD2 - OB2
        = 2AD2 - OD 2 OB 2
        = 2AD2 - (BD/2) 2 -(BD/2)2
        = 2AD2 - BD2/4 -BD2/4
        = 2AD2 - BD2/2 Proved


You may also Like These !


No comments:

Post a Comment

Creative Essays

Featured Post

50.अं ं और अः ः के बारे में और अंतर About Hindi ं and ः also D...

This video of Hindi is the most demanded one by commenters. Understanding ANG and AH [ ं और अः ः ] for many learner is difficu...