BdMO National Junior 2019/3
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- Posts:1007
- Joined:Sat Dec 09, 2017 1:32 pm
$ABCD$ is a rectangle. $E,F,G,H$ lies on $AB,BC,CA,AD$. $AE=CG=3,BE$ $=DG=6,AH=CF=4,$ $DH=BF=8$.$EFGH$ is a parallelogram. $DI\perp EF,DI=?$
Re: BdMO National Junior 2019/3
The length of HG = 10 (By 3-4-5 triangle ratio)
so DGH = 24= 10 ×0.5× DL
So, DL= 4.8
the area of EFGH= ABCD - 2 (AEB + EBF) = 108-60=48
it can be written as,
EFGH=LI×10
LI=4.8
SO,
DJ = 4.8×2 =9.6
so DGH = 24= 10 ×0.5× DL
So, DL= 4.8
the area of EFGH= ABCD - 2 (AEB + EBF) = 108-60=48
it can be written as,
EFGH=LI×10
LI=4.8
SO,
DJ = 4.8×2 =9.6
Oops, I forgot
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- Posts:1007
- Joined:Sat Dec 09, 2017 1:32 pm
Re: BdMO National Junior 2019/3
[quote=Arifa post_id=21128 time=1551690771 user_id=3998]
The length of $HG = 10$ (By $3-4-5$ triangle ratio)
so $DGH = 24= 10 ×0.5× DL$
So, $DL= 4.8$
the area of $[EFGH]= [ABCD - 2 (AEB + EBF)] = [108-60]=48$
it can be written as,
$[EFGH]=LI×10$
$LI=4.8$
SO,
\[DJ = 4.8×2 =9.6\]
[/quote]
The length of $HG = 10$ (By $3-4-5$ triangle ratio)
so $DGH = 24= 10 ×0.5× DL$
So, $DL= 4.8$
the area of $[EFGH]= [ABCD - 2 (AEB + EBF)] = [108-60]=48$
it can be written as,
$[EFGH]=LI×10$
$LI=4.8$
SO,
\[DJ = 4.8×2 =9.6\]
[/quote]