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Pandey Ji

· started a discussion

· 1 Months ago

last line (Question: 187395.0 )

explain last line how it become ac2+bd2

Question:
If ABCD is rhombus, where point O is the point of intersection of diagonals. Then AB2 + BC2 + CD2 + AD2 is equal to–
Options:
A) AD2 + BC2
B) AO2 + OC2
C) AC2 + BD2
D) 2(AO2 + OB2) 
Solution:

Ans: (C)


Knowledge Expert

· commented

· 1 Months ago

@vivek

AB^2 + AD^2 + DC^2 + BC^2 = 2(AO^2 + BO^2 + DO^2 + CO^2 )

= 4AO^2 + 4BO^2 [Since, AO = CO and BO =DO]
= (2AO)^2 + (2BO)^2 = AC^2 + BD^2

vivek

· commented

· 1 Months ago

you are right bro
option should be twice of ac^2+twice of bd^2
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