It is given that:
∆ABC ~ ∆DEF
We know that the ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.
∴      
              
               

                  

                    
Hence, BC = 11.2 cm.

Given: Triangle ABC and ∆PQR in which AD and PM are medians drawn on sides BC and QR respectively. It is given that

To Prove : ∆ABC ~ ∆PQR
Const : Produce AD to E such that AD = DE and PM to F such that PM = MF.

Proof : In ∆ABD and ∆CDE,
AD = DE    [by construction]
∠ADB = ∠CDE
[vertically opposite angles]
and    BD = DC    [AD is a median]
Therefore, by using SAS congruent condition
                
                   [by CPCT]
Similarly, we can prove
                  
                              [by CPCT]
It is given that:

    
          
Therefore, by using SSS congruent condition
                   
                                 ...(i)
Similarly,                         ...(ii)
Adding (i) and (ii), we get
∠1 + ∠3 = ∠2 + ∠4
∠A = ∠P
Now, in ∆ABC and ∆PQR
                     
and           
Therefore, by using SAS similar condition
∆ABC ~ ∆PQR Hence Proved.