Area-Of-Parallelogram

All the triangles are similar in the diagram above.  What is interesting about it is that

       A_parallelogram = sqrt( 2A_1 * 2A_2)

or in terms of the dotted / solid parallelograms:

       A_parallelogram = sqrt( P_1 * P_2)

 

Derivation

       (1)   The ratio of area of A1 to A2 is

           A_1/A_2 = ( L_1 / L_2 )^2     ……subscripts correspond to triangle depicted above

       (2)   The ratio of area A3 to A2 is:

           A_3/A_2 = ( L_3 / L_2 )^2 

       (3)   Base of the big overall triangle: 

          L_3 =  L_1 +  L_2 

       (4) Substitute #3 into #2

          A_3/A_2 = (( L_1+L_2) / L_2 )^2 

       (5) Expand right side

          A_3/A_2 = ( L_1^2+2L_1*L_2+L_2^2) / L_2^2         =         L_1^2/L_2^2+2L_1*L_2/L_2^2 +L_2^2/L_2^2          =             L_1^2/L_2^2+2L_1/L_2+1 

       (6) The area of the overall big triangle also is related as follows

           A_3 = A_p + A_1 + A_2 

       (7)  Substituting #6 in #5:

           (A_p + A_1 + A_2)/A_2   =  A_p/A_2  + A_1/A_2  + A_2/A_2    =   A_p/A_2  + A_1/A_2  +  1    =       L_1^2/L_2^2+2L_1/L_2+1    

       (8)  referring back to #1 we can cancel the term on both sides of the equation.  The one also cancels on both sides leaving:

          A_p/A_2    =     2L_1/L_2   

        (9) Moving terms around:

          A_p   =   2 A_2  * (L_1/L_2) 

        (10) #1 can be rearranged to

            sqrt A_1 =  sqrt A_2 * ( L_1 / L_2 ) 

        (11) Cutting a term out of #9 equal to #10 yields

          A_p   =   2  sqrt ( A_1 * A_2 ) 

Categories: Math

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