Question

ABCD is a rhombus whose diagonals intersect at O. E and F are mid points of AO and BO respectively. if AC=12cm and BD=16cm, then find the length of EF.

Answer 1

bro length of EF is 5cm as i think
because let abcd is a rhombus and diagonal of rhombus intersect each other so ac and md are bisected so AO=6 (because AC=12 so AO=AC/2=12/2=6) and BO=4(because BD=16cm so BO=BD/2=16/2=8) 
So AO=6 And e is mid point so EO=AE=3 and as same BO=4 and f is moid point so BF=FO=4 
so as diagonal of rhombus make 90 degree angle at the diagonals so in
right angle triangle FOE
FEsquare=OEsquare+OFsquare
FEsquare=3square+4square
FEsquare=9+16
FEsquare=25
FE=5
HOPE THIS HELP YOU IF HELP THEN PLS MARK ME AS BRAINLIST AND THANKS ME

Answer 2

Answer:

AC=16cm (given)

BD=12cm (given)

OA=OC (diagonals of parallelogram bisect each other)

OD=OB ( diagonals of parallelogram bisect each other)

OE=EA (E is the midpoint of OA)

OF=FB (F is the midpoint of OB)

So,

(EF)^2=(OE)^2+(OF)^2

EF^2=(4)^2+(3)^2

EF^2 = 16+9

EF =√25

EF = 5 cm.

hope this will help you!!!!