Perimeter = 56. Let the side of the rhombus be “a”, then 4a = 56 => a =14.
Area of Rhombus = Half the product of its diagonals. Let the diagonals be d_{1} and d_{2} respectively.
*d_{1} * d_{2} = 100 => d_{1} * d_{2} = 200. By Pythagoras theorem, (d_{1})^{2} + (d_{2})^{2} = 4a^{2} => (d_{1})^{2} + (d_{2})^{2} = 4*196 = 784.
(d_{1})^{2} + (d_{2})^{2} + 2d_{1} * d_{2} = (d_{1}+ d_{2})^{2} = 784 +2*200 = 1184 => (d_{1}+ d_{2}) = √1184 = 34.40
Therefore, sum of the diagonals is equal to 34.40 cm .
Correct Answer: 34.40