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### Dhaka Secondary 2009/12

Posted: **Tue Feb 01, 2011 11:51 pm**

by **BdMO**

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**

by **Shifat**

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**

by **Ridwan Abrar**

How did you find out that DG=3?

### Re: Dhaka Secondary 2009/12

Posted: **Mon Aug 01, 2016 9:38 am**

by **mdhasib**

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