• # question_answer A rhombus when divided by one diagonal gives two congruent triangles of perimeter 36 cm each. If both the diagonals divide it into 4 congruent triangles of perimeter 24 cm each, what is the side of the rhombus? A)  10 cm            B)  11 cm C)  12 cm            D)  None of these

We know that, perimeter of MBC = 36 cm   i.e.,          $a+a+AC=36$ i.e.,            $2a+2AO=36$ ?(i) Also, perimeter of $a+AO+BO=24\,cm$ ?(ii) Using Eq. (i), we have $a+AO=18$ From Eq. (ii),      $BO=6\,cm$ $\Rightarrow$ $BD=12\,cm$ Now, in $\Delta AOB,$ $A{{B}^{2}}=A{{O}^{2}}+O{{B}^{2}}$ $A{{B}^{2}}-A{{O}^{2}}=36$ $\Rightarrow$ $(AB-AO)=2$ i.e., $a-AO=2$ ?(iii) On solving Eqs. (i) and (ii), we get a = 10 = Side of rhombus