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### Dhaka Secondary 2009/12

Posted: Tue Feb 01, 2011 11:51 pm
ABCD is $4 \times 4$ square. $E$ lies on $AB$; $AE=1$. $F$ lies on $AD$ & $AF=AE$. $EFG$ is a right angled triangle where $F$ is the right angle. Find the radius of the circumcirle of the triangle $EFG$.

### Re: Dhaka Secondary 2009/12

Posted: Thu Aug 25, 2011 4:50 pm
the answer is $\sqrt{5}$

here is how I got it
$AF=AE=1$
from F, BC er upor FM and from G, AB er upor duita lombo (:P) eke dekhano jay je,
$DG=3$, so $EF=\sqrt{2}$ and $GF= 3.\sqrt{2}$
so $EG= 2.\sqrt{5}$
as $EFG$ right angled, so the midpoint of EG will be the centre of it's circumcircle.  ### Re: Dhaka Secondary 2009/12

Posted: Sat Feb 01, 2014 12:10 am
How did you find out that DG=3?

### Re: Dhaka Secondary 2009/12

Posted: Mon Aug 01, 2016 9:38 am
Look carefully triangle AEF and triangle FGD are similar and AF/DF =1/3 so DG=3AE=3 and then we can use Pythagoras and obtain the value of the hypotenuse and the triangle EFG is right so the so answer would be half the value of the hypotenuse