Question

can anyone who had brain can solve this sum.

Answer 1

may this helps you may be the ans will be wrong but process is correct maybe there would be some mistake.


pls Mark it if it helped u a little also..if it then plz plz plz

Answer 2

Hey user have you remember me so,you want he answer for this question right so,here is your answer it is in the following below:
Area o first triangle AFM = 1/2 x base x height 
                                        = 1/2 x 4 x5
      so,here 2 and 4 will cancel 2 1's  and 4 2's
area of trapezium = 10cm
so, now it is 1/2 x (some of parallel sides) x distance between 11 sides 
1/2 x (5+6) x MP += 8
                                      so,1/2 x 11 x 8
now here 2 and 8 will cancel  now then the 8 becomes 4 
MP = PA - MA        =    44cm²
MP = 12 - 4
MP = 8 cm
Area of triangle EQD = 1/2 x b x h 
                                                       =1/2 x 6 x 6
now here 6  cancel s and becomes as 3
   QD = AD - AP                 = 18cm
    QD = 18-12   = 12cm
Area of triangle BNA = 1/2 x b x h 
                                                     = 1/2 x 8 x 5  =    20cm²
                   8 will cancel and becomes as 4
         
Area of another trapezium = 1/2 x (5+4) x NQ = 1/2 x 9 x 6    = 27cm²
         here 6 becomes as 3   
 now here       NQ = AQ - AN = 14 - 8 = 6cm 
Area of last triangle = 1/2 x b x h 
                                                     = 1/2 x 6 x 4       =   12cm²
  because here 4 becomes as 2 
totally we will get 
                             10 + 44 + 18 + 20 + 27 + 12 
                                               we will get the answer 131cm²as the answer
  *THANK YOU I MADE HARD WORK TO SOLVE AND WRITE IN BRAINLY SO,PLEASE MARK ME AS ANSWER PLEASE I REQUES YOU*