# In the given figure triangle ABC is a right angled at A. DEFG is a square , BD =12 cm and EC=27 cm . What is the length of the side of the square.

RAJ KUMAR
29 Points
6 years ago
let angle ABC=x
so, angle ACB=90-x
the ∆GDB and ∆FEC is also right angled ∆
so, angle BGC= 90-x and angle CFE=x
now ∆BDG~∆FEC by AAA property
BD/DG=FE/EC
FE^2=BD×EC...(GD=FE)
FE=√12×27=18
Lalitkumar
21 Points
6 years ago
I found a fault int the previous answer. Now I will go by trial and error method so first of all we prove triangle AGF and triangle in CFE by AA criterion and then we will apply the property that the sides of similar triangles are in proportion and thus we get27/x=x/y where x=Sides of the square and y= AG and now by trial and error,checking each of the sides(given in options)squares divisible by 27 and if it turns out to be divisible by 27 we can conclude that we have got the answer but we can see in the question there are two options which are divisible by x squared which are 27 and 12 so as AG cannot be 27 as it doesn`t appear to. We conclude 12 is the correct optionA bit informal.Regards, Lalit 10th grader